What element is this?

Imagine a planet that has a similar atmospheric pressure to Earth (760 mmHg) but different concentrations of gases in its atmosp

RUDIKE [14]

Answer:

76 mmHg is the partial pressure of oxygen on this new planet.

Explanation:

Atmospheric pressure on air on earth = 760 mmHg

On an unknown planet :

Oxygen makes up 10% of the atmosphere.

The mole fraction of oxygen gas can be written as = $\chi_1=\frac{10}{100}=0.1$

Carbon dioxide makes up 15% of the atmosphere.

The mole fraction of carbon dioxide gas can be written as = $\chi_2=\frac{15}{100}=0.15$

Nitrogen makes up the remaining 75% of the atmosphere.

The mole fraction of nitrogen gas can be written as = $\chi_1=\frac{75}{100}=0.75$

Atmospheric pressure of air on unknown planet,p= 760 mmHg (given)

Partial pressure of gases can be calculated by the help o Dalton's law of partial pressure:

Partial pressure of oxygen gas :

$p_1=p\times \chi_1$

$p_1=760 mmHg\times 0.1=76 mmHg$

Partial pressure of carbon dioxide gas :

$p_2=p\times \chi_2$

$p_2=760 mmHg\times 0.15=114 mmHg$

Partial pressure of nitrogen gas :

$p_3=p\times \chi_3$

$p_3=760 mmHg\times 0.75=570 mmHg$

8 0

3 years ago

If a container were to have 24 molecules of C5H12 and 24 molecules of O2 initially, how many total molecules (reactants plus pro

frutty [35]

Answer:

81 molecules

Explanation:

The reaction between C5H12 and O2 is a combustion reaction and is represented by the following equation;

C5H12 + 8O2 --> 5CO2 + 6H2O

The ratio of C5H12 to O2 from the above equation is 1 : 8.

Aplying the conditins of the question; 24 molecules each of C5H12 and O2 we have;

3C5H12 + 24O2 --> 15CO2 + 18H2O

This means we have 24 - 3 = 21 molecules of C5H12 that are unreacted.

Total molecules is given as;

3(C5H12) + 24(O2) + 15(CO2) + 18(H2O) + 21(Unreacted C5H12) = 81 molecules

5 0

3 years ago

A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af

larisa86 [58]

Explanation:

The given reaction is as follows.

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Hence, number of moles of NaOH are as follows.

n = $0.05 L \times 0.1 M$

= 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

n = $0.025 L \times 0.1 M$

= 0.0025 mol

According to ICE table,

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Initial: 0.005 mol 0.0025 mol 0 0

Change: -0.0025 mol -0.0025 mol +0.0025 mol

Equibm: 0.0025 mol 0 0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

[HA] = $\frac{0.0025 mol}{V}$

[NaA] = $\frac{0.0025 mol}{V}$

$[A^{-}] = [NaA] = \frac{0.0025 mol}{V}$

Now, we will calculate the $pK_{a}$ value as follows.

pH = $pK_{a} + log \frac{A^{-}}{HA}$

$pK_{a} = pH - log \frac{[A^{-}]}{[HA]}$

= $3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}$

= 3.42

Thus, we can conclude that $pK_{a}$ of the weak acid is 3.42.

8 0

3 years ago

How many ml of a 0.50m solution of hno3 solution are needed to make 500 ml of 0.15m hno3

Anna71 [15]

Answer:

150.0 mL.

Explanation:

It is known that the no. of millimoles of HNO₃ before dilution = the no. of millimoles of HNO₃ after dilution.

∵ (MV) before dilution = (MV) after dilution.

<em>∴ V before dilution = (MV) after dilution / M before dilution</em> = (0.15 M)(500.0 mL)/(0.50 M) = <em>150.0 mL.</em>

7 0

3 years ago

Can someone please list one of the groups of representative elements in 1A-7A

Vadim26 [7]

Group 1A(1), the alkali metals, includes lithium, sodium, and potassium. Group 7A(17) the halogens, includes chlorine, bromine, and iodine. hope this helps:)

7 0

3 years ago