Answer:
endothermic
Explanation:
Heat is added to make the process possible.
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.
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.
Which of the following structures in the human body has the highest level of organization
Answer:
The brain
Explanation:
With all those instructions the body recqures to respond to it must be so
Hope it helps
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.
There are __________ moles of N atoms present in a 2.0 g C8H10O2N4.
Answer:
[tex]n_N=0.041molN[/tex]
Explanation:
Hello,
In this case, for this mole-mass relationship, we are able to compute the moles of nitrogen atoms by firstly obtaining the moles of the given compound, considering its molar mass that is 194 g/mol:
[tex]n_{C_8H_{10}O_2N_4}=2.0gC_8H_{10}O_2N_4*\frac{1molC_8H_{10}O_2N_4}{194gC_8H_{10}O_2N_4} =0.01molC_8H_{10}O_2N_4[/tex]
Then, by knowing that one mole of the given compound has four moles of nitrogen atoms, we apply the following relationship:
[tex]n_N=0.01molC_8H_{10}O_2N_4*\frac{4molN}{1molC_8H_{10}O_2N_4} \\\\n_N=0.041molN[/tex]
Best regards.
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
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.
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
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.
Identify whether each species functions as a Brønsted-Lowry acid or a Brønsted-Lowry base in this net ionic equation. HF (aq) + SO32- ⇌ F- + HSO3- Brønsted-Lowry _____ Brønsted-Lowry _____ Brønsted-Lowry _____ Brønsted-Lowry _____ In this reaction: The formula for the conjugate _____ of HF is The formula for the conjugate _____ of SO32- is
Explanation:
A Bronsted-Lowry base is a substance that accepts a proton in the form of a hydrogen (H) atom.
On the other hand;
Bronsted-Lowry acid is the substance that donates the proton.
HF (aq) + SO32- ⇌ F- + HSO3-
In the forward reaction;
Bronsted-Lowry acid : HF
Bronsted-Lowry base: SO32-
In the backward reaction;
Bronsted-Lowry acid : HSO3-
Bronsted-Lowry base: F-
The conjugate base of HF is F-
The conjugate acid of SO32- is HSO3-
How many significant figures are in 382.90?
Answer:
5
Explanation:
Answer:
5
Explanation:
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.
Learn more about diffusion,here:
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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.
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
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
What is the atomic mass of AlNO2?
Answer:
I am not sure, but I think this is the answer 72.987 g/mol
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.
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]
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 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.
What is it called when a gas changes into a liquid?
Answer:
Explanation:
Condensation is the word you seek.
Answer:
Condensation
Explanation:
When a gas is subjected to decrease in temperature it is condensed
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]
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.
To learn more about Electrochemical series here.
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During which stage of the water cycle could water enter the atmosphere as a gas? A. transpiration B. precipitation C. accumulation D. condensation
Answer: Transpiration---A
Explanation: Transpiration is the process in the water cycle whereby plant loose(excess) water by evaporation through the stomata of their leaves since not all water absorbed by the root is actually used for growth in plants.In order to allow the intake of carbon-dioxide, water must exit the leaves through transpiration which then provides the plant with cooling, rigidity and maintaining the overall water balance of the plant.
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.
which element causes burning when me mix it with oxygen
Answer:
Hydrogen peroxide is ans
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
what is the color of benzene and bromine
Explanation:
Benzene is colorless, with a sweet odour.
Color of Bromine is reddish brown .
Hope this helps.
