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atroni [7]
3 years ago
11

How many atoms are in 3.50 moles of copper (11) chloride

Chemistry
1 answer:
Usimov [2.4K]3 years ago
3 0

Answer:

6.02 x 1023

Explanation:

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scoundrel [369]
A. An Interdependent system of plants, animals, and land
8 0
3 years ago
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The question in the picture, I really a correct answer, no cheap answers​
Delvig [45]

Answer:

94.325 g

Explanation:

We'll begin by converting 350 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

350 mL = 350 mL × 1 L /1000 mL

350 mL = 0.35 L

Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:

Volume = 0.35 L

Molarity of KC₂H₃O₂ = 2.75 M

Mole of KC₂H₃O₂ =?

Molarity = mole /Volume

2.75 = Mole of KC₂H₃O₂ / 0.35

Cross multiply

Mole of KC₂H₃O₂ = 2.75 × 0.35

Mole of KC₂H₃O₂ = 0.9625 mole

Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:

Mole of KC₂H₃O₂ = 0.9625 mole

Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)

= 39 + 24 + 3 + 32

= 98 g/mol

Mass of KC₂H₃O₂ =?

Mass = mole × molar mass

Mass of KC₂H₃O₂ = 0.9625 × 98

Mass of KC₂H₃O₂ = 94.325 g

Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g

6 0
2 years ago
What would be the mass of 9.03*10^21 molecules of hydrobromic acid
Fynjy0 [20]

Answer:

2.11 g hydrobromic acid (correct to 3SF)

Explanation:

Molecular formula of hydrobromic acid = C2H5BrO2

mass of C2H5BrO2 = 140.96g

Beginning with what we're given, 9.03*10^21 we then make a conversion by using Avegadro's number which is 6.02*10^23 per mole (Oct. 23 at 6:02 am is national mole day :) Then, we need to convert out of moles, 140.96g hydrombromic acid per mole.

It looks like this:

9.03*10^21 molecules • (1 mol C2H5BrO2 / 6.02*10^23 molecules) • (140g C2H5BrO2 / 1 mol) = 2.1144 g C2H5BrO2

3 0
3 years ago
How many kilograms of solvent (water) must 0.71 moles of KI be dissolved in to produce a 1.93 m solution?
GalinKa [24]

Answer: kg= 0.37

Explanation:

Use the molality formula.

M= m/kg

6 0
3 years ago
What is the percent composition of Fluorine in CF5?
posledela
The answer is 100%
let me know if you want an explanation
6 0
3 years ago
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