E=MC(delta)T
=20.0g x 9.00J/g x (94.4-22.8) C
= 12,924.0 J
Answer:
- <em>The volume of 14.0 g of nitrogen gas at STP is </em><u><em>11.2 liter.</em></u>
Explanation:
STP stands for standard pressure and temperature.
The International Institute of of Pure and Applied Chemistry, IUPAC changed the definition of standard temperature and pressure (STP) in 1982:
- Before the change, STP was defined as a temperature of 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa).
- After the change, STP is defined as a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).
Using the ideal gas equation of state, PV = nRT you can calculate the volume of one mole (n = 1) of gas. With the former definition, the volume of a mol of gas at STP, rounded to 3 significant figures, was 22.4 liter. This is classical well known result.
With the later definition, the volume of a mol of gas at STP is 22.7 liter.
I will use the traditional measure of 22.4 liter per mole of gas.
<u>1) Convert 14.0 g of nitrogen gas to number of moles:</u>
- n = mass in grams / molar mass
- Atomic mass of nitrogen: 14.0 g/mol
- Nitrogen gas is a diatomic molecule, so the molar mass of nitrogen gas = molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol
- n = 14.0 g / 28.0 g/mol = 0.500 mol
<u>2) Set a proportion to calculate the volume of nitrogen gas:</u>
- 22.4 liter / mol = x / 0.500 mol
- Solve for x: x = 0.500 mol × 22.4 liter / mol = 11.2 liter.
<u>Conclusion:</u> the volume of 14.0 g of nitrogen gas at STP is 11.2 liter.
It would be carbon dioxide and water.
hope this helps you
Answer:
Approximately
, assuming that this gas is an ideal gas.
Explanation:
Look up the standard room temperature and pressure:
and
.
The question states that the volume of this gas is
.
Convert the unit of all three measures to standard units:
.
.
.
Look up the ideal gas constant in the corresponding units:
.
Let
denote the number of moles of this gas in that
. By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:
.
Rearrange this equation and solve for
:
.
In other words, there is approximately
of this gas in that
.