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Arlecino [84]
3 years ago
6

Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f

ollowing?Part A:Decrease the volume to one-fourth the original volume while holding the temperature constant.Express your answer in terms of the variable P initial.Part B:Reduce the Kelvin temperature to half its original value while holding the volume constant.Express your answer in terms of the variable P initial .Part C:Reduce the amount of gas to half while keeping the volume and temperature constant.Express your answer in terms of the variable P initial .
Chemistry
1 answer:
pishuonlain [190]3 years ago
3 0

Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>

<em />

<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>

T₂ = 0.5 T₁. According to Gay-Lussac's law,

\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}

P₂ in terms of the ideal gas equation is:

P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}

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riadik2000 [5.3K]

The fraction of Earth's radius (6371 km) relative to the thickness of the oceanic (7.5 km) and continental crust (35 km) is 0.12 and 0.55, respectively.    

What we know:

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  • The average thickness of oceanic crust (O) = 7.5 km
  • The average thickness of continental crust (C) = 35 km

We need to convert all the above units from kilometers to miles:

E = \frac{0.6214 mi}{1 km}*6371 km = 3958.9 mi

O = \frac{0.6214 mi}{1 km}*7.5 km = 4.7 mi

C = \frac{0.6214 mi}{1 km}*35 km = 21.7 mi

Now, we can calculate the fraction of Earth's radius relative to each type of crust, with the given equation:

X = \frac{avg. \: thickness}{avg. \: radius} \times 100

  • <u>For the oceanic crust (O)</u>:

X = \frac{4.7 mi}{3958.9 mi}\times 100 = 0.12

  • <u>For the continental crust (C)</u>:

X = \frac{21.7 mi}{3958.9 mi}\times 100 = 0.55

Therefore, the fraction of Earth's radius relative to the oceanic and continental crust is 0.12 and 0.55, respectively.

You can see another example of calculation of fractions of Earth's radius here: brainly.com/question/4675868?referrer=searchResults

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2 years ago
This is a sign of a chemical reaction that involves a new color being created during the reaction.
svetoff [14.1K]
Hey there!


When two or more substances combine, they create one or more new substances, which sometimes have different molecular structures from the original substances, meaning they absorb and radiate light in different ways, leading to a color change.


Hope this helps! :)
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3 years ago
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Answer:

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3 years ago
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
What is the ph of a solution that has a poh of 11.24
galina1969 [7]
Always remember that pH + pOH = 14 
Here, you have a pOH of 11.24, so  you replace it in the equation, and u get:
pH + 11.24 = 14

Then, You move 11.24 to the other part. and moving from a part to another change the sign of the equation. And you get:
pH = 14 - 11.24 = 2.76

So, the pH of a solution that has a pOH of 11.24 is pH = 2.76

Hope this Helps :)
7 0
3 years ago
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