A majority of the 118 elements on the periodic table are classified as what?

Chemistry

1 answer:

vredina [299]2 years ago

6 0

Answer:

Metals

Explanation:

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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

The equilibrium constant for the reaction of carbon monoxide with water is 1.845. if 1.00 mol of each reactant is placed in a 2.

SpyIntel [72]

[CO] = 1 mol / 2L = 0.5 M

[
According to the equation:

and by using the ICE table:

CO(g) + H2O(g) ↔ CO2(g) + H2(g)

initial 0.5 0.5 0 0

change -X -X +X +X

Equ (0.5-X) (0.5-X) X X

when Kc = X^2 * (0.5-X)^2

by substitution:

1.845 = X^2 * (0.5-X)^2 by solving for X

∴X = 0.26

∴ [CO2] = X = 0.26

4 0

2 years ago

A scuba diver knows that she needs 50.0mol of air for an upcoming dive. What size (volume) tank will she need to fill for this d

erica [24]

Answer:

1120 L.

Explanation:

Hello!

In this case, as no conditions of pressure of temperature are given for this problem, we can assume that the scuba diver dives at STP (1 atm and 273.15 K), which means that 1 mole of air would occupy a volume of 22.4 L.

In such a way, since she needs 50.0 moles of air, the following ratio is useful to compute the size (volume) of the tank she needs:

$V_2=\frac{V_1*n_2}{n_1}$