Answer:
u
uric acid is a useful diagnostic tool as screening for most of purine metabolic disorders. The importance of uric acid measurement in plasma and urine with respect of metabolic disorders is highlighted. Not only gout and renal stones are indications to send blood to the laboratory for uric acid examination
How many moles of PC15 can be produced from 51.0 g of Cl2 (and excess P4)?
Express your answer to three significant figures and include the appropriate units.
LIT....ITS NOT .227 or .228!!!!
Answer:
0.287 mole of PCl5.
Explanation:
We'll begin by calculating the number of mole in 51g of Cl2. This is illustrated below:
Molar mass of Cl2 = 2 x 35.5 = 71g/mol
Mass of Cl2 = 51g
Number of mole of Cl2 =..?
Mole = Mass /Molar Mass
Number of mole of Cl2 = 51/71 = 0.718 mole
Next, we shall write the balanced equation for the reaction. This is given below:
P4 + 10Cl2 → 4PCl5
Finally, we determine the number of mole of PCl5 produced from the reaction as follow:
From the balanced equation above,
10 moles of Cl2 reacted to produce 4 moles of PCl5.
Therefore, 0.718 mole of Cl2 will react to produce = (0.718 x 4)/10 = 0.287 mole of PCl5.
Therefore, 0.287 mole of PCl5 is produced from the reaction.
The reaction of hydrogen bromide(g) with chlorine(g) to form hydrogen chloride(g) and bromine(g) proceeds as follows: 2HBr(g) + Cl2(g)2HCl(g) + Br2(g) When 23.9 g HBr(g) reacts with sufficient Cl2(g), 12.0 kJ is evolved. Calculate the value of rH for the chemical equation given.
Answer:
The enthalpy of reaction per mole of HBr for this reaction = ΔrH = -40.62 kJ/mole.
Explanation:
2HBr(g) + Cl2(g) → 2HCl(g) + Br2(g)
When 23.9 g HBr(g) reacts with sufficient Cl2(g), 12.0 kJ of heat is evolved, calculate the value of ΔrH for the chemical reaction.
Note that ΔrH is the enthalpy per mole for the reaction.
Molar mass of HBr (g) = 80.91 g/mol.
Hence, 1 mole of HBr = 80.91 g
23.9 g of HBr led to the reaction giving off 12.0 kJ of heat
80.91 g of HBr will lead to the evolution of (80.91 × 12/23.9) = 40.62 kJ heat is given off.
Hence, 40.62 kJ of heat is given off per 80.91 g of HBr.
This directly translates to that 40.62 kJ of heat is given off per 1 mole of HBr
Hence, the heat given off per mole of HBr for this reaction is 40.62 kJ/mole.
But since the reaction liberates heat, it means the reaction is exothermic and the enthalpy change for the reaction (ΔHrxn) is negative.
Hence, ΔrH = -40.62 kJ/mole.
Hope this Helps!!!
Clues about the history of the earth have been obtained from the study of
fossil fuels
rain forest materials
soll samples
O synthetic plastics
Answer: Fossil fuels
Explanation:
Fossil fuels such as petroleum, oil, and natural gas, are non-renewable energy resources which are formed from the remains of prehistoric ancient plants and animals beneath layers of rock of the earth surface.
By analyzing and studying fossil fuels using Radiocarbon analyses by archaeologists, earth scientists etc, Information about the history of the earth can be obtained from the decomposition of dead organisms present in fossil fuels.
A chemistry student is given 600. mL of a clear aqueous solution at 37° C. He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 21° C. At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitate, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 0.084 kg. Using only the information from above, can you calculate the solubility of X at 21° C?If yes, calculate it. Be sure your answer has a unit symbol and the right number of significant figures.
