Answer:
\large \boxed{\text{2.0 atm}}
Explanation:
We can use Dalton's Law of Partial Pressures:
Each gas in a mixture of gases exerts its pressure separately from the other gases.
0.25 mol of O₂ exerts 0.50 atm.
If you add 0.75 mol of CO, the total amount of gas is
0.25 mol + 0.75 mol = 1.00 mol
[tex]p_{\text{total}} = \text{1.00 mol} \times \dfrac{\text{0.50 atm}}{\text{0.25 mol}}= \textbf{2.0 atm}\\\\\text{The total pressure in the flask is $\large \boxed{\textbf{2.0 atm}}$}[/tex]
The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.
Partial pressure can be defined as the pressure exerted by each gas in a given solution.
The total moles of gas in the container by the addition of CO has been:
Total moles = moles of oxygen + moles of CO
Total moles = 0.25 + 0.75
Total moles = 1 mol.
By using Dalton's law of partial pressure:
Total pressure = total moles [tex]\rm \times\;\dfrac{pressure\;of\;oxygen}{moles\;of\;oxygen}[/tex]
Total pressure = 1 [tex]\rm \times\;\dfrac{0.50}{0.25}[/tex]
Total pressure = 2 atm.
The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.
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Compounds A and BB are colorless gases obtained by combining sulfur with oxygen. Compound A results from combining 6.00 gg of sulfur with 5.99 gg of oxygen, and compound BB results from combining 8.60 gg of sulfur with 12.88 gg of oxygen. Show that the mass ratios in the two compounds are simple multiples of each other.
Answer:
Mass ratio of sulfur and oxygen in compounds A and B is 3:2 which confirms that the mass ratios in the two compounds are simple multiples of each other
Explanation:
This question seeks to establish/confirm the law of multiple proportions which posits that elements combine to form different substances which are whole number multiples of each other. Best example of this plays out in the formation of several oxides of the same element. Looking at the ratio in which the elements combine in each of the oxides, we can assume that these ratios are simple whole number multiples of each other.
Now back to the question.
In substance A, we have 6 g of sulfur combining with 5.99 g of oxygen
Now, lest us calculate the ratio of the mass of sulfur to that of oxygen = 6g/5.99g = 1
Now let us calculate the mass ratio of sulfur to oxygen in the second compound = 8.6/12.88 = 0.668
Now the ratios in both compounds are 1 to 0.668. 0.668 to fraction is approximately 1/1.5.
So therefore, the ratio we are having would be 1:1/1.5 or 1:0.668
This is same as 1/(1÷1.5) which is 1.5/1 or simply 3/2
This gives a ratio of approximately 1.5 to 1 or 3 to 2
The ratio 3 to 2 indicates that the mass ratios in both com pounds are simple multiples of each other
5. Rubbing alcohol is a commonly used disinfectant and has a cooling effect when applied to the skin. The active ingredient in rubbing alcohol is isopropanol. In drugstores, the most common concentration of rubbing alcohol sold contains 70% (vol/vol) isopropanol in water. Assuming the rubbing alcohol manufacturer uses a 100% isopropanol solution, what volume of pure isopropanol is required to produce a 200-mL bottle of rubbing alcohol
Answer:
Explanation:
70% (vol/vol) means
cotnaimns 70 %(vol/vol) 70 ml of isoprapnol is there in 100 ml of Rubbing sold alcohol.
if it is 200 ml then obvouly it has the 70*2 =140 ml of isoproanol required.
Alcohol is an organic compound that when rubbed on the skin it evaporates quickly leaving a cool effect on the skin. The reason why it evaporates is because it has loosely bound molecules and a low boiling temperature.
The volume of pure isopropanol required to produce a 200-ml bottle of rubbing alcohol is 140 ml
From the question:
Alcohol sold contains 70%(vol/vol). This means 70 ml of the solute of isopropanol can be found in 100 ml of solution.
Hence:
100ml of solution = 70ml of isopropanol
200ml of solution = ?
