Answer:
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Hrxn = Q reaction / mol of reaction
mol of reaction = M * V = 10 * 1 = 10 mmol = 0.01 mol
Q water = m * C * (Tf - Ti)
= (10 + 10) (4.184) (26-20) = 502.08 J
Q reaction = - Q water = -502.08 J
Hrxn = -502.08 / (0.01) = - 50208 J = - 50.21 kJ/mol
There are 3.98 × 10^23 atoms of oxygen in the sample.
Given that;
1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen
x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen
x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms
x = 0.0368 moles
The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18
Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.
Learn more: brainly.com/question/9743981
Answer: P= 1.64 atm
Explanation: solution attached.
Use Ideal gas law
PV= nRT
Derive for P
P= nRT/V R= 0.08205 L.atm/mol.K
Substitute the values.
I believe that the answer is 12 because there is already 3 O molecules and since its in parentheses with 3 outside it that means that there are 3 of those CO molecules meaning that for every 1 CO there will be 3 O’s so 3, four times Is 12