Answer:
The solubility is [tex]S = 0.0014 \ g[/tex]
Explanation:
From the question we are told that
The volume of the solution is [tex]V = 600 mL[/tex]
The initial temperature is [tex]T_i = 37 ^oC[/tex]
The final temperature is [tex]T_f = 21^oC[/tex]
The additional precipitate is [tex]m = 0.084 \ kg = 84 \ g[/tex]
Yes because the solubility of the substance X is the amount of X needed to saturate a unit volume of the solvent (for solubility of a solute to be calculated the solute must be able to saturate the solvent)
now we see that the substance X saturated the solvent because a precipitate was formed which the student threw away
The solubility at 21 ° C is mathematically represented as
[tex]S = \frac{m}{m_w * 100 g \ of water }[/tex]
Mass of water([tex]m_w[/tex]) in the solution is mathematically represented as
[tex]m_w = V * \rho_w[/tex]
Where [tex]\rho = 1 \frac{g}{mL}[/tex]
So
[tex]m_w =600 * 1[/tex]
[tex]m_w =600g[/tex]
So
[tex]S = \frac{84}{600 * 100 g \ of water }[/tex]
[tex]S = 0.0014 \ g[/tex]
Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . If of sodium bromide is produced from the reaction of of hydrobromic acid and of sodium hydroxide, calculate the percent yield of sodium bromide.
Answer:
The percentage yield is 50%
Use the formation reactions below such that when added together, they match the balanced equation for the combustion of methane.
Cgraphite(s)+ 2H2(g) → CH4(g) ΔH 1=−74.80kJ
Cgraphite(s)+ O2(g) → CO2(g) ΔH2=−393.5k
H2(g)+ 1/2O2(g) → H2O(g) ΔH3=−241.80kJ
Calculate ΔHrxn for the combustion of methane, CH4(g).
CH4(g)+ 2O2(g) → CO2(g)+ 2H2O(g) ΔHrxn =--------------kJ
Answer:
ΔH of the reaction is -802.3kJ.
Explanation:
Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.
Using the reactions:
(1) Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ
(2) Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k J
(3) H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ
The sum of (2) - (1) produce:
CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ
And the sum of this reaction with 2×(3) produce:
CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =
-802.3kJ
If g(x) is the inverse of f(x) and f(x)=4x+12, what is g(x)?
Answer: [tex]y = \frac{1}{4}x-3[/tex]
Explanation:
To find the inverse of an equation, follow these steps:
Replace every f(x) or y with x, and every x with y. Solve the equation for yWe are given the equation [tex]f(x) = 4x + 12[/tex] , so replace f(x) with x.
Then, replace x with y.
Your new equation:
[tex]x = 4y + 12[/tex]
Now, solve for y:
[tex]x = 4y + 12\\\\4y = x - 12\\\\y = \frac{1}{4}x-3[/tex]
This equation is the inverse of f(x), or g(x).
Solids in which the atoms have no particular order or pattern are called what solids
Answer:
Amorphous solids .
Explanation:
They have no particular order or pattern.Each particle is in a particular spot, but the particles are in no organized pattern.
A 5.00-L tank contains helium gas at 1.50 atm. What is the pressure of the gas in mmHg
Answer:
1140 mmHg
Explanation:
1 atmosphere is 760 mmHg, so 1.5 atmospheres is ...
1.5×760 mmHg = 1140 mmHg
Calculate the amount of ATP in kg that is turned over by a resting human every 24 hours. Assume that a typical human contains ~50g of ATP (Mr 505) and consumes ~8000 kJ of energy in food each day. The energy stored in the terminal anhydride bond of ATP under standard conditions is 30.6 kJmol-1. Assume also that the dietary energy is channeled through ATP with an energy transfer efficiency of ~50%.
Answer:
The correct answer is 66.35 kilograms.
Explanation:
Based on the data given in the question, the energy consumed by the body of a human being is 50%. Based on the given data, the energy consumed in a day is 8000 kJ, 50 percent is the energy transfer efficiency. Thus, the consumption of total energy is 4000 kJ, and for the transformation of ADP to ATP, the energy involved is 30.6 kJ per mole.
Hence, the total ATP produced in the process is,
ATP = 4000 kJ / 30.6 kJ/mol
= 130.7189 mol.
Thus, with the energy transfer efficiency of 50 percent, the total moles of ATP produced is 130.7 mol.
The mass of ATP can be calculated by using the formula,
moles = mass/molecular mass
The molecular mass of ATP is 507.18 g per mol
Now by putting the values we get,
mass of ATP = 130.7189 mol * 507.18 g/mol
= 66298.011 g or 66.298 kg
It is mentioned that human comprise 50 g of ATP or 0.05 kg of ATP. Therefore, the sum of the available ATP will be.
= Total production of ATP + Total ATP available
= 66.298 kg + 0.05 kg
= 66.348 kg
Hence, the sum of the ATP that is turned over by a resting human in a day is 66.35 kg.