Cross Multiply
200 ml x 70 ml / 100 ml
= 140 ml
Therefore, the volume of pure isopropanol required to produce a 200-ml bottle of rubbing alcohol is 140 ml
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A compound is known to be Na2CO3, Na2SO4, NaOH, NaCl, NaC2H3O2, or NaNO3. When a barium nitrate solution is added to a solution containing the unknown a white precipitate forms. No precipitate is observed when a magnesium nitrate solution is added to a solution containing the unknown. What is the identity of the unknown compound
Answer:
Na₂SO₄
Explanation:
Barium nitrate, Ba(NO₃)₂ produce precipitate with SO₄²⁻, CO₃²⁻. That means the precipitate could be obtained from Na₂SO₄ and Na₂CO₃.
Also, magnesium nitrate, Mg(NO₃)₂, produce precipitate just with CO₃²⁻. As the unknown solution produce no precipitate, the unknown compound is:
Na₂SO₄
What happens in a double replacement reaction
Answer: D
Explanation: The elements in two compunds switch places
The substances nitrogen monoxide and hydrogen gas react to form nitrogen gas and water. Unbalanced equation: NO (g) + H2 (g) N2 (g) + H2O (l) In one reaction, 76.2 g of H2O is produced. What amount (in mol) of H2 was consumed? What mass (in grams) of N2 is produced?
Answer:
H2 consumed 4.22 mol
N2 produced 59.107 g
Explanation:
Balanced equation:
2NO (g) + 2H2 (g) N2 (g) + 2H2O (l)
To perform the calculations, the molecular weights of the following compounds must be known:H2O MW = 18.02 g/mol
N2 MW = 28.01 g/mol
To determine the moles of H2O produced, the following formula should be used:
[tex]MW=\frac{mass}{mol}[/tex]
The value of moles is cleared:
[tex]mol=\frac{mass}{MW} =\frac{76.2g}{18.02\frac{g}{mol} } =4.22 mol[/tex]
Now, to calculate the grams of N2 consumed, we look at the balanced equation and note that 2 moles of H2 produce 1 mole of N2. Therefore, through said observation, the amount of moles of H2 consumed can be determined.2 mol H2 ⇒ 1 mol N2
4.22 mol H2 ⇒ X
[tex]X=\frac{4.22mol*1 mol}{2 mol} =2.11 mol[/tex]
To calculate the mass of H2 consumed, the molecular weight equation is used again:
[tex]mass=MW*mol=28.013\frac{g}{mol}*2.11mol=59.107g[/tex]
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank with of ammonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 13. mol. Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration equilibrium constant is [tex]K_c = 14.39[/tex]
Explanation:
The chemical equation for this decomposition of ammonia is
[tex]2 NH_3[/tex] ↔ [tex]N_2 + 3 H_2[/tex]
The initial concentration of ammonia is mathematically represented a
[tex][NH_3] = \frac{n_1}{V_1} = \frac{29}{75}[/tex]
[tex][NH_3] = 0.387 \ M[/tex]
The initial concentration of nitrogen gas is mathematically represented a
[tex][N_2] = \frac{n_2}{V_2}[/tex]
[tex][N_2] = 0.173 \ M[/tex]
So looking at the equation
Initially (Before reaction)
[tex]NH_3 = 0.387 \ M[/tex]
[tex]N_2 = 0 \ M[/tex]
[tex]H_2 = 0 \ M[/tex]
During reaction(this is gotten from the reaction equation )
[tex]NH_3 = -2 x[/tex](this implies that it losses two moles of concentration )
[tex]N_2 = + x[/tex] (this implies that it gains 1 moles)
[tex]H_2 = +3 x[/tex](this implies that it gains 3 moles)
Note : x denotes concentration
At equilibrium
[tex]NH_3 = 0.387 -2x[/tex]
[tex]N_2 = x[/tex]
[tex]H_2 = 3 x[/tex]
Now since
[tex][NH_3] = 0.387 \ M[/tex]
[tex]x= 0.387 \ M[/tex]
[tex]H_2 = 3 * 0.173[/tex]
[tex]H_2 = 0.519 \ M[/tex]
[tex]NH_3 = 0.387 -2(0.173)[/tex]
[tex]NH_3 = 0.041 \ M[/tex]
Now the equilibrium constant is
[tex]K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]
substituting values
[tex]K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}[/tex]
[tex]K_c = 14.39[/tex]
Which of the following is a conjugate acid-base pair in the reaction represented by the
equation below?
