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3 years ago

You have 40 g of a radioisotope. If the half-life of this radioisotope is 4 days, how many grams will remain after 12 days?

1 answer:

5 0

With reference to radioactive material, half-life is the time required to 50% depletion of initial amount of material.

Given: Initial amount of radioactive material = 40 g
Half life = 4 days.

Therefore, After 4 days, amount of compound left = 40/2 = 20 g
After 8 days, i.e 2 half-life, amount of compound left = 20/2 = 10 g
Finally after 12 days, i.e. 3 half-life, amount of compound left = 10/2 = 5 g

Thus, 5 grams will remain after 12 days

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44.90 g of reactants are placed in a beaker

sergiy2304 [10]

Answer:

D. -1882J

Explanation:

We can solve the energy released in a chemical reaction in an aqueous medium using the equation:

Q = -m*C*ΔT

Where Q is energy (In J),

m is mass of water (45.00g)

C is specific heat of water (4.184J/g°C)

And ΔT is change in temperature (25.00°C - 15.00°C = 10.00°C)

Replacing:

Q = -45.00*4.184J/g°C*10.00°C

Q = -1882J

Right answer is:

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2 years ago

Why is the temperature needed to freeze ocean water lower than the temperature needed to freeze the surface of a freshwater lake

shepuryov [24]

Ocean water freezes just like freshwater, but at Lower temperature. Fresh water freezes At 32°F but see water freezes at about 28.4°F because of the salt in it it can be melted down to use as drinking water

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3 years ago

You have a red ballooe that has a volume of 20 liters, a pressure of 1.5 atmospheres and a temperature of 28 C. What is the volu

ki77a [65]

Explanation:

The given data is as follows.

$P_{1}$ = 1.5 atm, $V_{1}$ = 20 L, $T_{1}$ = (28 + 273) K = 301 K

$P_{2}$ = 5 atm, $V_{2}$ = ?, $T_{2}$ = (50 + 273) K = 323 K

Formula to calculate the volume will be as follows.

$\frac{P_{1} \times V_{1}}{T_{1}}$ = $\frac{P_{2} \times V_{2}}{T_{2}}$

Putting the given values into the above formula as follows.

$\frac{P_{1} \times V_{1}}{T_{1}}$ = $\frac{P_{2} \times V_{2}}{T_{2}}$

$\frac{1.5 atm \times 20 L}{301 K}$ = $\frac{5 atm \times V_{2}}{323 K}$

$V_{2}$ = 0.64 L

Thus, we can conclude that the change in volume of the balloon will be 0.64 L.

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3 years ago

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves

hjlf

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

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide:

$2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)$

In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide:

$3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)$

Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.

Answer : The net chemical equation for the production of manganese is:

$6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)$