<h2>
Answer:</h2>
44.06 g/mol
<h3>
Explanation:</h3>
We are given;
- Number of moles of unidentified gas as 1.674×10^-4 mol
- Time of effusion of unidentified gas 86.6 s
- Number of moles of Argon gas as 1.715×10^-4 mol
- Time of effusion of Argon gas is 84.5 s
We are supposed to calculate the molar mass of unidentified gas
<h3>Step 1: Calculate the effusion rates of each gas</h3>
Effusion rate = Number of moles/time
Effusion rate of unidentified gas (R₁)
= 1.674×10^-4 mol ÷ 86.6 s
= 1.933 × 10^-6 mol/s
Effusion rate of Argon gas (R₂)
= 1.715×10^-4 mol ÷ 84.5 sec
= 2.030 × 10^-6 mol/s
<h3>Step 2: Calculate the molar mass of unidentified gas</h3>
- Assuming the molar mass of unidentified gas is x;
- We can use the Graham's law of effusion to find x;
- According to Graham's law of diffusion;
But, Molar mass of Argon is 39.948 g/mol
Therefore;
Solving for X
x = 44.06 g/mol
Therefore, the molar mass of the identified gas is 44.06 g/mol
Answer: 154.09 g
Explanation:
3 Mg + 2 FeCl3 = 3 MgCl2 + 2 Fe
The mass of MgCl2 produced is calculated:
m MgCl2 = 175 g FeCl3 * (1 mol FeCl3 / 162.2 g) * (3 mol MgCl2 / 2 mol FeCl3) * (95.21 g MgCl2 / 1 mol) = 154.09 g
Answer:
ok all you need is a little study
Explanation:
organic chemistry isn't something to worry about is something that is a very simple concept if you understand what you're doing so a little bit of studying will help you go through it and you really don't this is not the topics that you will really understand if you just go through it as once so there is no need to worry about organic chemistry it is very simple concept just learning you are a you are close just learn all those simple aspects alkane etc it's very easy
Amino acids join end to end to form proteins since they are monomers.
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Answer:
23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr
Explanation:
The balanced equation here is
6NaBr + 1AlO3 = 3Na2O + 2AlBr3
6 moles of NaBr are required to produce 2 moles of AlBr3
Mass of one mole of NaBr = 102.894 g/mol
Mass of one mole of AlBr3 = 266.69 g/mol
Mass of 6 moles of NaBr = 6*102.894 g/mol
Mass of two moles of AlBr3 = 2*266.69 g/mol
6*102.894 g NaBr produces 2*266.69 g of AlBr3
23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr
