Mercury may have ice trapped in the crater floors near the planet’s north pole.

a 160 milligram sample of a radioactive isotope decays to 10 kilograms in 12 years. what is the half life of this element

Shkiper50 [21]

Answer:

3 years

Explanation:

Given data:

Initial amount of sample = 160 Kg

Amount left after 12 years = 10 Kg

Half life = ?

Solution:

at time zero = 160 Kg

1st half life = 160/2 = 80 kg

2nd half life = 80/2 = 40 kg

3rd half life = 40 / 2 = 20 kg

4th half life = 20 / 2 = 10 kg

Half life:

HL = elapsed time / half life

12 years / 4 = 3 years

8 0

3 years ago

How is kinetic energy related to mass and velocity?

Andrew [12]

The amount of kinetic energy that an object has depends on its mass or how heavy it is and how fast it is moving, or velocity

4 0

3 years ago

WILL MARK BRANILEST

ahrayia [7]

Answer:

FeCl3 is the limiting reactant

O2 is in excess

Theoretical yield Cl2 = 9.84 grams

The % yield is 96.5 %

Explanation:

Step 1: Data given

Mass of FeCl3 = 15.0 grams

Moles O2 = 4.0 moles

Mass of Cl2 produced = 9.5 grams

Step 2: The balanced equation

4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

Step 3: Calculate moles FeCl3

Moles FeCl3 = mass FeCl3 / molar mass FeCl3

Moles FeCl3 = 15.0 grams / 162.2 g/mol

Moles FeCl3 = 0.0925 moles

Step 4: Calculate limiting reactant

FeCl3 is the limiting reactant. Because we have way more (more than ratio 3:4) moles O2 than FeCl3. It will completely be consumed (0.0925 moles). O2 is in excess. There will react = 0.069375 moles O2

There will remain 4.0 - 0.069375 = 3.930625 moles

Step 5: Calculate moles Cl2

For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2

For 0.0925 moles FeCl3 moles we'll have 6/4 * 0.0925 = 0.13875 moles Cl2.

Step 6: Calculate mass Cl2

Mass Cl2 = moles * molar mass

Mass Cl2 = 0.13875 moles * 70.9 g/mol

Mass Cl2 = 9.84 grams

Step 7: Calculate % yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (9.5 grams / 9.84 grams ) * 100%

% yield = 96.5 %

The % yield is 96.5 %

4 0

4 years ago

When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H

Taya2010 [7]

Answer:

- Empirical:

$C_3H_7$

- Molecular:

$C_6H_{14}$

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

$n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC$

Moreover, the moles of hydrogen:

$n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH$

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

$C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}$

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

$C_3H_7$

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

$C_6H_{14}$

Which is hexane.

Best regards.

6 0

3 years ago

One litre of hydrogen at STP weight 0.09gm of 2 litre of gas at STP weight 2.880gm. Calculate the vapour density and molecular w

expeople1 [14]

Answer:

we know, at STP ( standard temperature and pressure).

we know, volume of 1 mole of gas = 22.4L

weight of 1 Litre of hydrogen gas = 0.09g

so, weight of 22.4 litres of hydrogen gas = 22.4 × 0.09 = 2.016g ≈ 2g = molecular weight of hydrogen gas.

similarly,

weight of 2L of a gas = 2.88gm

so, weight of 22.4 L of the gas = 2.88 × 22.4/2 = 2.88 × 11.2 = 32.256g

hence, molecular weight of the gas = 32.256g

vapor density = molecular weight/2

= 32.256/2 = 16.128g

hence, vapor density of the gas is 16.128g.

Explanation:

4 0

3 years ago