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Pepsi [2]
4 years ago
13

4. Evaluate the information contained in this calculation and complete the

Chemistry
1 answer:
Neporo4naja [7]4 years ago
7 0

Since you did not give a calculation, I will just give an example. Suppose that you are to burn 5 kg of methane (CH4) from 0 to 10°C. The specific heat capacity of methane is 4.475 kJ/kg-K.

H = mCpT

H = (5kg)( 4.475 kJ/kg-K)(10-0)

H = 223.75 kJ

Because the enthalpy is positive in value, methane takes in heat. 

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A rectangle has a length of 5.50m and a width of 12.0m. What is the area of this rectangle?​
jolli1 [7]

Answer:

66m

Explanation:

To get the area of something you multiple the length (5.5) by the width (12) together. So the problem would look like 5.5×12 and if you multipe that you get 66

3 0
3 years ago
Sulfur dioxide has a vapor pressure of 462.7 mm Hg at –21.0 °C and a vapor pressure of 140.5 mm Hg at –44.0 °C. What is the enth
stealth61 [152]

Answer : The value of \Delta H_{vap} is 28.97 kJ/mol

Explanation :

To calculate \Delta H_{vap} of the reaction, we use clausius claypron equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = vapor pressure at temperature -21.0^oC = 462.7 mmHg

P_2 = vapor pressure at temperature -44.0^oC = 140.5 mmHg

\Delta H_{vap} = Enthalpy of vaporization = ?

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = -21.0^oC=[-21.0+273]K=252K

T_2 = final temperature = 45^oC=[-41.0+273]K=232K

Putting values in above equation, we get:

\ln(\frac{140.5mmHg}{462.7mmHg})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{252}-\frac{1}{232}]\\\\\Delta H_{vap}=28966.6J/mol=28.97kJ/mol

Therefore, the value of \Delta H_{vap} is 28.97 kJ/mol

4 0
3 years ago
Which of the following is a negatively charged particle?
kirza4 [7]

Answer:

B. electron

Explanation:

B. electron

electron is a negatively charged particle

proton is a positively charged particle

7 0
3 years ago
True or False: A theory is something that can never be proven correct or incorrect?
cestrela7 [59]

Answer:true

true

Explanation:

8 0
3 years ago
What is the pressure of 0.5 mol of nitrogen gas in a 5 L container at 203 K
vlada-n [284]

Answer:

=1.666 liters

Explanation:

1 mole of a has at standard temperature and pressure occupies a volume of 22.4 liters.

0.5 moles of nitrogen occupy a volume of (0.5 moles×22.4 dm³/mol)/ 1

=11.2 liters.

Standard pressure= 1 atmosphere (Atm)

Standard temperature = 273.15 Kelvin

According to Combined gas equation, P₁V₁/T₁=P₂V₂/T₂

Let us take the conditions under standard conditions as the reference, with the subscript 1 and the conditions under the 5L container to be scenario 2 with subscript 2.

Therefore P₂ =P₁V₁T₂/T₁V₂

Substituting for the values we get:

P₂= (1 atm× 11.2L ×203K)/ (273K×5L)

=1.666 atm

5 0
3 years ago
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