The pressure exerted by 0.400 moles of carbon dioxide in a 5.00 Liter container at 25 °C would be 1.9563 atm or 1486.788 mm Hg.
<h3>The ideal gas law</h3>
According to the ideal gas law, the product of the pressure and volume of a gas is a constant.
This can be mathematically expressed as:
pv = nRT
Where:
p = pressure of the gas
v = volume
n = number of moles
R = Rydberg constant (0.08206 L•atm•mol-1K)
T = temperature.
In this case:
p is what we are looking for.
v = 5.00 L
n = 0.400 moles
T = 25 + 273
= 298 K
Now, let's make p the subject of the formula of the equation.
p = nRT/v
= 0.400 x 0.08206 x 298/5
= 1.9563 atm
Recall that: 1 atm = 760 mm Hg
Thus:
1.9563 atm = 1.9563 x 760 mm Hg
= 1486.788 mm Hg
In other words, the pressure exerted by the gas in atm is 1.9563 atm and in mm HG is 1486.788 mm Hg.
More on the ideal gas law can be found here: brainly.com/question/28257995
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Answer:
See explanation
Explanation:
The rate of reaction depends on on the concentration of reactants. As the concentration of reactants increases, the rate of reaction increases likewise.
If we look at the table, we will discover that the concentration of the thiosulphate is increasing because the volume of water added is decreasing. As such, the rate of reaction increases simultaneously.
Since the rate of reaction increases, the time taken for the cross to disappear decreases steadily. Hence, the values in the last column of the table decreases steadily.
Chloroplasts is the answer
<span>200 moles
The balanced equation for creating water from hydrogen and oxygen gas is
2H2 + O2 => 2H2O
So for every mole of oxygen gas, you need two moles of hydrogen. So looking that the amount of oxygen and hydrogen you have, it's obvious that oxygen is the limiting reactant since 100 moles of oxygen will consume 200 moles of hydrogen. While 210 moles of hydrogen requires 105 moles of oxygen.
Now for each mole of oxygen gas you use, you create 2 moles of water. So
100 mol * 2 = 200 mol
So you can create 200 moles of water from the given amounts of reactants.</span>
