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Fed [463]
2 years ago
15

You are a NASA engineer. You are the chief engineer for the Apollo 13 mission to the moon. The astronauts are running out of oxy

gen and need to get rid of the excess carbon dioxide. You know that sodium hydroxide (NaOH) has been suggested as a means of removing carbon dioxide from the spacecraft cabin. The filter which they had been using is fully saturated and no longer works. You remember that the astronauts have a 5000. g container of sodium hydroxide on the ship. You also know that sodium hydroxide can be used to remove carbon dioxide.
Use the following reaction: NaOH + CO2 → Na2CO3 + H2O

The astronauts have 2 days left before they land on Earth. You know that there are three astronauts, and each astronaut emits roughly 500. g Of carbon dioxide each day. Is there enough sodium hydroxide in the cabin to cleanse the cabin air of the carbon dioxide, or are the astronauts doomed?
Chemistry
1 answer:
Naddik [55]2 years ago
8 0

Given the data from the question, the astronauts do not have enough sodium hydroxide in the cabin to cleanse the cabin air of the carbon dioxide.

<h3>How to determine the total CO₂</h3>
  • CO₂ emitted per astronaut per day = 500 g
  • CO₂ for 3 astronaut per day = 3 × 500 = 1500 g
  • CO₂ for 3 astronaut for 2 days = 2 × 1500 = 3000 g

<h3>Balanced equation </h3>

2NaOH + CO₂ —>Na₂CO₃ + H₂O

Molar mass of NaOH = 40 g/mol

Mass of NaOH from the balanced equation = 2 × 40 = 80 g

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY

From the balanced equation above,

44 g of CO₂ required 80 g of NaOH

<h3>How to determine the mass of NaOH needed </h3>

From the balanced equation above,

44 g of CO₂ required 80 g of NaOH

Therefore,

3000 g of CO₂ will require = (3000 × 80) / 44 = 5454.55 g of NaOH

From the calculation made above, we can see that 5454.55 g of NaOH is needed by the astronaut to clean the cabin air but unfortunately, they only have 5000 g of NaOH.

Thus, we can conclude that the astronauts do not have enough NaOH to clean the air

Learn more about stoichiometry:

brainly.com/question/14735801

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Briefly explain why the bulk temperature of the water remains low (at room temperature)
elena55 [62]
If the humidity of the room is low, the water that contacts the air directly could evaporate and takes some energy from the bulk thus decreasing its temperature steadily. This allows the water to have a lower temperature.
4 0
3 years ago
What are the amounts of molecules in 19 g flourine
olya-2409 [2.1K]

Answer:

molecules present in, 76 gram of fluorine

Explanation:

atomic weight of fluorine is 19 g per mole, molecular weight  is 38g per mole .

Therefore 76 gram fluorine contains 1976=4 gram atom

8 0
4 years ago
How many grams of C5H12 must be burned to heat 1.39 kg of water from 21.2 °C to 97.0 °C? Assume that all the heat released durin
faust18 [17]

Answer:

m = 8.9856 g

Explanation:

In order to do this, we need to write the expressions that are to be used. First, to calculate heat:

Q = m*C*ΔT (1)

Where C would be heat capacity of the substance.

The heat can also be relationed with the moles and enthalpy of a compound using the following expression:

Q = n*ΔH (2)

Finally for the mass of any compound, we use the following expression:

m = n*MM (3)

So, in order to calculate the grams of pentane (C5H12), we need to calculate the moles of the compound, and to do that, we need the heat exerted.

So, as we are using water, let's calculate the heat that is been exerted with the water. The C of the water is 4.186 J/g °C so:

Q = (1.39 * 1000) * 4.186 * (21.2 - 97)

Q = -441,045.33 J

This is the heat neccesary to burn pentane and heat water. Now, with this value, let's calculate the moles used of pentane with expression (2). The ΔH of the pentane is -3,535 045.kJ/mol or -3.535x10⁶ J/mol. Solving for n we have:

n = -441,045.3 / -3.535x10⁶

n = 0.1248 moles

Finally, we can calculate the grams needed with expression (3). The molar mass of pentane is 72 g/mol

m = 0.1248 * 72

m = 8.9856 g

This is the mass needed to heat 1.39 kg of water

6 0
4 years ago
9. Calculate the percentage composition of silver sulfate, Ag2SO4.
riadik2000 [5.3K]

Answer:

69% Ag, 10.5% S, 20.5% O.

Explanation:

Ag₂SO₄ → Molar mass = 311.8 g/mol

2 mol of Ag → 107.87g/m . 2m = 215.74 g

1 mol of S → 32.06g/m . 1m = 32.06 g

4 mol of O → 16g/m  .4m = 64 g

Percentage composition will be:

(Mass of Ag/ Total mass) . 100

(215.74 g / 311.8 g) . 100 = 69 %

(Mass of S/ Total mass) . 100

(32.06 g / 311.8 g) .100 = 10.5%

(Mass of O/ Total mass) . 100

(64 g / 311.8 g) . 100 = 20.5 %

7 0
3 years ago
How do erosion and deposition work together to form sand dunes?
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Answer:

Erosion occurs through deflation, and sand that was picked up is deposited against an obstruction.

Explanation:

Erosion can be defined as a geological process which involves the wearing away of earth material and transportation of these material  to another location by Natural agents such as wind and water. Erosion involves removal of top soil, rocks and other earth particles by action of wind, glacier and water.

Wind causes erosion by deflation. Deflation is the actual removal of grain of rocks . Generally, it s the lowering of land surface that happens as a result of removal of surface particles(sands grains) by wind. Deflation is usually prevalent in arid areas. This sand can later be picked up and deposited to form sand dunes . Sand dunes are mounds of sand formed by wind . This sands are sheltered by an obstacle. And sand dunes are prevalent in desert and or along the beach. The sand dune grows as sand accumulate.

8 0
3 years ago
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