Consider 10.0 g of helium gas (He) in a rigid steel container. If you add 10.0 g of neon gas (Ne) to this container, which of the following best describes what happens? (Assume the temperature is constant.)
a) The pressure in the container doubles.
b) The pressure in the container more than doubles.
c) The volume of the container doubles.
d) The volume of the container more than doubles.
e) The pressure in the container increases but does not double.
Answer: (e) The pressure in the container increases but does not double.
Explanation:
To solve this, we need to first remember our gas law, Boyle's law states that the pressure and volume of a gas have an inverse relationship. That is, If volume increases, then pressure decreases and vice versa, when temperature is held constant. Therefore, increasing the volume in this case does not double the pressure owning to out gas law, but an increase in pressure would be noticed if temperature is constant
The pressure in the container increases but does not double.
At constant temperature and volume, the pressure of a given mass of gas is directly proportional to the number of moles of gas present.
Number of moles of He = 10 g/4 g/mol = 2.5 moles
Number of moles of Ne = 10 g/20 g/mol = 0.5 moles
We can see that the number of moles only increases by 1/5 of its initial value therefore, the pressure in the container increases but does not double.
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Please help! (:
question above — how much money would you need to buy 7.0 lb of arugula? If 27lb of arugula cost $16
Answer:
$11.81
Explanation:
27 lb cost $16
27/16=$1.69 per pound
$1.69*7=$11.81 for 7 lbs
A gas has volume of 800.0mL at -23.0°c and 300.0torr. What would the volume of the gas be at 227.0°c and 600.0torr of pressure
Answer:
Explanation:
use gas law eqation
P1 * V1 / T1 = P2 * V2 /T2
600*V1/227 = 300*800/23
V1 = 300*800*227 / 23*600 = ............ can you solve this and get the answer?
C12H22O11 + 12O2 ---> 12CO2 + 11H2O
there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent?
Answer:
Oxygen is the limiting reactant.
Explanation:
Based on the reaction:
C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O
1 mole of sucrose reacts with 12 moles of oxygen to produce 12 moles of CO₂ and 11 moles of H₂O.
10.0g of sucrose (Molar mass: 342.3g /mol) are:
10.0g C₁₂H₂₂O₁₁ × (1mole / 342.3g) = 0.0292 moles of C₁₂H₂₂O₁₁
And moles of 10.0g of oxygen (Molar mass: 32g/mol) are:
10.0g O₂ × (1mole / 32g) = 0.3125 moles of O₂
For a complete reaction of 0.0292 moles of C₁₂H₂₂O₁₁ you need (knowing 12 moles of oxygen react per mole of sucrose):
0.0292 moles of C₁₂H₂₂O₁₁ × (12 moles O₂ / 1 mole C₁₂H₂₂O₁₁) = 0.3504 moles of O₂
As you have just 0.3125 moles of O₂, oxygen is the limiting reactant.
What can be known about the salt sample that Gerry is looking at?
Answer:
That its small pointed. Pink(Himalayan salt)or white(normal salt)
Explanation:
Summa dees questions are so stupid, deys makin me salty.
According to the following reaction, what amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? A few of the molar masses are as follows: Al2S3 = 150.17 g/mol, H2O = 18.02 g/mol.
Answer:
Mass of Al2S3 remaining is 17.212 g
Explanation:
Equation of the reaction is given below:
Al2S3 + 6H2O -----> 2Al(OH)3 + 3H2S
From the balanced equation above
6 mole of H20 reacts with 1 mole of Al2S3
i.e. 6 * 18.02 g of H2O reacts with 1 * 150.71 g of Al2S3
= 108.12 g of H2O reacts with 150.71 g of Al2S3
Therefore 2.0 g of water will react with 2.0 * (150.71/108.12) g of Al2S3
= 2.788 g of Al2S3
Mass of Al2S3 remaining = 20.0 g - 2.788 g = 17.212 g
According to the properly balanced chemical equation, the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction is 17.22 grams.
Given the following data:
Mass of [tex]AL_2S_3[/tex] = 20.00 gramsMass of [tex]H_2O[/tex] = 2.00 gramsMolar mass of [tex]AL_2S_3[/tex] = 150.17 g/molMolar mass of [tex]H_2O[/tex] = 18.02 g/mol.To calculate the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction:
First of all, we would write a properly balanced chemical equation for this chemical reaction.
