All categories

2 years ago

If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm?

2 answers:

8 0

Since the temperature is constant, therefore, this problem can be solved based on Boyle's law.
Boyle's law states that: " At constant temperature, the pressure of a certain mass of gas is inversely proportional to its pressure".

This can be written as:
P1V1 = P2V2
where:
P1 is the initial pressure = 1 atm
V1 is the initial volume = 3.6 liters
P2 is the final pressure = 2.5 atm
V2 is the final volume that we need to calculate

Substitute with the givens in the above mentioned equation to get the final volume as follows:
P2V1 = P2V2
1(3.6) = 2.5V2
3.6 = 2.5V2
V2 = 3.6 / 2.5 = 1.44 liters

yaroslaw [1]2 years ago

6 0

(1.00 atm) (3.60 liters) =(2.5 atm ) (x); x= 1.44 L

I hope that helped

You might be interested in

Determine whether the statement is true or false, and why? “Climate change could cause many habitats to be destroyed, leading to

77julia77 [94]

Answer:

False. It should read that both plant and animal species are in danger of extinction, and climate change can destroy habitats.

Explanation:

8 0

2 years ago

At 50 degree C pK_w = 13.26. What is the pH of pure water at this temperature?

Ksivusya [100]

Answer:

6.63

Explanation:

From the relationship;

pH = pKw/2

Pkw = 13.26

Then it follows that;

pH = 6.63

Hence if. pKw = 13.26, the pH = 6.63

8 0

2 years ago

A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net

Margarita [4]

Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

4 0

2 years ago

You are eating a pizza. What type of mixture are you consuming?

Hitman42 [59]

Answer:

Explanation:

Cheese, Meat, dough, Sauce

3 0

2 years ago

Read 2 more answers

Diborane, B2H6 a possible rocket propellant, can be made by using lithium hydride (LiH): 6 LiH+ 2 BCl2àB2H6+ 6 LiCl . If you mix

Genrish500 [490]

Answer :

(a) Limiting reactant = $LiH$

(b) The excess reactant = $BCl_3$

(c) The percent of excess reactant is, 50.87 %

(d) The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

(e) The mass of $LiCl$ produced is, 1066.42 lb

Explanation : Given,

Mass of $LiH$ = 200 lb = 90718.5 g

conversion used : (1 lb = 453.592 g)

Mass of $BCl_3$ = 1000 lb = 453592 g

Molar mass of $LiH$ = 7.95 g/mole

Molar mass of $BCl_3$ = 117.17 g/mole

Molar mass of $B_2H_6$ = 27.66 g/mole

Molar mass of $LiCl$ = 42.39 g/mole

First we have to calculate the moles of $LiH$ and $BCl_3$ .

$\text{Moles of }LiH=\frac{\text{Mass of }LiH}{\text{Molar mass of }LiH}=\frac{90718.5g}{7.95g/mole}=11411.13moles$

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{453592g}{117.17g/mole}=3871.23moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$6LiH+2BCl_3\rightarrow B_2H_6+6LiCl$

From the balanced reaction we conclude that

As, 6 moles of $LiH$ react with 1 mole of $BCl_3$

So, 11411.13 moles of $LiH$ react with $\frac{11411.13}{6}=1901.855$ moles of $BCl_3$

From this we conclude that, $BCl_3$ is an excess reagent because the given moles are greater than the required moles and $LiH$ is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 3871.23 - 1901.855 = 1969.375 moles

Total excess reactant = 3871.23 moles

Now we have to determine the percent of excess reactant $(BCl_3)$ .

$\% \text{ excess reactant}=\frac{\text{Moles of remaining excess reactant}}{\text{Moles of total excess reagent}}\times 100$

$\% \text{ excess reactant}=\frac{1969.375}{3871.23}\times 100=50.87\%$

The percent of excess reactant is, 50.87 %

Now we have to calculate the moles of $B_2H_6$ .

As, 6 moles of $LiH$ react to give 1 mole of $B_2H_6$

So, 11411.13 moles of $LiH$ react to give $\frac{11411.13}{6}=1901.855$ moles of $B_2H_6$

Now we have to calculate the mass of $B_2H_6$ .

$\text{Mass of }B_2H_6=\text{Moles of }B_2H_6\times \text{Molar mass of }B_2H_6$

$\text{Mass of }B_2H_6=(1901.855mole)\times (27.66g/mole)=52605.3093g$

Now we have to calculate the percent yield of $B_2H_6$ .

$\%\text{ yield of }B_2H_6=\frac{\text{Actual yield of }B_2H_6}{\text{Theoretical yield of }B_2H_6}\times 100=\frac{20411.7g}{52605.3093g}\times 100=38.80\%$

The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

Now we have to calculate the moles of $LiCl$ .

As, 6 moles of $LiH$ react to give 6 mole of $LiCl$

So, 11411.13 moles of $LiH$ react to give 11411.13 moles of $LiCl$

Now we have to calculate the mass of $LiCl$ .

$\text{Mass of }LiCl=\text{Moles of }LiCl\times \text{Molar mass of }LiCl$

$\text{Mass of }LiCl=(11411.13mole)\times (42.39g/mole)=483717.8007g=1066.42lb$

The mass of $LiCl$ produced is, 1066.42 lb

8 0

2 years ago