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zvonat [6]
2 years ago
9

At stp,a certain mass of gas occupies a volume of 790cm3. find the temperature at which the gas occupies 1000cm3 amd has a press

ure of 726mmHg​
Chemistry
1 answer:
Luden [163]2 years ago
3 0

The temperature : 332.75 K

<h3>Further explanation</h3>

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

volume of gas = 790 cm³ = 0.79 L

mol of gas at STP :

\tt \dfrac{0.79}{22.4}=0.035

Use ideal gas :

\tt PV=nRT

P=726 mmHg=0.955 atm

V= 1000 cm³ = 1 L

\tt T=\dfrac{PV}{nR}\\\\T=\dfrac{0.955\times 1}{0.035\times 0.082}\\\\T=332.75~K

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Three groups

gases, metals, metalliods/nonmetals
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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!

3 0
2 years ago
In the polypeptide Phe-Tyr-Glu-Asp-Ser-Ile-Leu-Ser what is the N-terminal amino acid?
avanturin [10]

Answer:

N-terminal Phe, C-terminal Ser

Explanation:

Amino acids connect like

NH2 -CH(R1) -CO -NH-CH(R2)-CO-.....-NH-CH(Rn)COOH

So, 1st amino acid is N -terminal , and it is Phe.

Last amino acid is C- terminal, and it is Ser.

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Which of the following is true of the Sun? *
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Read 2 more answers
A researcher studying the nutritional value of a new candy places a 4.90 g sample of the candy inside a bomb calorimeter and com
snow_lady [41]

Answer:

449730.879 cal/g

Explanation:

Given data:

Mass of sample = 4.9 g

Change in temperature  = 2.08 °C  (275.23 k)

Heat capacity of calorimeter = 33.50 KJ . K⁻¹

Solution:

C(candy) = Q/m

Q = C (calorimeter) × ΔT

C(candy) = C (calorimeter) × ΔT / m

C(candy) =  33.50 KJ . K⁻¹ × 275.23 K / 4.90 g

C(candy) = 9220.205 KJ / 4.90 g

C(candy) =  1881.674 KJ / g

It is known that,

1 KJ /g = 239.006 cal/g

1881.674 × 239.006 = 449730.879 cal/g

8 0
3 years ago
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