There are six kingdoms of living things. Archaebacteria is one of these kingdoms.

Assume the volume of the crew cabin is 74,000 L. The pressure is maintained at around 1.00 atm, ideally with an 80% nitrogen and

hjlf

Answer:

The number of moles of oxygen present in the crew cabin at any given time is 615.309 moles

Explanation:

The given parameters are;

Volume of the crew cabin = 74,000 L

Pressure of the crew cabin = 1.00 atm

Percentage of nitrogen in the mixture of gases in the cabin = 80%

Percentage of oxygen in the mixture of gases in the cabin = 20%

Temperature of the cabin = 20°C = 293.15 K

Therefore, volume of oxygen in the crew cabin = 20% of 74,000 L

Hence, volume of oxygen in the crew cabin = $\frac{20}{100} \times 74,000 \, L = 14,800 \, L$

From the universal gas equation, we have;

$n = \frac{P \times V}{R \times T}$

Where:

n = Number of moles of oxygen

P = Pressure = 1.00 atm

V = Volume of oxygen = 14,800 L

T = Temperature = 293.15 K

R = Universal Gas Constant = 0.08205 L·atm/(mol·K)

Plugging in the values, we have;

$n = \frac{1 \times 14,800 }{0.08205 \times 293.15 } = 615.309 \, moles$

The number of moles of oxygen present in the crew cabin at any given time = 615.309 moles.

3 0

3 years ago

Which description of salt is a physical property?

Nonamiya [84]

Answer:

White being the color and coming in small grains.

Explanation:

Physical properties are something you can clearly see about the object.

6 0

2 years ago

Read 2 more answers

Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o

Gnom [1K]

Answer: The empirical formula for the given compound is $C_5H_7N$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

$C_xH_yN_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of $CO_2=21.363mg=21.363\times 10^3g=21363g$

Mass of $H_2O=6.125g=6.125\times 10^3g=6125g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, $\frac{12}{44}\times 21363=5826.27g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, $\frac{2}{18}\times 6125=680.55$ of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = $\frac{485.52}{97.73}=4.96\approx 5$

For Hydrogen = $\frac{680.55}{97.73}=6.96\approx 7$

For Nitrogen = $\frac{97.73}{97.73}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is $C_5H_7N_1=C_5H_7N$

7 0

2 years ago

How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them? (answ

Jet001 [13]

Answer:

7.12 mm

Explanation:

From coulomb's law,

F = kqq'/r².................... Equation 1

Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.

Make r the subject of the equation,

r = √(kqq'/F).......................... Equation 2

Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute into equation 2

r = √[ (75×10⁻⁹ )²9.0×10⁹/1]

r = 75×10⁻⁹.√(9.0×10⁹)

r = (75×10⁻⁹)(9.49×10⁴)

r = 711.75×10⁻⁵

r = 7.12×10⁻³ m

r = 7.12 mm

Hence the distance between the point charge = 7.12 mm

3 0

3 years ago

1. Cp is smaller than Hx

andreev551 [17]

1 has a higher ionization dismal aoa

8 0

2 years ago