Answer:
de lapora en ciesta coma bodre sin una jacsinis loadera comastora jingla belaa
Explanation:
Answer:
Explanation:
The law of multiple proportions states that if two elements X and Y combine together to form more than one compound, then the several masses of X which chemically combine with a fixed mass of Y is in simple ratio.T
FIRST CASE ;
- Molar mass of Nitrogen gas = 28 g/mol
- Number of moles of Nitrogen gas = Mass/molar mass = 3.50/28 = 0.125 moles
- Number of moles of Oxygen gas = Mass/molar mass = 2.00/32 = 0.0625 moles
- Hence the ratio of number of moles of N2 to O2 will be 0.1295 : 0.0625 or 2:1
- Similarly for the SECOND CASE ;
- Molar mass of Nitrogen gas = 28 g/mol
- Number of moles of Nitrogen gas = Mass/molar mass = 0.875/28 = 0.03125 moles
- Number of moles of Oxygen gas = Mass/molar mass = 1/32 = 0.03125 moles
- Hence the ratio of number of moles of N2 to O2 will be 0.03125 : 0.03125 or 1:1
<span>Mass of KCl= 88.4g
</span>Molar mass of KCl=39+35.5
<span> =74.5
</span><span> The Moles of KCl= 88.4/74.5
</span> =1.186
<span>If for the 2 moles of KCl there are 3 moles of oxygen
</span>so
<span>1 mole of KCl will give=3/2=1.5 moles of oxygen
</span>
<span>1.186 moles of KCl will give =1.5*1.186=1.779 oxygen
</span><span>32 the molar mass of oxygen
</span>so its mass will be
<span>1.779 *32=56.928
</span>enjoy
1 mol of Xe has a mass of 131 grams
x mol of Xe has a mass of 12 grams.
1/x = 131/12 Set up proportion from the information above. Cross multiply
131x = 12 Divide by 13`
x = 12/131
x = 0.0916 mol <<<< ====
Just for fun
I don't know if a calculator would even tell me the number of miles in 12 grams of Xe
The diameter is 216 * 10^-12 meters. Maybe.
1 mol of anything is 6.02 * 10^23
0.0912 mols contain x atoms
1/0.0912 = 6.02 * 10^23 / x
x = 0.0912 * 6.02 * 10^23
x = 5.49 * 10 ^ 22 This thing is going to stretch out quite a ways.
1 atom = 312 * 10^- 12 meters.
5.049 * 10^22 = x meters.
x = 312 * 10^-12 * 5.49 * 10^22
x = 1.7 * 10 ^10 km = 1.07 * 10 ^10 miles We're talking about beyond Pluto.
I'm sorry I should delete this. If you want to you can. Take my answer for the number of mols. It's likely correct.
