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3 years ago

How many atoms of K are present in 195.49 grams of K? 3.0110 × 1024 6.0220 × 1024 1.1772 × 1026 4.5797 × 1027

Chemistry

1 answer:

Ierofanga [76]3 years ago

3 0

<h2>Hello!</h2>

The answer is: $3.0110x10^{24}$

First, we need to look for the K molecular mass:

$K=39,098 g/mol$

Also, we need to remember that 1 mol of a compound or element is equal to $6,022x10^{23}$ of that compound.

From the statement, we are asked to calculate how many atoms of K are present in 195,49 of K.

So, doing the calculations, we have:

$195,49g*\frac{1mol}{39,08}*\frac{6,022x10^{23} atoms}{1 mol K} =3,0110x10^{24} \\$

So, we have $3.0110x10^{24}$ atoms of K in 195,49 grams of K.

Have a nice day!

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John dissolves .5g of a white powder in 25g of benzene (FP 5oC) (kf benzene is 5.1) and finds the solution freezes at 3.7oC. Det

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Answer:

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Explanation:

Step 1: data given

Mass of a sample = 0.5 grams

Mass of benzene = 25 grams

Freezing poing = 5 °C

Kf of benzene = 5.1 °C/m

Freezing point solution = 3.7 °C

Step 2: Calculate molality

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 5.0 - 3.7 = 1.3 °C

⇒with i = the can't hoff factor = 1

⇒with Kf = the freezing point depression constant of benzene = 5.1 °C/m

⇒with m = the molality

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m = 1.3 / 5.1

m = 0.255 moles /kg

Step 3: Calculate moles

Molality = moles / mass benzene

0.255 molal = moles / 0.025 kg

Moles = 0.255 molal * 0.025 kg

Moles = 0.006375 moles

Step 4: Calculate molar mass of the compound

Molar mass compund = mass / moles

Molar mass compound = 0.5 grams / 0.006375 moles

Molar mass compound = 78.4 g/mol

The compound has a molar mass of 78.4 g/mol

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<h3>Further explanation </h3>

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• 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index (subscript) between reactant and product

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