H2PO4 + H20 H3PO, + OH
H2PO, and H2O
b) H,PO, and OH
c) H2PO, and H3PO,
None of the above
Answer: [tex]H_2PO_4[/tex] and [tex]H_3PO_4[/tex]
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
For the given reaction:
[tex]H_2PO_4^-+H_2O\rightleftharpoons H_3PO_4+OH^-[/tex]
Here, [tex]H_2O[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]OH^-[/tex] which is a conjugate base.
Similarly , [tex]H_2PO_4^-[/tex] is gaining a proton, thus it is considered as an base and after gaining a proton, it forms [tex]H_3PO_4[/tex] which is a conjugate acid.
Thus [tex]H_2PO_4[/tex] and [tex]H_3PO_4[/tex] is a conjugate acid-base pair in the reaction represented by the equation below
The Bronsted-Lowry conjugate acid-base hypothesis defines an acid as a substance that loses protons and donates them to another chemical to produce conjugate base, and a base as a substance that takes protons to generate conjugate acid.
Thus, a proton is being lost, making it an acid, and once a proton is lost, a conjugate base is formed. Similar to that, is gaining a proton, making it a base, and then it produces a conjugate acid after gaining a proton.
The Brnsted-Lowry hypothesis, often known as the proton theory of acids and bases, is an independent theory of acid-base reactions that was put forth in 1923 by Johannes Nicolaus Brnsted and Thomas Martin Lowry.
This theory's central idea is that when an acid and a base interact, the acid creates its conjugate base and the base creates its conjugate acid by exchanging a proton (the hydrogen cation, or H+).
Thus, The Bronsted-Lowry conjugate acid-base hypothesis defines an acid as a substance that loses protons and donates them to another chemical to produce conjugate base, and a base as a substance that takes protons to generate conjugate acid.
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A filtration system continuously removes water from a swimming pool, passes the water through filters, and then returns it to the pool. Both pipes are located near the surface of the water. The flow rate is 15 gallons per minute. The water entering the pump is at 0 psig, and the water leaving the pump is at 10 psig.
A. The diameter of the pipe that leaves the pump is 1 inch. How much flow work is done by the water as it leaves the pump and enters the pipe?
B. The water returns to the pool through an opening that is 1.5 inches in diameter, located at the surface of the water, where the pressure is 1 atm. How much work is done by the water as it leaves the pipe and enters the pool?
C. "The system" consists of the water in the pump and in the pipes that transport water between the pump and the pool. Is the system at steady state, equilibrium, both, or neither?
Answer:
A . [tex]\mathbf{W = 7133.2 \dfrac{ft. lb_f}{min} }[/tex]
B. [tex]\mathbf{W = 4245.24 \dfrac{ft. lb_f}{min} }[/tex]
C. The system is at steady state but not at equilibrium
Explanation:
Given that:
The volumetric flow rate of the water = 15 gallons per minute
The diameter of the pipe that leaves the pump is 1 inch.
A. The objective here is to determine how much work flow is done by the water as it leaves the pump and enters the pipe
The work flow that is said to be done can be expressed by the relation :
W = P × V
where;
P = pressure
V = volume
Also the given outlet pressure is the gauge pressure
The pressure in the pump P is can now be expressed by the relation:
[tex]P_{absolute} = P_{guage} + P_{atmospheric}[/tex]
[tex]P_{absolute}[/tex] = 10 psig + 14.7 psig
[tex]P_{absolute}[/tex] = 24.7 psig
W = P × V
W = 24.7 psig × 15 gal/min
[tex]W = (24.7 \ psig * \dfrac{\frac{lb_f}{in^2}}{psig}) * ( 15 \frac{gal}{min}* \dfrac{0.1337 \ ft^3}{1 \ gal }* \dfrac{144 \ in^2}{1 \ ft^2})[/tex]
[tex]\mathbf{W = 7133.2 \dfrac{ft. lb_f}{min} }[/tex]
Thus ; the rate of flow of work is said to be done by the water at [tex]\mathbf{W = 7133.2 \dfrac{ft. lb_f}{min} }[/tex]
B.
