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Papessa [141]
3 years ago
15

Suppose "Element X" has a molar mass of 38.5 g/mol. How many atoms of Element X would be present in 2.78 mol?

Chemistry
1 answer:
slava [35]3 years ago
3 0

Answer:

16.74×10²³ atoms

Explanation:

Given data:

Number of moles of X = 2.78 mol

Number of atoms = ?

Solution:

According to Avogadro number.

1 mole of any substance contain 6.022×10²³ atoms. Thus,

2.78 mol × 6.022×10²³ atoms / 1 mol

16.74×10²³ atoms

The number 6.022 × 10²³ is called Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

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Hydrazine (N2H4) emits a large quantity of
azamat

Answer:

              1.75 moles of H₂O

Solution:

The Balance Chemical Equation is as follow,

                                   N₂H₄  +  O₂    →    N₂  +  2 H₂O

Step 1: Calculate the Limiting Reagent,

According to Balance equation,

             32.04 g (1 mol) N₂H₄ reacts with  =  32 g (1 mol) of O₂

So,

                   28 g of N₂H₄ will react with  =  X g of O₂

Solving for X,

                       X  =  (28 g × 32 g) ÷ 32.04 g

                       X  =  27.96 g of O₂

It means 29 g of N₂H₄ requires 47.96 g of O₂, while we are provided with 73 g of O₂ which is in excess. Therefore, N₂H₄ is the limiting reagent and will control the yield of products.

Step 2: Calculate moles of Water produced,

According to equation,

            32.04 g (1 mol) of N₂H₄ produces  =  2 moles of H₂O

So,

                        28 g of N₂H₄ will produce  =  X moles of H₂O

Solving for X,

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4 years ago
Calculate the pH of a 0.10 M HCN solution that is 0.0070% ionized.
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Answer:

D) 5.15

Explanation:

Step 1: Write the equation for the dissociation of HCN

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

Step 2: Calculate [H⁺] at equilibrium

The percent of ionization (α%) is equal to the concentration of one ion at the equilibrium divided by the initial concentration of the acid times 100%.

α% = [H⁺]eq / [HCN]₀ × 100%

[H⁺]eq = α%/100% × [HCN]₀

[H⁺]eq = 0.0070%/100% × 0.10 M

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3 0
3 years ago
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