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Sidana [21]
3 years ago
6

Find the number of moles of argon in 364 grams of argon.

Chemistry
1 answer:
serg [7]3 years ago
3 0

Answer:D. 9.1 mole Ar

Explanation:

364 g Ar x 1 mole Ar / 40 g Ar

= 9.1 moles Ar

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Does each phrase describe photosynthesis, cellular respiration, or both? Drag and drop the phrases into the correct locations.
arlik [135]

Answer:

Anabolic Reaction : Photosynthesis

Occurs in plant life : Both

a way to store energy : Both

Catabolic Reaction : Cellular Respiration

a way to release energy : Both

involves oxygen and carbon dioxide : oxygen releases by the process of photosynthesis. Oxygen is involved in areobic respiration and not in anarobic respiration.

5 0
3 years ago
What is the concentration of a solution of HCl in which a 10.0 mL sample of acid required 50.0 mL of 0.150 M NaOH for neutraliza
puteri [66]

Answer:

The answer is 0.75M HCl

Explanation:

To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:

To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:

C1V1 = C2V2

C1 = concentration of acid

C2 = concentration of base

V1 = volume of acid

V2 = volume of base

From the information supplied in the question:

concentration of acid (HCl) is the unknown

volume of acid (HCl) = 10ml

concentration of base (NaOH) = 0.15M

volume of base (NaOH) = 50ml

C1 x 10ml = 0.15M x 50ml

C1 x 10 = 7.5

divide both side by 10

C1 = 0.75M

concentration of acid (HCl) is 0.75M

5 0
3 years ago
Mass x height x gravity is the formula to calculate:
Sindrei [870]

Answer:

Potential energy

Explanation:

p.e = mgh

8 0
3 years ago
A mixture of C3H8C3H8 and C2H2C2H2 has a mass of 2.8 gg . It is burned in excess O2O2 to form a mixture of water and carbon diox
julsineya [31]

Answer:

gghhgyuhuyyugffygcggyy

8 0
3 years ago
How can one kg of iron melt more ice than 1 kg lead at 100 °C
Vanyuwa [196]

Answer:

Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)

Explanation:

The given parameters are;

The metals provided to melt the ice and their temperature includes;

One kg (1000 g) of iron;

Specific heat capacity = 0.444 J/(g·°C)

Temperature = 100°C

1 kg (1000 g) of lead

Specific heat capacity = 0.160 J/(g·°C)

Temperature = 100°C

Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;

For iron, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ_{iron} = Heat obtained from the iron by the ice

ΔQ_{iron} = 0.444 m × 1000 × (100 - 0) = 44400 J

Heat absorbed by the ice for melting, H_l = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H_l = Mass of ice × Latent heat of ice

H_l = Mass of ice × 334 J/g = 44400 J

Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g

Mass of ice melted by the iron ≈ 132.9 g

For lead, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ_{lead} = Heat obtained from the iron by the ice

ΔQ_{lead} = 0.160 m × 1000 × (100 - 0) = 16000 J

Heat absorbed by the ice for melting, H_l = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H_l = Mass of ice × Latent heat of ice

H_l = Mass of ice × 334 J/g = 16000 J

Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g

Mass of ice melted by the lead ≈ 47.9 g

Therefore, mass of  ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.

3 0
3 years ago
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