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tatiyna
3 years ago
13

A small hole in the wing of a space shuttle requires a 16.2 cm2 patch. (b) If the patching material costs NASA $3.58/in2, what i

s the cost of the patch to the nearest cent?
Chemistry
1 answer:
Tju [1.3M]3 years ago
6 0

The cost of the patch : = 887.84 cents

<h3>Further explanation</h3>

Given

16.2 cm² patch

cost = $3.58/in²

Required

The cost of the patch

Solution

We must pay attention to the conversion factor when changing units

1 cm² = 0.155 in²

16 cm² = 2.48 in²

Cost :

= $3.58/in² x 2.48 in²

= $8.8784

1 dollar is equal to 100 cents, so :

= $8.8784 x 100 cents/$

= 887.84 cents

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Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel
n200080 [17]

The question is incomplete, here is the complete question:

Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the following. The mole fraction of gas B?

A mixture of gases contains 10.25 g of N₂, 2.05 g of H₂, and 7.63 g of NH₃.

<u>Answer:</u> The mole fraction of gas B (hydrogen gas) is 0.557

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For nitrogen gas:</u>

Given mass of nitrogen gas = 10.25 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen gas}=\frac{10.25g}{28g/mol}=0.366mol

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 2.05 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrogen gas}=\frac{2.05g}{2g/mol}=1.025mol

  • <u>For ammonia gas:</u>

Given mass of ammonia gas = 7.63 g

Molar mass of ammonia gas = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia gas}=\frac{7.63g}{17g/mol}=0.449mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B+n_C}

Moles of gas B (hydrogen gas) = 1.025 moles

Total moles = [0.366 + 1.025 + 0.449] = 1.84 moles

Putting values in above equation, we get:

\chi_{(H_2)}=\frac{1.025}{1.84}=0.557

Hence, the mole fraction of gas B (hydrogen gas) is 0.557

6 0
3 years ago
Find ΔHrxn for the following reaction: <br><br> 2PbS(s)+3O2(g)→2PbO(s)+2SO2(g)
ch4aika [34]

Answer:

ΔH°rxn = -827.5 kJ

Explanation:

Let's consider the following balanced equation.

2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)

We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]

ΔH°rxn = -827.5 kJ

5 0
3 years ago
6. Which of the following could be a base in a water solution?
Elza [17]

Answer:

h20

Explanation:

H2O is an acid and a base, and H20 is water.

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Dennis_Churaev [7]

Answer: The mowed lawn is the one from where the grasses are removed by using the machines or tools.

Explanation:

The mowed lawn is expected to have low number of species as the grasses may be few or scanty thus can support the population of few species like insects, mice, birds, and small number of grazing animals. On the other hand the weedy field can be hub of insects, reptiles like snakes, small mammals, and large mammals. Large weed field can provide food, and habitat to the large number of species. This will support the increase in biodiversity as compared to the mowed lawn.

5 0
2 years ago
An element has the mass number 12 and atomic number 6. The number of neutron in it is?
slamgirl [31]

Answer:

12-6=6

the answer is A)6

8 0
3 years ago
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