Given that :
The water returns to the pool through an opening that is 1.5 inches in diameter.
where the pressure is 1 atm.
Then ; the rate of work done by the water as it leaves the pipe and enter the pool is as follows:
W = P × V
W = 1 atm × 15 gal/min
[tex]W = 1 \ atm * ( 15 \frac{gal}{min}* \dfrac{0.1337 \ ft^3}{1 \ gal }* \dfrac{144 \ in^2}{1 \ ft^2})[/tex]
[tex]\mathbf{W = 4245.24 \dfrac{ft. lb_f}{min} }[/tex]
Thus ; the rate of flow of work done by the water leaving the pipe and enters into the pool is at [tex]\mathbf{W = 4245.24 \dfrac{ft. lb_f}{min} }[/tex]
C.
We can consider the system to be at steady state due to the fact that; the data given for the flow rate and pressure doesn't reflect upon the change in time in the space between the pump and the pool.
On the other-hand the integral factor why the system is not at equilibrium is that :
the pressure leaving the pipe is different from that of the water at the surface of the pool as stated in the question.
A vegetable soup recipe requires one teaspoonful of salt. A chef accidentally puts in one tablespoonful. Now the soup is much too salty.
a) What can the chef do to reduce the salty taste of the soup?
b) What effects would your suggestion in a) have on the soup?
Answer:
a. Put a piece of fresh sliced yam with a bore into it into the soup.
Explanation:
b. Osmosis may occur
The chef can put a slice of yam in the soup with a hole in it as it will absorb excess of salt by process of diffusion.
What is diffusion?
Diffusion is defined as the process of movement of molecules which takes place under concentration gradient. It helps in movement of substances in and out from the cell.The molecules move from lower concentration region to a higher concentration region till the concentration becomes equal.
There are 2 main types of diffusion:
1) simple diffusion-process in which substances move through a semi-permeable membrane without the aid of transport proteins.
2) facilitated diffusion- It is a passive movement of molecules across cell membrane from higher concentration region to lower concentration.
There are 2 types of facilitated diffusion one is osmosis and dialysis.
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Smooth muscle myosin is a motor protein that plays a crucial role in the contraction of smooth muscle. If this protein has a molar mass of 480,000 grams/mol, what is the mass, in grams, of 27 moles of smooth muscle myosin
Answer: Thus the mass, in grams, of 27 moles of smooth muscle myosin is 12960000 grams
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass, occupies 22.4 L at STP contains avogadro's number [tex](6.023\times 10^{23})[/tex] of particles.
Molecular mass of protein = 480,000 g/mol
Thus 1 mole of protein weighs = 480,000 g
So 27 moles of protein weighs = [tex]\frac{480,000}{1}\times 27=12960000g[/tex]
Thus the mass, in grams, of 27 moles of smooth muscle myosin is 12960000 grams
The major source of aluminum in the world this bauxite (mostly aluminum oxide). It’s thermal decomposition can be represented by:
Al2 O3 (s) —> 2 Al (s) + 3/2 O2 (g)
ΔH rxn = 1676
If aluminum is produced this way, how many grams of aluminum can conform when 1.000×10^3 kJ of heat is transferred?
Answer:
The correct answer is 32.2 grams.
Explanation:
Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,
ΔHrxn = 1676/2 = 838 kJ/mol
Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,
(1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum
The grams of aluminum produced can be obtained by using the formula,
mass = moles * molecular mass
= 1.19 * 26.98
= 32.2 grams.
In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.
What is a thermochemical equation?A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change.
Step 1: Write the thermochemical equation.Al₂O₃(s) ⇒ 2 Al(s) + 3/2 O₂(g) ΔH rxn = 1676 kJ
Step 2: Calculate the moles of Al formed when 1.000 × 10³ kJ of heat is transferred.According to the thermochemical equation, 2 moles of Al are formed when 1676 kJ of heat is transferred.
1.000 × 10³ kJ × (2 mol Al/1676 kJ) = 1.193 mol Al
Step 3: Calculate the mass corresponding to 1.193 moles of AlThe molar mass of Al is 26.98 g/mol.
