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kifflom [539]
3 years ago
14

What change in volume results if 60 mL of a gas is cooled from 33 C to 5 C?

Chemistry
1 answer:
Eva8 [605]3 years ago
6 0

Answer:

Change in volume on changing temperature from 33^{\circ}C to 5^{\circ}C is 5.49 mL

Explanation:

Initial volume of gas = V = 60 mL

Assuming final volume of gas to be V' mL

Initial temperature = T = 33^{\circ}C = 306 K

Final temperature = T' = 5^{\circ}C = 278 K

The relationship between volume and temperature of gas at constant pressure is shown below

\displaystyle \frac{V}{V'}=\displaystyle  \frac{T}{T'} \\\displaystyle \frac{60\textrm{ mL}}{V'} = \displaystyle \frac{306\textrm{ K}}{T} \\V' = 54.51 \textrm{ mL} \\\textrm{Change in volume} = \left ( V-V' \right ) \\\textrm{Change in volume} = \left ( 60-54.51 \right )\textrm{ mL} \\\textrm{Change in volume} = 5.49 \textrm{ mL}

Change in volume on changing temperature = 5.49 mL

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My name is Ann [436]

Answer: It's the anode broski (B)

Explanation: I'm taking the Chem summer course too broski, this was the correct answer. Cheers broski

8 0
3 years ago
How many moles of hydrogen sulfide are contained in a 77.7-8 sample of this gas?
valentina_108 [34]

Answer:

its either C or D my guess is C though because 35.0g of hyddrogen is =1.03 mol

Explanation:

7 0
3 years ago
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


4 0
3 years ago
450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS
ExtremeBDS [4]

Answer:

1597.959 g  

Explanation:

Given Data:

Amount of Cr₂(SO₄)₃ = 450 g

Amount of potassium phosphate K₃PO₄ = in Excess

grams of potassium sulfate K₂SO₄= ?

Solution

The Reaction will be

                 Cr₂(SO₄)₃ + 2K₃PO₄  ----------> 3K₂SO₄ + 2CrPO₄

Information that we have from reaction

                Cr₂(SO₄)₃ + 2K₃PO₄  ----------> 3K₂SO₄ + 2CrPO₄

                    1 mol          2 mol                   3 mol

we come to know from the above reaction that

1 mole of chromium(iii) sulfate (Cr₂(SO₄)₃) react with 2 mole of potassium phosphate (K₃PO₄) to produce 3 mole of K₂SO₄

We also know that

molar mass of Cr₂(SO₄)₃ = 147 g/mol

molar mass of K₃PO₄ = 212 g/mol

molar mass of K₂SO₄ = 174 g/mol

if we represent mole in grams then

      Cr₂(SO₄)₃             +       2K₃PO₄          ----------> 3K₂SO₄ + 2CrPO₄

       1 mol (147 g/mol)         2 mol  (212 g/mol)      3 mol  (174g/mol)

So, Now we have the following details

          Cr₂(SO₄)₃  +  2K₃PO₄          ----------> 3K₂SO₄ + 2CrPO₄

              147 g         424 g                           522 g

So,

we come to know that 147 g of Cr₂(SO₄)₃ combine with 424 g of 2K₃PO₄ produce  522 g of K₂SO₄

So now we calculate that how many grams of potassium sulfate will be produced

Apply unity formula

              147 g of  Cr₂(SO₄)₃  ≅ 522 g of K₂SO₄

              450 g of  Cr₂(SO₄)₃  ≅ ? g of K₂SO₄

by doing cross multiplication

g of K₂SO₄ =522 g x 450 g / 147 g

g of K₂SO₄ =  1597.959 g

So the write answer is  1597.959 g  

***Note: By calculation it is obvious that the correct answer is  1597.959 g  

8 0
3 years ago
If the same amount of heat is added to 25.0 g of each of the metals, which are all at the same initial temperature, which metal
AVprozaik [17]

Answer:

The bismuth sample.

Explanation:

The specific heat c of a substance (might not be a metal) is the amount of heat required for heating a unit mass of this substance by unit temperature (e.g., \rm 1\; ^{\circ}C.) The formula for specific heat is:

\displaystyle c = \frac{Q}{m \cdot \Delta T},

where

  • Q is the amount of heat supplied.
  • m is the mass of the sample.
  • \Delta T is the increase in temperature.

In this question, the value of Q (amount of heat supplied to the metal) and m (mass of the metal sample) are the same for all four metals. To find \Delta T (change in temperature,) rearrange the equation:

\displaystyle c \cdot \Delta T = \frac{Q}{m},

\displaystyle \Delta T = \frac{Q}{c \cdot m}.

In other words, the change in temperature of the sample, \Delta T can be expressed as a fraction. Additionally, the specific heat of sample, c, is in the denominator of that fraction. Hence, the value of the fraction would be the largest for sample with the smallest specific heat.

Make sure that all the specific heat values are in the same unit. Find the one with the smallest specific heat: bismuth (\rm 0.123 \; J \cdot g\cdot \,^{\circ}C^{-1}.) That sample would have the greatest increase in temperature. Since all six samples started at the same temperature, the bismuth sample would also have the highest final temperature.

3 0
3 years ago
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