[tex]Al_2S_3 + 6H_2O ---> 2Al(OH)_3 + 3H_2S[/tex]
By stoichiometry:
1 mole of [tex]AL_2S_3[/tex] reacts with 6 moles of [tex]H_2O[/tex]
Next, we would calculate the mass of each compound.
For [tex]AL_2S_3[/tex]:
[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 1 \times 150.17[/tex]
Mass = 150.17 grams
For [tex]H_2O[/tex]:
[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 6 \times 18.02[/tex]
Mass = 108.12 grams
108.12 grams of [tex]H_2O[/tex] = 150.17 grams of [tex]AL_2S_3[/tex]
2.00 grams of [tex]H_2O[/tex] = X grams of [tex]AL_2S_3[/tex]
Cross-multiplying, we have:
[tex]108.12 \times X = 150.17 \times 2\\\\108.12X = 300.34\\\\X = \frac{300.34}{108.12}[/tex]
X = 2.78 grams of [tex]AL_2S_3[/tex]
Remaining mass = [tex]20.00 - 2.78[/tex]
Remaining mass = 17.22 grams of [tex]AL_2S_3[/tex]
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From the unbalanced reaction: B2H6 + O2 ---> HBO2 + H2O
How many grams of O2 (32g/mol) will be needed to burn 36.1 g of B2H6 (Molar mass = 27.67g/mol)? ______g
Include the correct number of significant figures in your final answer
Answer: 125 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles[/tex]
The balanced reaction is:
[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]
According to stoichiometry :
1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 1.30 moles of [tex]B_2H_6[/tex] will require=[tex]\frac{3}{1}\times 1.30=3.90moles[/tex] of [tex]O_2[/tex]
Mass of [tex]O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g[/tex]
Thus 125 g of [tex]O_2[/tex] will be needed to burn 36.1 g of [tex]B_2H_6[/tex]
Could someone please help me with this chemistry question I will mark the correct answer as brainliest
An organic chemistry student was studying the solubility of Methyl-N-acetyl-α-D-glucosaminide (1-O-methyl-GlcNAc), a derivative of glucosamine, in water but inadvertently added 1 equiv. of periodic acid instead. Based on your understanding of the reactions of monosaccharides with periodates, draw the organic product that the student obtained.
Complete Question
The diagram for this question is shown on the second uploaded image
Answer:
The organic product obtained is shown on the first uploaded image
Explanation:
The process that lead to this product formation is known as oxidative cleavage which is a reaction that involves the cleavage of a carbon to carbon bond at the same time this carbon which formed the carbon bond are oxidized i.e oxygen is been added to them
6. To isolate benzoic acid from a bicarbonate solution, it is acidified with concen- trated hydrochloric acid, as in experiment 1. What volume of acid is needed to neutralize the bicarbonate
Answer:
For our assumed experiment; the expected volume of Hcl acid needed to neutralize the bicarbonate is 0.13 mL
Explanation:
We are going attempt this question experimentally.
We know that benzoic acid originate from the relationship between benzene and a carboxylic group. So basically , the functional group of a carboxylic acid (-COOH) joins with a benzene ring(C₆H₆) to form a simple aromatic carboxylic acid known as Benzoic acid. (C₇H₆O₂)
However, it is possible to isolate benzoic acid from a bicarbonate solution in the presence of an acidified concentrated hydrochloric acid.
Let assume that ;
0.20 g of benzoic acid was reacted with 2 mL of a 20% solution of NaHCO₃, the amount of the excess NaHCO₃ can be determined by subtracting the amount of benzoic acid from the amount of NaHCO₃.