1.193 mol × 26.98 g/mol = 32.19 g
In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.
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Which of the following reactions would be predicted by the activity series list
A. A metal ion reacts with another ion to form a precipitate.
B. A metal replaces a metallic ion below it on the list.
C. A metal replaces a metallic ion above it on the list.
D. A metal reacts with oxygen in a combustion reaction.
Answer:
The answer is B) A metal replaces a metallic ion below it on the list.
Explanation:
I just did it and got it correct, luckily I didn't use the other answer posted for this question.
A metal replaces a metallic ion below it on the list give reaction which would be predicted by the activity series list.
So, option B is correct one.
What is Electrochemical series?The list in which elements arranged in the increasing order of their electrode potential values is called Electrochemical series.
The Electrochemical series is also called activity series.
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A solution of pentane and ethanol (CH3CH2OH)that is 50.% pentane by mass is boiling at 57.2°C. The vapor is collected and cooled until it condenses to form a new solution.
Calculate the percent by mass of pentane in the new solution. Here's some data you may need:
normal boiling point density vapor pressure at
57.2°C
pentane 36.°C 0.63gmL 1439.torr
ethanol 78.°C 0.79gmL 326.torr
Be sure your answer has 2 significant digits.
dont round during math only for answer!
Note for advanced students: you may assume the solution and vapor above it are ideal.
Answer:
The correct answer is 81.52 percent.
Explanation:
Based on the given information, the boiling point of pentane is 36 degree C and the boiling point of ethanol is 78 degree C. The density of pentane and ethanol is 0.63 g/ml and 0.79 g/ml. The vapor pressure of pentane at 57.2 degree C is 1439 torr and the vapor pressure of ethanol at 57.2 degree C is 326 torr.
In the given case, 50 percent pentane by mass signifies that mass of pentane is 50 grams. Thus, the mass of ethanol will be 100-50 = 50 grams.
The moles or n can be calculated by using the formula,
n = weight/molecular mass
The molecular mass of pentane is 72.15 g per mol and the molar mass of ethanol is 46.07 g/mol.
The moles of pentane is,
= 50 g/72.15 g/mol = 0.6930 mol
The moles of ethanol is,
= 50 g/46.07 g/mol = 1.0853 mol
The mole fraction of pentane is,
= 0.6930 mol / (0.6930 + 1.0853) mol = 0.3897
The mole fraction of ethanol is,
= 1.0853 mol / (0.6930 + 1.0853) mol = 0.6103
Now the vapor pressure of solution will be,
= pressure of pentane * mole fraction of pentane + pressure of ethanol * mole fraction of ethanol
= (1439 * 0.3897) + (326 * 0.6103)
= 759.736 torr
The vapor pressure of pentane within the solution,
= vapor pressure of pentane * mole fraction of pentane
= 1439 torr * 0.3897
= 560.778 torr
The fraction of pentane is,
= 560.778 / 759.736 = 0.738
Let us assume that the total mole is 1, the mole fraction of pentane is 0.738, so the mole fraction of ethanol will become, 1-0.738 = 0.262
The mass of pentane = 0.738 * 72.15 = 53.2467
The mass of ethanol = 0.262 * 46.07 = 12.07034
The percent by mass of pentane in new solution will be,
Mass% = mass of pentane/Total mass * 100%
= 53.2467/(53.2467 + 12.07034) * 100%
= 53.2467/65.31704 * 100 %
= 81.52 %
how many grams of NH3 can be produced from 2.51 mil of N2 and excess H2 ?
please help! due in a bit
Answer:
85.34g of NH3
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Step 2:
Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.
Therefore, 5.02 moles of NH3 is produced from the reaction.
Step 3:
Conversion of 5.02 moles of NH3 to grams. This is illustrated below:
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Number of mole of NH3 = 5.02 moles
Mass of NH3 =..?
Mass = mole x molar Mass
Mass of NH3 = 5.02 x 17
Mass of NH3 = 85.34g
Therefore, 85.34g of NH3 is produced.
Propane (C3H8) burns in a combustion reaction. How many grams of C3H8 are needed to produce 80.3 mols CO2 ?
