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2 years ago

Ways to reduce your carbon footprint.

Chemistry

1 answer:

sergejj [24]2 years ago

5 0

Answer:

Avoid driving as much as you can
Avoid plastic in general (Plastic is usually not recycled, despite what people say. It ends up in landfills, the ocean, or it is burned)
Don't buy fast fashion (Clothes that are made in bulk by factories usually in Bangladesh. They are trendy but bad for the environment. Instead, shop at thrift stores.)
Don't waste food! Stop throwing out those leftovers, and make sure you eat everything on your plate.
Try to switch to renewable energy. (Use solar panels or find lights that conserve energy. Also, these options most likely cost less :)
Try to cut back on your meat intake. (I am a pescatarian who loves meat. I started this diet in 4th or 5th grade because it was a dare. I know, it was a crazy dare, but hey I don't back down! It's been about 5 years now and I'm still going strong. If me - a meat lover - can do it, you can to :)

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An artifact contains one-fourth as much carbon-14 as the atmosphere. how old is the artifact

Nikolay [14]

<h3>Answer;</h3>

= 11,460 years

<h3>Explanation;</h3>

The half life of Carbon-14 is 5,730 years . Half life is the time taken by a radioactive material to decay by half of its original mass. Therefore, it would take a time of 5730 years for a sample of 100 g of carbon-14 to decay to 50 grams

The initial amount of carbon-14 in this case was 1 whole; thus; 

1 → 1/2 →1/4

To contain 1/4 of the value, 2 half-lives have passed. 

But, 1 half life = 5,730 years

Therefore; The artifact is is therefore: 2 x 5,730 

 = 11,460 years 

4 0

2 years ago

Read 2 more answers

A scientist has 0.12 miles of a gas at a pressure of 4.06 kPa and a volume of 14 liters in an enclosed container. What is the te

natka813 [3]

Answer:

56972.17K

Explanation:

P = 4.06kPa = 4.06×10³Pa

V = 14L

n = 0.12 moles

R = 8.314J/Mol.K

T = ?

We need ideal gas equation to solve this question

From ideal gas equation,

PV = nRT

P = pressure of the ideal gas

V = volume the gas occupies

n = number of moles

R = ideal gas constant

T = temperature of the gas

PV = nRT

T = PV / nR

T = (4.06×10³ × 14) / (0.12 × 8.314)

T = 56840 / 0.99768

T = 56972.17K

Note : we have a large number for temperature because we converted the value of pressure from kPa to Pa

7 0

3 years ago

Given the balanced equation, 2 H2 + O2 -> 2 H2O, answer the following

9966 [12]

Answer: $O_2$ is the limiting reagent

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} H_2=\frac{10.0g}{2g/mol}=5.00moles$

$\text{Moles of} O_2=\frac{20.0g}{32g/mol}=0.625moles$

The balanced chemical reaction is :

$2H_2+O_2\rightarrow 2H_2O$

According to stoichiometry :

1 moles of $O_2$ require = 2 moles of $H_2$

Thus 0.625 moles of $O_2$ will require= $\frac{2}{1}\times 0.625=1.25moles$ of $H_2$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent as it is present more than the required amount.

8 0

3 years ago

Melted wax has a temperature of 54 deg. Celsius. Physical or chemical property?

zalisa [80]

The melting of solid wax to form liquid wax and the evaporation of liquid wax to form wax vapor are physical changes. The burning of the wax vapor is a chemical change.

7 0

3 years ago

Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta

scoray [572]

Answer:

70.77 g/mol is the molar mass of the unknown gas.

Explanation:

Effusion is defined as rate of change of volume with respect to time.

Rate of Effusion= $\frac{Volume}{Time}$

Effusion rate of oxygen gas after time t = $E=\frac{4.64 mL}{t}$

Molar mass of oxygen gas = M = 32 g/mol

Effusion rate of unknown gas after time t = $E'=\frac{3.12 mL}{t}$

Molar mass of unknown gas = M'

The rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{E}{E'}=\sqrt{\frac{M'}{M}}$

$\frac{\frac{4.64 mL}{t}}{\frac{3.12 mL}{t}}=\sqrt{\frac{M'}{32 g/mol}}$

M' = 70.77 g/mol

70.77 g/mol is the molar mass of the unknown gas.

8 0

2 years ago