Let first calculate the number of moles in 0.20 g of benzoic acid
we know that the standard molar mass of benzoic acid is 122.12 g/mol
number of moles of benzoic acid = mass of benzoic acid/molar mass of benzoic acid =
number of moles of benzoic acid = 0.20/ 122.12
number of moles of benzoic acid = 0.0016 mol
number of moles of bicarbonate solution = mass of bicarbonate solution/ molar mass of bicarbonate solution
number of moles of bicarbonate solution = 0.2/84.00654 g/mol
number of moles of bicarbonate solution = 0.00238 mol
∴
(0.00238 - 0.0016) mol
= 7.8 × 10⁻⁴ mol
Let assume that the concentrated HCl is 12 M
Also. HCl and NaHCO₃ react together at the ratio of 1:1; thus the volume of Hcl acid needed to neutralize the bicarbonate is:
[tex]= ( 7.8 * 10^{-4} \ \ mol )* ( \dfrac{2\ L}{ 12 \ M})*( \dfrac{10^3 \mL}{1 \ L})[/tex]
= 0.13 mL
Thus; for our assumed experiment; the expected volume of Hcl acid needed to neutralize the bicarbonate is 0.13 mL
How do the particles in plasmas compare with the particles in solids?
Answer:
Plasmas and solids are both made up of cation-anion pairs. Solids and plasmas are both made up of electrons and cations. Solids are made up of cation-anion pairs, but plasmas are not.
Answer:
Solids are made up of cation-anion pairs, but plasmas are not.
Explanation:
A rule of thumb is that a reaction rate roughly doubles for every 10 °C increase in temperature. What is the activation energy of a reaction whose rate exactly doubles between 25.0 °C and 35.0 °C
Answer:
FOR EVERY 10 DEGREE CELSIUS INCREASE IN TEMPERATURE, THE ACTIVATION ENERGY THAT SHOWS THIS IS 52.4 KJ/MOL
Explanation:
From Arrhenius equation, the relationship between the rate constant and the temperature is as shown below:
k = Ae^ -Ea/RT
At initial temperature T1, the initial rate constant is (k1)
At final temperature T2, the final rate constant is k2
For the reaction rate to be doubled, we must double the rate constant which shows that the ratio of k2 / k1 must be equal to 2.
That is, k2 / k1 = 2 (rate is doubled)
Equating this into the Arrhenius equation, we have:
k2 / k1 = Ae^ (-Ea / R ) (1/ T2 - 1/T1)
2 = e^ (-Ea / R) (1 / T2 = 1 / T1)
Taking the natural logarithm of both sides:
ln 2 = - (Ea / R) (1 / T2 - 1 / T1)
Making Ea the subject of the formula, we obtain:
Ea = - (ln 2 R / (1 / T2- 1 / T1))
Let T1 = 25 C = 25 + 273 K = 298 K
T2 = 35 C = 35 + 273 K = 308 K
R = 8.314
So,
Ea = - (ln 2 * 8.314 / ( 1/308 - 1 / 298))
Ea = - (0.693 * 8.314 / 0.00324 - 0.00335)
Ea = - 5.7616 / -0.00011
Ea = 52 378,18 J / mol
So therefore, the activation energy Ea is 52.4 kJ/mol.
For some hypothetical metal the equilibrium number of vacancies at 750°C is 2.8 × 1024 m−3. If the density and atomic weight of this metal are 5.60 g/cm3 and 65.6 g/mol, respectively, calculate the fraction of vacancies for this metal at 750°C.
Answer:
The correct answer is 5.447 × 10⁻⁵ vacancies per atom.
Explanation:
Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,
Fv = Nv/N------ (i)
Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,
N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,
N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol
N = 5.14073 × 10²⁸ atoms/m³
Now putting the values of Nv and N in the equation (i) we get,
Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3
Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.
An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.35×10-2 M CH2Cl2, 0.173 M CH4 and 0.173 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.155 mol of CH4(g) is added to the flask?
Answer:
[CH₂Cl₂] = 7.07x10⁻² M
[CH₄] = 0.319 M
[CCl₄] = 0.164 M
Explanation:
The equilibrium reaction is the following:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
The equilibrium constant of the above reaction is:
[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]
When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:
[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]
[tex]C_{CH_{4}} = 0.328 M[/tex]
Now, the concentrations at the equilibrium are:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
5.35x10⁻² - 2x 0.328 + x 0.173 + x
[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]
[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]
Solving the above equation for x:
x₁ = 0.076 and x₂ = -0.0086
Hence, the concentration of the three gases once equilibrium has been reestablished is:
[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M
[CH₄] = 0.328 + (-0.0086) = 0.319 M
[CCl₄] = 0.173 + (-0.0086) = 0.164 M
We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.
I hope it helps you!
7.Which one of the following statements is not true?