Answer:
1177.88g of C3H8
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
C3H8 + 5O2 —> 3CO2 + 4H2O
Next we shall determine the number of mole of C3H8 required to produce 80.3 moles of CO2. This is illustrated below:
From the balanced equation above,
1 mole of C3H8 reacted to produce 3 moles of CO2.
Therefore, Xmol of C3H8 will react to produce 80.3 moles of CO2 i.e
Xmol of C3H8 = 80.3/3
Xmol of C3H8 = 26.77 moles
Finally, we shall convert 26.77 moles of C3H8 to grams.
Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol
Mole of C3H8 = 26.77 moles
Mass of C3H8 =..?
Mass = mole x molar mass
Mass of C3H8 = 26.77 x 44
Mass of C3H8 = 1177.88g
Therefore, 1177.88g of C3H8 are needed for the reaction
If 196L of air at 1.0 atm is compressed to 26000ml, what is the new pressure
Answer:
7.5 atm
Explanation:
Initial pressure P1 = 1.0 ATM
Initial volume V1= 196 L
Final pressure P2= the unknown
Final volume V2= 26000ml or 26 L
From Boyle's law we have;
P1V1= P2V2
P2= P1V1/V2
P2= 1.0 × 196/26
P2 = 7.5 atm
Therefore, as the air is compressed, the pressure increases to 7.5 atm.
A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Litre volumetric flask. What is the molarity (M) of the resulting solution relative to the nitrate ion, NO3 1-
Answer:
[tex]M=0.213M[/tex]
Explanation:
Hello,
In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:
[tex]n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-[/tex]
[tex]n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-[/tex]
[tex]n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-[/tex]
We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:
[tex]n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-[/tex]
Finally, we compute the molarity:
[tex]M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M[/tex]
Regards.
Which of these statements gives a correct reason as to why our body needs water?
(1 Point)
1. It provides us with energy.
2. It helps us to eliminate waste.
3. It helps regulate our body temperature.
Answer:
2. It helps us to eliminate waste
3. It helps regulate our body temperature
Explanation:
In addition to the function of bringing nutrients to the cells, water provides the elimination of substances out of the body. This occurs, for example, through urine, which is basically formed by water and toxic or excess substances dissolved.
Water also helps in regulating body temperature. This occurs when the heat becomes exaggerated, sweat is released, which has water in its composition. When in contact with the medium, the sweat evaporates on the surface of the skin, causing the body to cool.
Asbestosis is a lung disease caused by inhaling asbestos fibers. The US Department of Health and Human Services considers a particular form of asbestos to be a carcinogen. The composition of this form of asbestos is 26.31% Mg, 20.20% Is, 1.45% H and the rest of the mass is due to oxygen. The molar mass of the compound is 277 g/mol. What is the molecular formula for the carcinogenic form of asbestos
Answer: The molecular formula for the carcinogenic form of asbestos [tex]Mg_3Si_2H_4O_9[/tex]
Explanation:
a) If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of Mg = 26.31 g
Mass of Si= 20.20 g
Mass of H= 1.45 g
Mass of O= (100-(26.31+ 20.20+ 1.45)) = 52.04 g
Step 1 : convert given masses into moles
Moles of Mg=[tex]\frac{\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac{26.31g}{24g/mole}=1.10moles[/tex]
Moles of Si=[tex]\frac{\text{ given mass of Si}}{\text{ molar mass of Si}}= \frac{20.20g}{28g/mole}=0.72moles[/tex]
Moles of H=[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.45g}{1g/mole}=1.45moles[/tex]
Moles of O=[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{52.04g}{16g/mole}=3.25moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Mg = [tex]\frac{1.10}{0.72}=1.5[/tex]
For Si =[tex]\frac{0.72}{0.72}=1[/tex]
For H=[tex]\frac{1.45}{0.72}=2[/tex]
For O =[tex]\frac{3.25}{0.72}=4.5[/tex]
The ratio of Mg : Si: H: O = 1.5 : 1 : 2 : 4.5
Converting them into whole numbers :
The ratio of Mg : Si: H: O = 3 : 2 : 4 : 9
Hence the empirical formula is [tex]Mg_3Si_2H_4O_9[/tex]
Empirical mass =[tex]3\times 24+2\times 28+4\times 1+9\times 16=276g[/tex]
Molecular mass = 277 g
[tex]n= \frac{\text {Molecular mass}}{\text {Empirical mass}}=\frac{277}{276}=1[/tex]
Thus molecular formula =[tex]1\times Mg_3Si_2H_4O_9=Mg_3Si_2H_4O_9[/tex]
Indicate whether the following represents a Chemical or Physical change: Milk sours
Answer:
Chemical Change
Explanation:
Physical change normally mean that the change can revert back to its orginal state, which in this case that is not possible therfore it is a chemical change.
