If 45 g of O2 gas in a 500 mL container is exerting a pressure of 5.2 atm, 22.53K is the temperature in the gas.
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
First, calculate the moles of the gas using the gas law,
PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.
Given data:
P= 5.2 atm
V= 500 mL =0.5 L
n=?
R= 
T=?
Moles = 1.40625
Putting value in the given equation:
T= 22.53167034 K= 22.53K
Hence, If 45 g of O2 gas in a 500 mL container is exerting a pressure of 5.2 atm, 22.53K is the temperature in the gas.
Learn more about the ideal gas here:
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Answer:
=8 atoms
Explanation:
In (NH4)2C2O4 there are four moles of Hydrogen in the compound (NH4), but there two molecules of (NH4) in this compound. That's what the 2 in (NH4)2 means, so multiply 4 x 2 = 8.
Hope this helps (:
LiCO3-> LiO+ CO2
It is already balanced as there is one Li on each side, one C on each side and three Os on each side.
Answer:
48 grams
Explanation:
The chemical equation for the reaction is the following:
2 H₂ + O₂ → 2 H₂O
That means that 2 moles of H₂ react with 1 mol of O₂ to produce 2 moles of H₂O. We convert the moles of oxygen (O₂) by using the molecular weight (MW) as follows:
MW(O₂) = 16 g/mol x 2 = 32 g/mol
mass of O₂ = 1 mol x 32 g/mol = 32 g
So, we have the following stoichiometric ratio: 32 g O₂/2 moles H₂. We have 3 moles of hydrogen (H₂), so we multiply the moles by the stoichiometric ratio to calculate how many grams are needed:
3 moles H₂ x 32 g O₂/2 moles H₂ = 48 g O₂
<em>Therefore, 48 grams of O₂ are needed to react with 3 moles of H₂.</em>
