When the water level of a stream or river exceeds its natural banks, it has reached its

A gas with 4.0 atmospheres of pressure has a temperature of 27°C.

Gala2k [10]

Answer:

7.462

Explanation:

Well, every time that the tempurature is increased, the atmspheric pressure is increased by 0.574%. This would then mean that you would have 0.574 times

13. That would then equal 7.462. I hope this helps.

6 0

2 years ago

If you can run 3 miles in 19 minutes. What is your pace in kilometers per hour?

Norma-Jean [14]

Answer:

10.1389

Explanation:

5 0

2 years ago

Is BH2F a gas or a liquid and what is the boiling point

hram777 [196]

Hey I tried but could not find your answer, but maybe you could try to calculate the average of bpts of the different elements in the compound to find the boiling point. And when you know the bpt, you can find out whether it is a liquid or gas. But after searching a lot, I found it was a solid. Good luck!

7 0

2 years ago

A certain gas is present in a 10.0 LL cylinder at 4.0 atmatm pressure. If the pressure is increased to 8.0 atmatm the volume of

ICE Princess25 [194]

Answer:

The gas obeys Boyle’s law and the value of $k_i\&k_f$ both are equal to 40.0 atm L.

Explanation:

Initial volume of the gas = $V_1=10.0 L$

Initial pressure of the gas = $P_1=4.0 atm$

Final volume of the gas = $V_2=5.0 L$

Final pressure of the gas = $P_2=8.0 atm$

This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

$PV=k$

The equation given by this law is:

$P_1V_1=P_2V_2$

$P_1\propto \frac{1}{V_1}$

$P_1V_1=k_i$

$k_i=4.0 atm\times 10.0 L = 40.0 atm L$

$P_2\propto \frac{1}{V_2}$

$P_V_2=k_f$

$k_f=8.0 atm\times 5.0 L = 40.0 atm L$

$k_i=k_f=40.0 atm L$

The gas in the cylinder is obeying Boyle's law.

The gas obeys Boyle’s law and the value of $k_i\&k_f$ both are equal to 40.0 atm L.

6 0

2 years ago

Consider the balanced equation for the following reaction:

Bad White [126]

<u>Answer:</u> The theoretical yield of the lithium chlorate is 1054.67 grams

<u>Explanation:</u>

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Actual moles of lithium chlorate = 9.45 moles

Molar mass of lithium chlorate = 90.4 g/mol

Putting values in above equation, we get:

$9.45mol=\frac{\text{Actual yield of lithium chlorate}}{90.4g/mol}\\\\\text{Actual yield of lithium chlorate}=(9.45mol\times 90.4g/mol)=854.28g$

To calculate the theoretical yield of lithium chlorate, we use the equation:

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield of lithium chlorate = 854.28 g

Percentage yield of lithium chlorate = 81.0 %

Putting values in above equation, we get:

$81=\frac{854.28g}{\text{Theoretical yield of lithium chlorate}}\times 100\\\\\text{Theoretical yield of lithium chlorate}=\frac{854.28\times 100}{81}=1054.67g$

Hence, the theoretical yield of the lithium chlorate is 1054.67 grams

7 0

2 years ago