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3 years ago

Explain why the pbcl2 dissolved when water was added

1 answer:

4 0

PbCl2 would not dissolve because it is insoluble based on the solubility rules for substances that will dissolve in water. This compound would instead form a solid precipitate at the bottom of the container.

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A student was asked to determine the densities of the four metals listed. The student was asked to determine the density of an u

Iteru [2.4K]

Answer:

D is correct

Explanation:

because

we know that

density of lead is 11.36 g/cm3

and

density of tin is 7.31 g/cm3

so..

density of alloy by mixing 50/50

=(11.36+7.31)/2 g/cm3

=18.67/2 g/cm3

=9.33 g/cm3

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3 years ago

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creativ13 [48]

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2 years ago

What is the maximum mass of h2o that can be produced by combining 68.6 g of each reactant?

Sauron [17]

The answer will be 47.5

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2 years ago

How does a sciences form a hypothesis

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2 years ago

Read 2 more answers

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

3 0

3 years ago