1 point
O The molecules in a solid vibrate about a fixed position
O The molecules in a liquid are arranged in a regular pattern
The molecules in a gas exert negligibly small forces on each other, except during
collisions
The molecules of a gas occupy all the space available
Answer:
B. the molecules in liquid are loosely packed and scattered thus, they cannot be arranged
Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of the visible spectrum. Suppose a particular cone cell absorbs light with a wavelength of 434.nm. Calculate the frequency of this light. Round your answer to 3 significant digits.
Answer:
6.91 × 10¹⁴ s⁻¹
Explanation:
Step 1: Given data
Wavelength of the radiation absorbed by the cone (λ): 434 nm
Step 2: Convert the wavelength to meters
We will use the relationship 1 m = 10⁹ nm.
[tex]434nm \times \frac{1m}{10^{9}nm } =4.34 \times 10^{-7} m[/tex]
Step 3: Calculate the frequency (ν) of the radiation
We will use the following expression.
[tex]c = \lambda \times \nu[/tex]
where,
c is the speed of light (3.00 × 10⁸ m/s)
[tex]c = \lambda \times \nu\\\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^{8}m/s }{4.34 \times 10^{-7}m }= 6.91 \times 10^{14} s^{-1}[/tex]
What are the relations between Electrochemistry and Cancer?
Answer: if im not wrong the relations are that the electrochemistry can detect the cancer and any other sickness
just like it does with chemical phenomena
=)
Draw every stereoisomer for 1-bromo-2-chloro-1,2-difluorocyclopentane. Use wedge-and-dash bonds for the substituent groups, and be sure that they are drawn on the outside of the ring, adjacent to each other. The skeletal structure of one molecule is included to indicate the proper format.
Answer:
Explanation:
The objective here is mainly drawing the diagrams of every stereoisomer for 1-bromo-2-chloro-1,2-difluorocyclopentane.
Stereoisomerism is the difference of the spatial arrangement of atoms in a molecule or a compound with the same molecular formula.
For 1-bromo-2-chloro-1,2-difluorocyclopentane.
We have the stereoisomers as follows:
(1R,2S)-1-bromo-2-chloro-1,2-difluorocyclopentane.
(1S,2R)-1-bromo-2-chloro-1,2-difluorocyclopentane.
(1S,1S)-1-bromo-2-chloro-1,2-difluorocyclopentane.
(1R,1R)-1-bromo-2-chloro-1,2-difluorocyclopentane.
Their diagrams are drawn and shown in the attached file below in the order with which they are listed above.
What type of bond will be formed for atoms that have a +1 or -1 charge?
Determine the limiting reactant in a mixture containing 95.7 g of B2O3, 75.7 g of C, and 369 g of Cl2. Calculate the maximum mass (in grams) of boron trichloride, BCl3, that can be produced in the reaction. The limiting reactant is:
Answer:
[tex]B_2O_3[/tex]
Explanation:
First, we have to find the reaction:
[tex]B_2O_3~+~C~+~Cl_2~->~BCl_3~+~CO[/tex]
The next step is to balance the reaction:
[tex]B_2O_3~+~3C~+~3Cl_2~->~2BCl_3~+~3CO[/tex]
Now, we have to calculate the molar mass for each compound, so:
[tex]B_2O_3=~69.62~g/mol[/tex]
[tex]C=~12~g/mol[/tex]
[tex]Cl_2=~70.96~g/mol[/tex]
With these values, we can calculate the moles of each compound:
[tex]95.7~g~B_2O_3\frac{1~mol~B_2O_3}{69.62~g~B_2O_3}=1.37~mol~B_2O_3[/tex]
[tex]75.7~g~C\frac{1~mol~C}{112~g~C}=6.30~mol~C[/tex]
[tex]369~g~Cl_2\frac{1~mol~Cl_2}{70.96~g~C}=5.20~mol~Cl_2[/tex]
Now we can divide by the coefficient of each compound in the balanced equation:
[tex]\frac{1.37~mol~B_2O_3}{1}=~1.37[/tex]
[tex]\frac{6.30~mol~C}{3}=~2.1[/tex]
[tex]\frac{5.20~mol~Cl_2}{3}=~1.73[/tex]
The smallest values are for [tex]B_2O_3[/tex], so this is our limiting reagent.
I hope it heps!
