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ivanzaharov [21]
3 years ago
5

Please help me it’s just that I can’t remember how to do this it’s about the Mole and Stoichiomerty

Chemistry
1 answer:
joja [24]3 years ago
8 0

Answer:

6.52moles of H2O

Explanation:

First, let us calculate the number of moles of O2 containing 1.96x10^24 molecules. This is illustrated below:

1mole of O2 contains 6.02x10^23 molecules.

Therefore Xmol of O2 contain 1.96x10^24 molecules i.e

Xmol of O2 = 1.96x10^24/6.02x10^23 = 3.26moles

Now let us generate a balanced equation for the formation of water. This is illustrated below:

2H2 + O2 —> 2H2O

From the equation,

1mole of O2 produced 2moles of H2O.

Therefore, 3.26moles of O2 will produce = 3.26 x 2 = 6.52moles of H2O

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bonufazy [111]

Answer:

region 2 and region 3

Explanation:

you can tell by the color of the land my friends^^

6 0
3 years ago
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Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
s2008m [1.1K]

Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

4 0
3 years ago
1. What are commonly called as "shooting stars"?
nydimaria [60]

Answer:

B

B

A

C

D

Explanation:

I think dont take my word for though

6 0
3 years ago
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velikii [3]
To solve this question you need to calculate the number of the gas molecule. The calculation would be:
PV=nRT
n=PV/RT
n= 1 atm * 40 L/ (0.082 L atm mol-1K-<span>1 * 298.15K)
</span>n= 1.636 moles

The volume at bottom of the lake would be:
PV=nRT
V= nRT/P
V= (1.636 mol * 277.15K* 0.082 L atm mol-1K-1 )/ 11 atm= <span>3.38 L</span>
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3 years ago
Which of the following are testable?
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you forgot the well answers

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