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2 years ago

9

A sample of polonium-210 has an initial mass of 390 milligrams (mg). If the half-life of polonium-210 is 36 days, how many mg of

the sample remains after 72 days?
A.
392 mg

B.
195 mg

C.
97.5 mg

D.
48.75 mg

2 answers:

kondaur [170]2 years ago

5 0

Answer: D. 48.75

Explanation: just took the test

maw [93]2 years ago

5 0

Answer:

D

Explanation:

I got this question right on my test.

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A solution is made by mixing equal masses of methanol, CH4O, and ethanol, C2H6O. Determine the mole fraction of each component t

Serggg [28]

Answer: mole fraction of methanol = 0.590

mole fraction of ethanol = 0.410

Explanation:

We are given:

Equal masses of methanol $CH_4O$ and ethanol $C_2H_6O$ are mixed.

let the mass be x g.

Calculating the moles of methanol in the solution, by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{32.04g/mol}$

Calculating the moles of ethanol in the solution, by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{46.07g/mol}$

To calculate the mole fraction of methanol, we use the equation:

$\chi_{methanol}=\frac{n_{methanol}}{n_{methaol}+n_{ethanol}}$

$\chi_{methanol}=\frac{\frac{xg}{32.04g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.590$

To calculate the mole fraction of ethanol, we use the equation:

$\chi_{ethanol}=\frac{n_{ethanol}}{n_{methaol}+n_{ethanol}}$

$\chi_{ethanol}=\frac{\frac{xg}{46.07g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.410$

Thus mole fraction of methanol is 0.590 and mole fraction of ethanol 0.410 in three significant figures.

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2 years ago

You have 100 grams of potassium fluoride (KF) .

laiz [17]

Answer:

Option C = 1.72 mol

Explanation:

Given data:

Mass of KF = 100 g

Moles of KF = ?

Solution:

First of all we have to calculate the molar mass of KF.

Molar mass of KF = 39.0983 g/mol + 18.998403 g/mol

Molar mass of KF = 58. 0967 g/mol

Formula:

Number of moles = mass/molar mass

Number of moles = 100 g/ 58.0967 g/mol

Number of moles = 1.72 mol

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2 years ago

The empirical formula of a compound is CH. At 200 degree C, 0.145 g of this compound occupies 97.2 mL at a pressure of 0.74 atm.

Vladimir [108]

Answer:

The molecular formula = $C_{6}H_{6}$

Explanation:

Given that:

Mass of compound, m = 0.145 g

Temperature = 200 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (200 + 273.15) K = 473.15 K

V = 97.2 mL = 0.0972 L

Pressure = 0.74 atm

Considering,

$n=\frac{m}{M}$

Using ideal gas equation as:

$PV=\frac{m}{M}RT$

where,

P is the pressure

V is the volume

m is the mass of the gas

M is the molar mass of the gas

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the values in the above equation as:-

$0.74\times 0.0972=\frac{0.145}{M}\times 0.0821\times 473.15$

$M=78.31\ g/mol$

The empirical formula is = $CH$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12 + 1 = 13 g/mol

Molar mass = 78.31 g/mol

So,

Molecular mass = n × Empirical mass

78.31 = n × 13

⇒ n ≅ 6

The molecular formula = $C_{6}H_{6}$

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