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3 years ago

What is the mass of 6.02 x 10^23 atoms of potassium

Chemistry

1 answer:

guajiro [1.7K]3 years ago

5 0

So we know that this amount of atoms is Avogadro's number - the number of atoms in 1 mole of a substance.

So then, using dimensional analysis, we can find the mass, in grams, of potassium.

First we must find the molar weight of potassium as found on the periodic table: 39.098g/mol

Then we can solve directly for mass:

$\frac{6.022x10^{23}atoms}{1}*\frac{1mol}{6.022x10^{23}atoms}*\frac{39.098g}{1mol}=39.098g$

Therefore, the amount of atoms you have given, in terms of mass, will equal 39.098g - if the answer requires significant figures, then the mass would be 39.1g of potassium.

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3 years ago

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4) In water, Vanillin, C8H8O3, has a solubility of 0.070 moles of vanillin per liter of solution at 25C. What will be produced i

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Answer:

An unsaturated solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly realize we need to calculate the grams of vanillin in 0.070 moles by using its molar mass of 152.15 g/mol:

$m=0.070mol*\frac{152.15 g}{1mol} =10.65g$

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3 years ago

How many milligrams of magnesium should you take a day?.

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3 years ago

What is the h oh ph and poh of a 0.005m solution of calcium hydroxide?

fiasKO [112]

Answer: $[OH^{-}]= 0.01M or 1.0\times 10^{-2}M$

$[H^{+}]= 1.0\times 10^{-12}M$

pH = 12

pOH = 2

Explanation: Calcium hydroxide ( $Ca(OH)_{2}$ ) is a strong base that dissociates completely.

Dissociation equation of Calcium hydroxide is :

$Ca(OH)_{2} \rightarrow Ca^{+2} + 2OH^{-}$

1. Concentration of [OH-]

1 mol $Ca(OH)_{2}$ produces 2 mol OH- ions.

The given solution is 0.005M $Ca(OH)_{2}$ , then concentration of OH- would be twice the concentration of $Ca(OH)_{2}$

$[OH^{-}] = 0.005\times 2$ = 0.01M or $1.0\times 10^{-2}M$

2.Concentration of [H+]

Concentration of [H+] can be calculated by the formula: $[H^{+}] = \frac{Kw}{[OH^{-}]}$

kw = ionic product of water and its values is $(1\times 10^{-14})$

[OH-] = 0.01 M or $1.0\times 10^{-2}M$

$[H^{+}] = \frac{(1\times 10^{-14})}{[OH^{-}]}$

$[H^{+}] = \frac{(1\times 10^{-14})}{[0.01]}$

$[H^{+}] = 1.0\times 10^{-12}M$

3. pH value

pH is calculated by the formula : $pH = -log[H^{+}]$

$pH = -log[1.0\times 10^{-12}]$

pH = 12

4. pOH value

pOH is calculated by the formula : pOH = 14 - pH

pOH = 14 - 12

pOH = 2

pOH can also be calculated by using a different formula which is :

$pOH = -log(OH^{-})$

$pOH = -log(0.01)$

pOH = 2.

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3 years ago

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