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Yuri [45]
4 years ago
7

A tank containing liquid butane and one containing liquid propane are sitting adjacent to one another. Each tank is equipped wit

h a regulator with a pressure gauge. Classify these statements as true or false. The pressure reading for the butane tank is higher than that for the propane tank. The pressure readings will change when more liquid is added to the tanks. If the tanks are placed outside and the sun warms each tank to 46 , the pressure readings will be higher than they were initially. If an equal number of moles of gas is allowed to escape rapidly from each tank, the temperature of the butane tank will be lower than that of the propane tank.
Chemistry
1 answer:
-BARSIC- [3]4 years ago
5 0

Explanation:

The pressure reading for the butane tank is higher than that for the propane tank is False.  

The pressure readings will change when more liquid is added to the tanks is False

If the tanks are put outside and the sun warms each tank to 46, the pressure levels will be greater than they were originally True When an equal number of moles of gas will easily exit from each tank, the butane tank temperature will be smaller than that of the propane tank True.

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Rudiy27
Answered above

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3 0
3 years ago
which term refers to the amount of water vapor in the air? A) dew point B) Precipitation C) humidity or D) relative humidity ?
Fudgin [204]
C) Humidity.
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4 0
3 years ago
Read 2 more answers
Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

6 0
4 years ago
If 7.6 moles of H2O are produced from the reaction how many moles of O2 were used?
elena-14-01-66 [18.8K]

Answer:

Number of moles of oxygen used=13.5moles

Explanation:

To know the number of moles of oxygen used, first calculate the molar mass of water.

H2O=(2*1)+(1*16)

=2+16

18g/mol of H2O

The moles of H2O is 7.6 moles

So first, find the molar mass of oxygen

O2=2*16

=32g/mol

Then, number of moles of oxygen is equal to molar mass of oxygen divide by the molar mass of water times the number of moles of water

No. moles of O2=[32g/mol]/[18g/mol] *7.6moles

Moles of O2=1.778*7.6moles

No.moles of O2=13.5 moles

Therefore, the number of moles of oxygen used was 13.5moles

7 0
3 years ago
Determine the expected diffraction angle for the first-order diffraction from the (111) set of planes for FCC nickel (Ni) when m
faust18 [17]

Answer:

56°

Explanation:

First calculate a:

a=2 R \sqrt{2}=2(0.1246) \sqrt{2}=0.352 \mathrm{nm}

The interplanar spacing can be calculated from:

d_{111}=\frac{a}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{0.352}{\sqrt{3}}=0.203 \mathrm{nm}

The diffraction angle is determined from:

\sin \theta=\frac{n \lambda}{2 d_{111}}=\frac{1(0.1927)}{2(0.2035)}=0.476

Solve for \theta

\theta=\sin ^{-1}(0.476)=28^{\circ}

The diffraction angle is:

2 \theta=2\left(28^{\circ}\right)=56^{\circ}

4 0
4 years ago
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