25mL if water as the highest average of the kinetic energy
<span>9.40x10^19 molecules.
The balanced equation for ammonia is:
N2 + 3H2 ==> 2NH3
So for every 3 moles of hydrogen gas, 2 moles of ammonia is produced. So let's calculate the molar mass of hydrogen and ammonia, starting with the respective atomic weights:
Atomic weight nitrogen = 14.0067
Atomic weight hydrogen = 1.00794
Molar mass H2 = 2 * 1.00794 = 2.01588 g/mol
Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol
Moles H2 = 4.72 x 10^-4 g / 2.01588 g/mol = 2.34140921086573x10^-4 mol
Moles NH3 = 2.34140921086573x10^-4 mol * (2/3) = 1.56094x10^-4 mol
Now to convert from moles to molecules, just multiply by Avogadro's number:
1.56094x10^-4 * 6.0221409x10^23 = 9.400197448261x10^19
Rounding to 3 significant figures gives 9.40x10^19 molecules.</span>
To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:
Clearing the specific heat of the unknown substance:
Answer:
5.37 L
Explanation:
To solve this problem we need to use the PV=nRT equation.
First we <u>calculate the amount of CO₂</u>, using the initial given conditions for P, V and T:
- P = 785 mmHg ⇒ 785/760 = 1.03 atm
- T = 18 °C ⇒ 18 + 273.16 = 291.16 K
1.03 atm * 4.80 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 291.16 K
We <u>solve for n</u>:
Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).
- 0.98 atm * V = 0.207 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 310.16 K
And we <u>solve for V</u>:
