moles of CO gas : 1.545
<h3>Further explanation</h3>
Standard conditions for temperature and pressure are used as a reference in certain calculations or conditions
There are 2 conditions that are usually used as a reference : STP and RTP
Assuming the STP state :
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
Then for 34.6 L of CO gas :
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Answer:
58.0 g of MgO
Explanation:
in a perfect world, 70 g, however we don't live in a perfect world
The equation of reaction
2Mg + O₂ --> 2MgO
first find which element is limiting:
35 g x 1 mol/24.3 g of Mg x 2 mol of MgO/ 2 mole of Mg = 1.44 moles of MgO
35 g x 1 mol/32g of Mg x 2 mol of MgO/ 1 mole of O₂ = 2.1875 moles of MgO
This means Mg is the limiting factor, so you will be using this moles to find grams of MgO
1.44 mols of MgO x 40.3 g of MgO/ 1 mol = 58.0 g of MgO
Answer:
98.3 gradius Celsius
Explanation:
This problem is solved using the Ideal Gas Equation
pV = nRT
...
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
