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2 years ago

In a mixture of 3 gases what is the total pressure of the gases if Gas A exerts a pressure

Chemistry

1 answer:

kirill [66]2 years ago

4 0

Answer:

760 mm of Hg

Explanation:

If the gases A , B and C are non reacting , then according to Dalton's Law of Partial Pressure the total pressure exerted is equal to sum of individual partial pressure of the gases .

If there are n , number of gases then ,

$P_{total}= P_1+P_2+P_3+\dots +P_n$

Here ,

Partial pressure of Gas A = 400mm of Hg
Partial pressure of Gas B = 220 mm of Hg
Partial pressure of Gas C = 140mm of Hg

Hence the total pressure exerted is ,

$P_{total}= P_{Gas\ A }+P_{Gas\ B }+P_{Gas\ C }$

Substitute ,

$P_{total}=( 400 + 220 + 140 )mm\ of \ Hg$

Add ,

$P_{total}= 760\ mm\ of \ Hg$

Hence the total pressure exerted by the gases is 760 mm of Hg.

I hope this helps.

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Questions Answer all please

dybincka [34]

Answer:

Both,

Vertical

Explanation:

When you push a lawn mower, you can be moving up and down or across, when your pushing a wheel barrel you can do the same, when you do a push up your going up and down which is vertical.

6 0

3 years ago

A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the tempe

Vlad1618 [11]

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: $V_{1}$ = 571 mL, $T_{1} = 26^{o}C$

(a) $T_{2} = 5^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL$

(b) $T_{2} = 95^{o}F$

Convert degree Fahrenheit into degree Cesius as follows.

$(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL$

(c) $T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL$

8 0

3 years ago

Unstable isotopes undergo radioactive decay. what occurs during radioactive decay?

mel-nik [20]

Here is what radioactive decay is:
Radioactive decay is the spontaneous breakdown of an atomic nucleus resulting in the release of energy and matter from the nucleus. Remember that a radioisotope has unstable nuclei that does not have enough binding energy to hold the nucleus together.

5 0

3 years ago

Identify the correct coefficients to balance the redox reaction with the lowest possible integer coefficients.

Monica [59]

Answer:

$\rm 3\; Ag^{1+} + 1\; Al \to 1\; Al^{3+} + 3\; Ag$ .

Explanation:

Electrons are conserved in a chemical equation.

The superscript of $\rm Ag^{1+}$ indicates that each of these ions carries a charge of $+1$ . That corresponds to the shortage of one electron for each $\rm Ag^{+}$ ion.

Similarly, the superscript $+3$ on each $\rm Al^{3+}$ ion indicates a shortage of three electrons per such ion.

Assume that the coefficient of $\rm Ag^{+}$ (among the reactants) is $x$ , and that the coefficient of $\rm Al^{3+}$ (among the reactants) is $y$ .

$\rm \mathnormal{x}\; Ag^{1+} + ?\; Al \to \mathnormal{y}\; Al^{3+} + ?\; Ag$ .

There would thus be $x$ silver ( $\rm Ag$ ) atoms and $y$ aluminum ( $\rm Al$ ) atoms on either side of the equation. Hence, the coefficient for $\rm Al\!$ and $\rm Ag\!$ would be $y\!$ and $x\!$ , respectively.

$\rm \mathnormal{x}\; Ag^{1+} + \mathnormal{y}\; Al \to \mathnormal{y}\; Al^{3+} + \mathnormal{x}\; Ag$ .

The $x$ $\rm Ag^{1+}$ ions on the left-hand side of the equation would correspond to the shortage of $x$ electrons. On the other hand, the $y$ $Al^{3+}$ ions on the right-hand side of this equation would correspond to the shortage of $3\, y$ electrons.

Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of $x$ electrons, the right-hand side should also be $x\!$ electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of $3\, y$ electrons. These two expressions should have the same value. Therefore, $x = 3\, y$ .