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lys-0071 [83]
3 years ago
15

How many molecules are in 1.52 mol of dinitrogen monoxide

Chemistry
1 answer:
Natalka [10]3 years ago
3 0

Answer:

9.15 × 10²³ molecules of N₂O present.

Explanation:

Given data:

number of moles = 1.52 mol

molar mass of N₂O = 44.013 g/mol

number of molecules = ?

Solution:

First of all we will calculate the mass of 1.52 mol.

mass = number of moles × molar mass

mass = 1.52 × 44.013 g/mol

mass = 66.9 g

Now we will find the number of molecules by using Avogadro number,

The number 6.022×10∧23 is called Avogadro number and it is the number of atoms or molecule in one gram atom of an element or one gram molecule of a compound.

44.013 g of N₂O = one mole of N₂O= 6.02 × 10²³ molecules

For 66.9 g:

66.9 g/ 44.013 g × 6.02 × 10²³ molecules

9.15 × 10²³ molecules of N₂O present

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How many moles are there in 3.9 grams of potassium
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1.77 so A. I don’t need to explain this just helping out haha
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What is the pH of a buffer that consists of 0.254 M CH3CH2COONa and 0.329 M CH3CH2COOH? Ka of propanoic acid, CH3CH2COOH is 1.3
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Answer:

4.77 is the pH of the given buffer .

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=-\log[K_a]+\log(\frac{[salt]}{[acid]})

pH=-\log[K_a]+\log(\frac{[CH_3CH_2COONa]}{[CH_3CH_2COOH]})

We are given:

K_a = Dissociation constant of propanoic acid = 1.3\times 10^{-5}

[CH_3CH_2COONa]=0.254 M

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pH = ?

Putting values in above equation, we get:

pH=-\log[1.3\times 10^{-5}]+\log(\frac{[0.254 M]}{[0.329]})

pH = 4.77

4.77 is the pH of the given buffer .

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