C + O2 ——> CO2
Carbon plus oxygen forms carbon dioxide
<u>Answer:</u> The value of 
for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K
<u>Explanation:</u>
Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate entropy change is of a reaction is:
![\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the entropy change of the above reaction is:
![\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20S%5Eo_%7B%28HF%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28H_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28F_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28173.78%29%29%5D-%5B%281%5Ctimes%20%28130.68%29%29%2B%281%5Ctimes%20%28202.78%29%29%5D%5C%5C%5C%5C%5CDelta%20S%5Eo_%7Brxn%7D%3D14.1J%2FK)
Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K
We are given:
Moles of hydrogen gas reacted = 2.20 moles
By Stoichiometry of the reaction:
When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K
So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = 
Hence, the value of for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K
Answer:
The most abundant isotope is 1.007 amu.
Explanation:
Given data:
Average atomic mass = 1.008 amu
Mass of first isotope = 1.007 amu
Mass of 2nd isotope = 2.014 amu
Most abundant isotope = ?
Solution:
First of all we will set the fraction for both isotopes
X for the isotopes having mass 2.014 amu
1-x for isotopes having mass 1.007 amu
The average atomic mass is 1.008 amu
we will use the following equation,
2.014x + 1.007 (1-x) = 1.008
2.014x + 1.007 - 1.007 x = 1.008
2.014x - 1.007x = 1.008 - 1.007
1.007 x = 0.001
x= 0.001/ 1.007
x= 0.0009
0.0009 × 100 = 0.09 %
0.09 % is abundance of isotope having mass 2.014 amu because we solve the fraction x.
now we will calculate the abundance of second isotope.
(1-x)
1-0.0009 = 0.9991
0.9991 × 100= 99.91%
Answer:
E) All three samples have the same number of hydrogen atoms
Explanation:
The statements are:
A) Sample A has more hydrogen atoms than sample B or sample C.
B) Sample B has more hydrogen atoms than sample A or sample C.
C) Sample C has more hydrogen atoms than sample A or sample B
D) Both samples A and C have the same number of hydrogen atoms, but more than in sample B.
E) All three samples have the same number of hydrogen atoms
Hydrogens in sample A are:
180g × (1mol Glucose / 180.156g) × (12mol H / 1 mol Glucose) =<em> 12 moles of H</em>
Hydrogens in sample B are:
90g Glucose × (1mol Glucose / 180.156g) × (12mol H / 1 mol Glucose) =<em> 6 moles of H</em>
90g mannose× (1mol Mannose / 180.156g) × (12mol H / 1 mol Glucose) =<em> 6 moles of H</em>
<em>Total moles: 12</em>
Hydrogens in sample C are:
180g × (1mol Mannose / 180.156g) × (12mol H / 1 mol Glucose) =<em> 12 moles of H</em>
<em />
Thus, right answer is:
<em>E) All three samples have the same number of hydrogen atoms</em>
<em></em>
I hope it helps!
Answer:
D
Explanation:
You are looking for kg / L
3.188 kg / 3.23 L = .987 kg/L
