What is the heat required to vaporize 650 grams of a liquid with a heat of vaporization of 723 joules/grams?

What is the empirical formula for the compound P4O6??

777dan777 [17]

The empirical formula is P₂O₃

5 0

2 years ago

Nitrogen gas is introduced into a large deflated plastic bag. No gas is allowed to escape, but as more and more nitrogen is adde

Natalija [7]

Answer:

0.9307 moles have been introduced into the bag.

Explanation:

Pressure of the gas within the bag,P = 1.00 atm

Temperature of the gas remains at room temperature,T=20.0 °C = 293.15 K

Volume of the gas in the bag = V = 22.4 L

Number of moles of gas = n

Using an ideal gas equation:

$PV=nRT$

$1.00 atm\times 22.4 atm=n\times 0.0821 atm l/mol K\times 293.15 K$

n = 0.9307 moles

0.9307 moles have been introduced into the bag.

4 0

3 years ago

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coeffici

Brut [27]

Question:

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K

Answer:

The mean activity coefficient for HCl solution is 0.78.

Explanation:

Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is

$E=E_{0}-0.0514 \mathrm{V} \ln \gamma$

As we know that $E_{0}$ = 0.22 V and E = 0.342 V, the equation will become

$0.342 V+0.0514 V \ln (0.124)=0.22 V-0.0514 V \ln \gamma$

$0.342 V-0.222 V=-0.0514 V(\ln \gamma+\ln 0.124)$

$0.12 \mathrm{V}=-0.0514 \mathrm{V}(\ln \gamma+\ln 0.124)$

$\frac{0.12}{0.0514}=-\ln (0.124 \gamma)$

$-2.3346=\ln (0.124 \gamma)$

$e^{-2.3346}=0.124 \gamma$

$0.0968=0.124 \gamma$

$\gamma=\frac{0.0968}{0.124}=0.78$

So, the mean activity coefficient is 0.78.

6 0

2 years ago

How much water would need to be added to 1092 mL of a 54.7 M NaCl solution to make a 0.25 M solution?

babunello [35]

Answer:

237.8L of water would need to be added.

Explanation:

The first thing to do is to identify that the equation to be used is M1V1=M2V2. (This equation works because it turns everything into moles which can then be compared).

Then figure out what information you have and what is being found. In this case:

M1 = 54.7 M

V1 = 1092 mL = 1.092 L

M2 = 0.25 M

V2 = unknown

Then solve the equation for whatever you are trying to find.

M1V1=M2V2

V2=M1V1/M2

Now you need to plug everything in.

V2=(54.7M*1.091L)/0.25M

V2=238.93L

That means that the solution needs a volume of 238.7L to gain a molarity of 0.25M but the starting solution already had a volume of 1.092 L meaning that to find the amount of solvent that needs to be added you just subtract the starting volume by the volume that the solution needs to be.

238.93L - 1.091L = 237.8L

Therefore the answer is that 237.8L needs to be added to a 1.092L 54.7M NaCl solution to make the concentration 0.25M.

I hope this helps. Let me know if anything is unclear.

8 0

2 years ago

The unfolding of a protein by heat or chemical treatment is referred to as

aksik [14]

Answer:

14

Explanation:

6 0

2 years ago

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