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3 years ago

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Plutonium-240 decays according to the function \[Q(t)=Q _{o}e ^{-kt}\] where Q represents the quantity remaining after t years a

nd k is the decay constant, 0.00011.. How long will it take 36 grams of plutonium-240 to decay to 12 grams?. A. 18,900 years. B. 1.44 years. C. 9,990 years. D. 2,100 years.

2 answers:

enyata [817]3 years ago

8 0

<span>C. 9,990 years is the correct answer

</span>

marishachu [46]3 years ago

4 0

To answer the question above, substitute the given values to the given equation,
Q(t) = Q x e^-kt

12 grams = (36 grams) x e^(-0.00011)(t)

Solving for t gives t = 9,987.38 years or approximately equal to 9,990 years. Thus, the answer is letter C.

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Thanks for the help :) Question attached below:

Alekssandra [29.7K]

• Take a look at the steps below to see how to balance this equation. Let's start by writing the unbalanced equation given the information.

Unbalanced Equation : C₃H₈ (g) + O₂ (g) → CO₂ (g) + H₂O (g) ,

Start by Balancing the Carbons : C₃H₈ (g) + O₂ (g) → 3CO₂ (g) + H₂O (g)

Now let's balance the Hydrogen : C₃H₈ (g) + O₂ (g) → 3CO₂ (g) + 4H₂O (g)

Balancing the Oxygen : C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

Balanced Equation : C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

• Let's apply dimensional analysis here,

0.7 L propane × (5 liters Oxygen / 1 liter Propane) = 3.5 Liters of Oxygen

• Similarly we can identify the liters of carbon dioxide produced in the reaction,

0.7 L propane × (3 liters Carbon Dioxide / 1 liter Propane) = 2.1 Liters of Carbon Dioxide

• 0.7 L propane × (4 liter water vapor / 1 liter propane ) = 2.8 Liters of Water Vapor

7 0

3 years ago

gypsum is insoluble in water. you are asked to purify a sample of gypsum that is contaminated with a soluble salt.

katen-ka-za [31]

Answer:

a. The apparatus required to purify gypsum sample are: Bunsen burner, beaker, Filter Funnel, stirring rod, the filter paper.

b. Gypsum is a sulfate mineral that is made up of calcium sulfate dihydrate. Step-by-step instruction to purify gypsum sample is as follows:

1. Add water to the gypsum sample in a beaker.

2. Use the stirring rod to mix the mixture well.

3. Filter off the excess solid from the mixture using the filter paper and filter funnel.

4. Put the filtered mixture over the bunsen burner and evaporate the excess water from the mixture.

5. Allow the hot liquid to cool down and filter it again through the filter paper to get the pure gypsum.

7 0

2 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Determine the number of grams of carbon-14 remaining in this sample at 17,100 years.

earnstyle [38]

Answer: I need help

Explanation:

8 0

3 years ago

1. What mass of Mgo will I make from 48 g of Mg?<br> 2Mg + O2 + 2MgO

statuscvo [17]

Answer:

80.6 g

Explanation:

MgO=24.3+16=40.3 g

48.6/24.3= 2 moles

64/2x16= 2 moles

2x40.3= 80.6g

8 0

2 years ago