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3 years ago

A gas occupied 23.55 L at 118.2 Celsius. At what temp would the gas occupied 43.8 L

Chemistry

1 answer:

pantera1 [17]3 years ago

4 0

Answer:

The gas occupies 43.8 L at a temperature of 219.8 °C

Explanation:

Step 1: Data given

Volume of the gas = 23.55L

Temperature = 118.2 °C

New volume = 43.8 L

Step 2: Calculate changed temperature

(P1*V1)/T1 = (P2*V2)/T2

Let's assume the pressure stays constant, then we have:

V1/T1 = V2/T2

⇒ with V1 = the initial volume = 23.55L

⇒ with T1 = the initial temperature = 118.2 °C

⇒ with V2 = the changed volume = 43.8 L

⇒ with T2 = the changed temperature = TO BE DETERMINED

23.55L/118.2°C = 43.8 L / T2

T2 = 219.8 °C

The gas occupies 43.8 L at a temperature of 219.8 °C

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2 years ago

Excess magnesium reacts with 165.0 grams of hydrochloric acid in a single displacement reaction.

JulsSmile [24]

Answer:

The volume of hydrogen gas produced will be approximately 50.7 liters under STP.

Explanation:

Relative atomic mass data from a modern periodic table:

H: 1.008;
Cl: 35.45.

Magnesium is a reactive metal. It reacts with hydrochloric acid to produce

Hydrogen gas $\rm H_2$ , and
Magnesium chloride, which is a salt.

The chemical equation will be something like

$\rm ?\;Mg\;(s) + ?\;HCl \;(aq)\to ?\;H_2 \;(g)+ [\text{Formula of the Salt}]$ ,

where the coefficients and the formula of the salt are to be found.

To determine the number of moles of $\rm H_2$ that will be produced, first find the formula of the salt, magnesium chloride.

Magnesium is a group 2 metal. The oxidation state of magnesium in compounds tends to be +2.

On the other hand, the charge on each chloride ion is -1. Each magnesium ion needs to pair up with two chloride ions for the charge to balance in the salt, magnesium chloride. The formula for the salt will be $\rm MgCl_2$ .

$\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ ?\;MgCl_2\;(aq)$ .

Balance the equation. $\rm MgCl_2$ contains the largest number of atoms among all species in this reaction. Start by setting its coefficient to 1.

$\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ {\bf 1\;MgCl_2}\;(aq)$ .

The number of $\rm Mg$ and $\rm Cl$ atoms shall be the same on both sides. Therefore

$\rm {\bf 1\;Mg}\;(s) + {\bf 2\;HCl}\;(aq) \to ?\;H_2 \;(g)+ {1\;\underset{\wedge}{Mg}\underset{\wedge}{Cl_2}}\;(aq)$ .

The number of $\rm H$ atoms shall also conserve. Hence the equation:

$\rm {1\;Mg}\;(s) + {2\;\underset{\wedge}{H}Cl}\;(aq) \to {\bf 1\;H_2 \;(g)}+ {1\;MgCl_2}\;(aq)$ .

How many moles of HCl are available?

$M(\rm HCl) = 1.008 + 35.45 = 36.458\;g\cdot mol^{-1}$ .

$\displaystyle n({\rm HCl}) = \frac{m(\text{HCl})}{M(\text{HCl})} = \rm \frac{165.0\;g}{36.458\;g\cdot mol^{-1}} = 4.52576\;mol$ .

How many moles of Hydrogen gas will be produced?

Refer to the balanced chemical equation, the coefficient in front of $\rm HCl$ is 2 while the coefficient in front of $\rm H_2$ is 1. In other words, it will take two moles of $\rm HCl$ to produce one mole of $\rm H_2$ . $\rm 4.52576\;mol$ of $\rm HCl$ will produce only one half as much $\rm H_2$ .

Alternatively, consider the ratio between the coefficient in front of $\rm H_2$ and $\rm HCl$ is:

$\displaystyle \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}$ .

$\displaystyle n(\text{H}_2) = n(\text{HCl})\cdot \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}\;n(\text{HCl}) = \rm \frac{1}{2}\times 4.52576\;mol = 2.26288\;mol$ .

What will be the volume of that many hydrogen gas?

One mole of an ideal gas occupies a volume of 22.4 liters under STP (where the pressure is 1 atm.) On certain textbook where STP is defined as $\rm 1.00\times 10^{5}\;Pa$ , that volume will be 22.7 liters.

$V(\text{H}_2) = \rm 2.26288\;mol\times 22.4\;L\cdot mol^{-1} = 50.69\; L$ , or

$V(\text{H}_2) = \rm 2.26288\;mol\times 22.7\;L\cdot mol^{-1} = 51.37\; L$ .

The value "165.0 grams" from the question comes with four significant figures. Keep more significant figures than that in calculations. Round the final result to four significant figures.

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Hi,

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3 years ago

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3 years ago

Read 2 more answers

A large pot (10 kg) of beef stew is left out on the kitchen counter to cool. The rate of cooling, in kJ/min, equals 1.955 (T-25)

olga_2 [115]

Answer:

16 minutes

Explanation:

First, we need to calculate the amount of heat needed to cool the beef stew:

Q = mcΔT

Where m is the mass, c is the heat capacity and ΔT is the variation of the temperature.

Q = 10x4x(40 - 90)

Q = -2000 kJ

So, the beef stew needs to lost 2000 kJ to cool.

With the initial temperature at 90ºC, the rate of cooling(r) will be:

r = 1.955x(90 - 25)

r = 127.075 kJ/min

So, to lose 2000 kJ, will be necessary:

t = Q/r

t = 2000/127.075

t = 16 minutes

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3 years ago