2. In a paper chromatography analysis, three pigments, A, B, and C, were dissolved in a polar solvent. A is slightly polar, B is highly polar, and C is moderately polar. List in order how these will appear on the surface of the chromatography
what is the equation for "acid dissociation constant" of "carbonic acid"
Answer:
H2CO3 = 2H+ + CO3-
Explanation:
It is simply what carbonic acid breaks down into when placed in water. Since carbonic acid is made up of H and CO3, these are the products.
given a k value of 0.43 for the following aqueous equilibrium suppose sample z is placed into water such that its original concentration is 0.033M assume there was zero initial concentration of either A(aq) or B(ag) once equilibrium has occured what will be the equilibrium concentration of z? K=0.43
Answer:
Less than 0.033 M:
[tex][Z]_{eq}=2.4x10^{-3}M[/tex]
Explanation:
Hello,
In this case, the described equilibrium is:
[tex]2A+B\rightarrow 2Z[/tex]
Thus, the law of mass action is:
[tex]K=\frac{[Z]^2}{[A]^2[B]}=0.43[/tex]
Nevertheless, given the initial concentration of Z that is 0.033 M, we should invert the equilibrium since the reaction will move leftwards:
[tex]\frac{1}{K}=\frac{[A]^2[B]}{[Z]^2}=\frac{1}{0.43}=2.33[/tex]
Know, by introducing the change [tex]x[/tex] due to the reaction extent, we can write:
[tex]2.33=\frac{(2x)^2*x}{(0.033-2x)^2}[/tex]
Which has the following solution:
[tex]x_1=2.29M\\x_2=0.0181M\\x_3=0.0153M[/tex]
But the correct solution is [tex]x_3=0.0152M[/tex] since the other solutions make the equilibrium concentration of Z negative which is not possible. In such a way, its concentration at equilibrium is:
[tex][Z]_{eq}=0.033M-2(0.0153M)[/tex]
[tex][Z]_{eq}=2.4x10^{-3}M[/tex]
Which is clearly less than 0.033 M since the addition of a product shift the reaction leftwards in order to reestablish equilibrium (Le Chatelier's principle).
Regards.
You are trying to recrystallize compound X. You consider using ethyl acetate as your recrystallizing solvent and test a small amount of compound X with ethyl acetate. You find that compound X is soluble in ethyl acetate at room temperature and at boiling. Is ethyl acetate a good recrystallization solvent? No, the sample needs to be insoluble or sparingly soluble at room temperature so that the maximum amount of purified crystals form at room temperature and in the ice bath. Yes, you want the sample to fully dissolve at room temperature and boiling so that it will crystallize in the ice bath. Yes, you can only be sure that all the impurities dissolved if the sample is soluble at room temperature
Answer:
No, the sample needs to be insoluble or sparingly soluble at room temperature so that the maximum amount of purified crystals form at room temperature and in the ice bath.
Explanation:
For a solvent to be adequate it must completely dissolve the substance to be purified when it is hot, that is, at boiling temperature only. It should be practically insoluble when the solvent is cold or at room temperature. This must occur in this way since impurities must be removed by hot filtering or dissolved in the mother liquor.
The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Determine how much heat (in kJ) is produced by the decomposition of 1.71 mol of H2O2 under standard conditions.
