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nalin [4]
2 years ago
14

Mercury turns to a vapor at 629.88 K. How much heat is lost when 75.0 g of mercury vapor at 650 K condenses to a liquid at 297 K

?
Chemistry
1 answer:
notsponge [240]2 years ago
4 0

Answer:

- 3706.5 J.

Explanation:

The heat lost by the copper can be calculated using the formula:

Q = m.C.ΔT, where,

Q is the amount of heat lost by the mercury,

m is the mass of the copper <em>(m = 75.0 g)</em>,

C is the specific heat of Hg <em>(C = 0.140 J/g.°C)</em>,

ΔT is the temperature difference (the final temperature - the initial temperature<em> (ΔT = (297 K - 273) °C - (650 K - 273) °C = - 353 °C)</em>.

∴ The amount of heat lost by the mercury Q = m.C.ΔT = (75.0 g) (0.140 J/g.°C) (- 353 °C) = - 3706.5 J.

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5 0
3 years ago
What is the theoretical yield of methanol (CH3OH) when 12.0 grams of H2 is mixed with 74.5 grams of CO? CO + 2H2 CH3OH
AlladinOne [14]
1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2:    M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO:   M(CO) = 12.0 +16.0 = 28.0 g/mol

</span>12.0 g  H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO

<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover. 

                                  CO       +         2H2   ------->      CH3OH  
                                 1 mol              2 mol
given                        2.66 mol          6 mol (excess)

How much
we need  CO?           3 mol              6 mol

We see that H2 will be leftover, because for 6 moles H2  we need 3 moles CO, but we have only 2.66 mol  CO.
So, CO will react completely, and we are going to use CO to find  the mass of CH3OH.

3)                              </span>CO       +         2H2   ------->      CH3OH  
                                 1 mol                                        1 mol
                                2.66 mol                                    2.66 mol

4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 +  4*1.0 + 16.0 = 32.0 g/mol

2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH =  85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
3 0
3 years ago
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nikitadnepr [17]

Answer: The statement which could possibly not be true is C -" Liquid X can exist as a stable phase at 25°C, 1atm."

Explanation:

Triple point is the point where a substance co-exist as solid liquid and gas. At any point other than the triple point, the substance exist as a single phase substance.

As shown in the diagram, Liquid cannot exist as a stable phase at 1atm( below the the triple point pressure of 2atm) as the liquid can only exist beyond the pressure of triple point.

3 0
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