Answer:
The answer to your question is maybe letter D, but the last oxygen needs a number 6.
Explanation:
The empirical formula gives the actual elements that form part of a molecule but not the total numbers.
The molecular formula gives the total number of atoms of each element in a molecule.
We must factor the molecular formula to know if a formula is the empirical formula of that.
A. CH₄ C₂H₆ = 2(CH₃) these are not empirical molecular formulas
B. CH₂O C₄H₆O these are not empirical-molecular formulas
C. O₂ O₃ these are not empirical-molecular formulas
D. C₃H₄O₃ C₆H₈O these are not empirical-molecular formulas
the last oxygen needs a number 6 to be
the answer.
Answer:
5.83 mol.
Explanation:
- From the balanced reaction:
<em>2Al + 3Ag₂S → 6Ag + Al₂S₃,</em>
It is clear that 2 mol of Al react with 3 mol of Ag₂S to produce 1 mol of Ag and 1 mol of Al₂S₃.
Al reacts with Ag₂S with (2: 3) molar ratio.
<em>So, 2.27 mol of Al reacts completely with 3.4 mol of Ag₂S with (2: 3) molar ratio.</em>
<em />
- The limiting reactant is Ag₂S.
- The excess "left over" reactant is Al.
The reamining moles of excess reactant "Al" = 8.1 mol - 2.27 mol = 5.83 mol.
Answer:
(a) 
(b) 
(c) 
Explanation:
Hello,
(a) In this case, since entropy remains unchanged, the constant 
should be computed for air as an ideal gas by:
Next, we compute the final temperature:
Thus, the work is computed by:
(b) In this case, since 
is given, we compute the final temperature as well:
And the isentropic work:
(c) Finally, for isothermal, final temperature is not required as it could be computed as:
Regards.
Ammonia is formed by a reaction between hydrogen and nitrogen as shown by the equation below.
N2(g) + 3H2(g) = 2NH3(g)
1 mole of ammonia contains 17 g
Therefore 10.78 g of ammonia are equivalent to 10.78/17 = 0.6341 moles
The mole ratio of hydrogen to ammonia is 3 : 2
Therefore, moles of hydrogen used will be 0.6341 × 3/2 = 0.9512 moles
1 mole of hydrogen is equivalent to 2 g
Thus, the mas of hydrogen will be 0.9512 moles × 2 = 1.9023 g