Answer:
[tex]Q=-361.56kJ[/tex]
Explanation:
Hello,
In this case, the decomposition of hydrogen peroxide is given by:
[tex]2H_2O_2\rightarrow 2H_2O+O_2[/tex]
Which occurs in gaseous phase, therefore the enthalpy of reaction is:
[tex]\Delta _rH=2\Delta _fH_{H_2O}-2\Delta _fH_{H_2O_2}[/tex]
Oxygen is not included as it is a pure element. The enthalpies of formation for both hydrogen peroxide and water are -136.11 and -241.83 kJ/mol respectively, so we compute the enthalpy of reaction:
[tex]\Delta _rH=2(-241.83kJ/mol)-2(-136.11kJ/mol)=-211.44kJ/mol[/tex]
Then, the total heat that is released for 1.71 mol of hydrogen peroxide is:
[tex]Q=n*\Delta _rH=1.71mol*-211.44kJ/mol\\\\Q=-361.56kJ[/tex]
Whose sign means a released heat.
Regards.
AHP for the formation of rust (Fe2O3) is -826 kJ/mol. How much energy is
involved in the formation of 5.00 grams of rust?
A 25.9 kJ
B 25.9 J
C 66.0 kJ
D 66.0)
Answer:
A- 25.9 kJ
Explanation:
ΔH of formation is defined as the amount of energy that is involved in the formation of 1 mole of substance.
ΔH of rust is -826kJ/mol, that means when 1 mole of rust is formed, there are released -826kJ.
Moles of 5.00g of Fe₂O₃ (Molar mass: 159.69g/mol) are:
5.00g ₓ (1 mole / 159.69g) = 0.0313 moles of Fe₂O₃.
If 1 mole release -826kJ, 0.0313 moles release:
0.0313 moles ₓ (-826kJ / 1 mole) = -25.9kJ
Thus, heat involved is:
A- 25.9 kJwhat is the color of benzene and bromine
Explanation:
Benzene is colorless, with a sweet odour.
Color of Bromine is reddish brown .
Hope this helps.
Consider the reaction in a commercial heat pack: 4 Fe (s) + 3 O2(g) ® 2 Fe2O3 (s) DH = -1652 kJ a) How much heat is released when 1.00 g iron is reacted with excess O2? b) What mass of O2 must react with iron in order to generate 2150 kJ of heat?
Answer:
a) -7.395kJ of energy are released.
b) 125g of O₂ must react.
Explanation:
Based on the reaction:
4 Fe (s) + 3 O₂(g) → 2 Fe₂O₃ (s) ΔH = -1652 kJ
4 moles of iron with an excess of oxygen release -1652kJ of energy
a) The heat released is:
1.00g Fe (molar mass: 55.845g/mol)
1.00g × (1mol / 55.845g) = 0.0179 moles de Fe.
As 4 moles release -1652kJ, 0.0179 moles release:
0.0179 mol Fe × (-1652kJ / 4mol Fe) = -7.395kJ of energy are released.
b) As 3 moles of oxygen produce -1652kJ, 2150kJ are released when react:
2150kJ × (3 mol O₂ / 1652kJ) = 3.9 moles of O₂
As molar mass of O₂ is 32g/mol, mass of 3.9 moles of O₂ is:
3.9 mol O₂ × (32g / mol) = 125g of O₂ must react.
A maple tree could be studied in many fields of science. What aspects of a maple tree might be studied in chemistry?
Answer:
Chemical reactions, kinetics, organic chemistry
Explanation:
You might study the chemical reaction, learn about the differences between products and reactants, about delta H and exothermic and endothermic reactions. You may also study Kinetics by studying the rates of reactions with certain chemicals in a maple's enzymatic processes.
Another thing that you might learn about is organic chemistry. The glucose molecules, carbohydrates, lipids, nucleic acids, all have a structure based on the Carbon atom. You can learn about the specific structures of some chemicals that are involved in photosynthesis and simple hydrocarbons that are involved in photosynthetic/bio-synthetic pathways.
There's probably a lot more - but these are the most basic things I could think of.
In which of these statements are protons, electrons, and neutrons correctly compared?
Quarks are present in protons and neutrons but not in electrons.
Quarks are present in protons, neutrons, and electrons.
Quarks are present in neutrons and electrons but not in protons.
Quarks are present in protons and electrons but not in neutrons
the second statement is the correct one quarks are needed to balance charges in all subatomic particles such as neutrons, protons and electrons
