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3 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]3 years ago

5 0

Answer: D. 48.75

Explanation: just took the test

maw [93]3 years ago

5 0

Answer:

D

Explanation:

I got this question right on my test.

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A photon has a frequency of 2.68 x 10^6 HZ calculate it's energy

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Describe the main difference between the Bohr model of the atom and the Rutherford model

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Read 2 more answers

1. What is the molarity of a solution which contains 0.256 mol of a substance dissolved in 143 mL of solution.

LenaWriter [7]

Answer:

The equation for molarity is moles/liter for the first question you would do 0.256/0.143 liters to get 1.790 mol/L

Explanation:

The second problem you would do need to find the moles of NaCl which you would do by doing 4.89 g/58.44g/mol= 0.08367 then do 0.08367/0.600= 0.139 mol/L

The third problem would be the same steps as the second one.

The fourth problem would be (0.460M)(5.50L)= 2.53 moles

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3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

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3 years ago