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solong [7]
3 years ago
14

95.8 L of fluorine gas is being held at a temperature of 24.5oC. If the temperature were raised to 46.9 oC, what would the new v

olume be?
Chemistry
1 answer:
Xelga [282]3 years ago
6 0

Answer: 103 litres

Explanation:

Given that,

Original volume of fluorine gas (V1) = 95.8L

Original temperature of fluorine gas (T1) = 24.5°C

[Convert 24.5°C to Kelvin by adding 273

24.5°C + 273 = 297.5K]

New volume of gas (V2) = ?

New temperature of gas (T2) = 46.9°C

[Convert 46.9°C to Kelvin by adding 273

46.9°C + 273 = 319.9K]

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

95.8L/297.5K = V2/319.9K

To get the value of V2, cross multiply

95.8L x 319.9K = 297.5K x V2

30646.42L•K = 297.5K•V2

Divide both sides by 297.5K

30646.42L•K/297.5K = 297.5K•V2/297.5K

103L = V2

Thus, the new volume of fluorine gas will be 103 litres

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natali 33 [55]

Answer:

There are 0.93 g of glucose in 100 mL of the final solution

Explanation:

In the first solution, the concentration of glucose (in g/L) is:

15.5 g / 0.100 L = 155 g/L

Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.

  • 30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)

The concentration of the second solution is:

155 \frac{g}{L} *\frac{0.030L}{0.500L}=9.3\frac{g}{L}

So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:

1 L --------- 9.3 g

0.1 L--------- Xg

Xg = 9.3 g * 0.1 L / 1 L = 0.93 g

8 0
3 years ago
Hello<br>Thank you for not answering my question
julsineya [31]
**Visible confusion** no problem
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What mass of sodium hydroxide must be heated in order to produce 4.5 g of water?
inysia [295]

Answer:

10 g

Explanation:

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What functions of cells can be affected by a mutation?
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It changes the rate of growth that cells usually undergo.
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3 years ago
BY ANSWERING THIS QUESTION UR PUTTING IT ON UR MOM's LIFE THAT U WON'T STEAL MY POINTS.
Yakvenalex [24]

Answer:

T_2=-125.58\°C

Explanation:

Hello!

In this case, considering the Gay-Lussac's law which describes the pressure-temperature behavior as a directly proportional relationship by holding the volume as constant, we write:

\frac{T_1}{P_1} =\frac{T_2}{P_2}

Whereas solving for the final temperature T2, we get:

T_2=\frac{T_1P_2}{P_1}

Thus, we plug in the given data (temperature in Kelvins) to obtain:

T_2=\frac{(22+273.15)K*1.75atm}{3.50atm} \\\\T_2=147.58K-273.15\\\\T_2=-125.58\°C

Best regards!

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3 years ago
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