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3 years ago

How many atoms of nickel can be found in a 1.00 g sample?

1 answer:

7 0

The atoms of a substance can be determined by dividing the given mass of the substance by the molar mass. This results to moles of the substance. Then we multiply by the Avogadro's number. In this case, we divide 1 g by 58. 69 g/ mol and multiply by 6.022 x10^23 atoms / mole. The answer is 1.03 x10^22 atoms

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What is the pH of a 5.09 x 10-5 M solution of NaOH?

Varvara68 [4.7K]

<u>Answer:</u> The pH of the solution is 9.71

<u>Explanation:</u>

1 mole of NaOH produces 1 mole of sodium ions and 1 mole of hydroxide ions.

We are given:

pOH of the solution = 7.2

To calculate the pH of the solution, we need to determine pOH of the solution. To calculate pOh of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=5.09\times 10^{-5}M$

Putting values in above equation, we get:

$pOH=-\log(5.09\times 10^{-5})\\\\pOH=4.29$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-4.29=9.71$

Hence, the pH of the solution is 9.71

7 0

2 years ago

Who was the first president of Jamaica

Lyrx [107]

Sir William Alexander Clarke Bustamante.

Hope This Helps!

8 0

2 years ago

Read 2 more answers

Sodium (Na) A Homogeneous Mixture B Element C Compound D Heterogeneous Mixture

laiz [17]

Sodium is categorized as a B. Element.

3 0

2 years ago

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Diano4ka-milaya [45]

Answer:

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5 0

2 years ago

A solution is prepared at that is initially in benzoic acid , a weak acid with , and in sodium benzoate . Calculate the pH of th

Kisachek [45]

Answer:

$pH=4.1$

Explanation:

Hello,

In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:

$pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1$

Regards.

3 0

2 years ago