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zhuklara [117]
3 years ago
8

If a gas sample has a pressure of 30.7 kPa at 0.00*C, by how much does the temperature have to decrease to lower the pressure to

28.4 kPa? (volume is constant)
Chemistry
1 answer:
Scrat [10]3 years ago
8 0

Answer:

                      252.68 K  or   -20.46 °C

Explanation:

                    According to Gay-Lussac's Law, "Pressure and Temperature at given volume are directly proportional to each other".

Mathematically,

                                              P₁ / T₁  =  P₂ / T₂   ---- (1)

Data Given:

                  P₁  =  30.7 kPa

                  T₁  =  0.00 °C  =  273.15 K

                  P₂  =  28.4 kPa

                  T₂  =  <u>???</u>

Solving equation for T₂,

                  T₂  =  P₂ T₁ / P₁

Putting values,

                  T₂  =  28.4 kPa × 273.15 K / 30.7 kPa

                  T₂  =  252.68 K  or   -20.46 °C

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Mandarinka [93]

Answer:

Weak electrolytes are

HgCl2 and NH₃

Explanation:

  • Electrolytes are compounds that conduct electricity while in molten or aqueous form.
  • Electrolytes may therefore be decomposed by passing electric current through them.
  • They may be either strong or weak electrolytes.
  • Strong electrolytes are those that ionize completely to form mobile ions while weak electrolytes ionize partially to generate ions.
  • Examples of weak electrolytes  are HgCl2, NH₃
  • Examples of strong electrolytes are LiOH, HClO₄, and HI.

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What is it called when the moon blocks the view from the sun?
defon

Answer:

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Explanation:

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Convert 9.90km to mm
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9900000

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3 0
3 years ago
A flask with a volume of 3.16 l contains 9.33 grams of an unknown gas at 32.0°c and 1.00 atm. What is the molar mass of the gas?
Inessa05 [86]

Answer:

73.88 g/mol

Explanation:

For this question we have to keep in mind that the unknown substance is a <u>gas</u>, therefore we can use the <u>ideal gas law</u>:

PV=nRT

In this case we will have:

P= 1 atm

V= 3.16 L

T = 32 ªC = 305.15 ºK

R= 0.082 \frac{atm*L}{mol*K}

n= ?

So, we can <u>solve for "n"</u> (moles):

1~atm*3.16~L~=~n*0.082~\frac{atm*L}{mol*K}*305.15~K

n=\frac{1~atm*3.16~L~}{0.082~\frac{atm*L}{mol*K}*305.15~K}

n=0.126~mol

Now, we have to remember that the <u>molar mass value has "g/mol"</u> units. We already have the grams (9.33 g), so we have to <u>divide</u> by the moles:

molar~mass=\frac{9.33~grams}{0.126~mol}

molar~mass=73.88\frac{grams}{mol}

7 0
3 years ago
You have 8 moles of a gas at 250 K in a 6 L container. What is the pressure of the gas? Show your work
Bogdan [553]

Hello:

In this case, we will use the Clapeyron equation:

P = ?

n = 8 moles

T = 250 K

R = 0.082 atm.L/mol.K

V = 6 L

Therefore:

P * V = n *  R * T

P * 6 = 8 * 0.082* 250

P* 6 = 164

P = 164 / 6

P = 27.33 atm


Hope that helps!

7 0
3 years ago
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