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Aleks [24]
3 years ago
6

If a gas furnace releases 420 kcal of heat energy, what is the energy in kilojoules? (1 cal = 4.184 J)

Chemistry
1 answer:
Igoryamba3 years ago
8 0
Answer:
The furnace releases 1757280 J 

Explanation:
We will do the conversion on two steps:
1- convert the kcal to cal
2- convert the cal to J

Step 1: converting kcal to cal
1 kcal is equivalent to 1000 cal. Therefore:
420 kcal is equivalent to 420*1000 = 420000 cal

Step 2: converting cal to J
We are given that:
<span>1 cal = 4.184 J
</span>Therefore:
420000 cal is equivalent to 420000 * 4.184 = 1757280 J 

Hope this helps :)
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therefore  for  HF is
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Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a
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Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

  • Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT
  • Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g
  • Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT
  • Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g
  • Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) <u>THIS IS THE RIGHT ANSWER</u> (the more negative= the more released)

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How many grams of solid nickel will be plated if an aqueous nickel(II) sulfate solution is electroplated over 10.5 min with a co
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Answer:

613 mg

Explanation:

$Ni^{2+} + 2e \rightarrow Ni(s)$

Number of fargday's  $=\frac{It}{96500}$

Here, I = 9.20 A

        t = 10.5 min

          = 10.5 x 60 seconds

So, $\frac{It}{96500}$

  $=\frac{3.20 \times 10.5 \times 60}{96500}$

 = 0.0208 F

Here, 2e, 2F

2F = 1 mol of Ni

$0.0208 \ F = \frac{0.0208}{2} = 0.0104 \ mol \ Ni$

1 mol = 59 gm of Ni

0.0104 mol = 59 x0.0104 gm Ni

                    = 0.613 gm Ni

                    = (0.613 x 1000 ) mg of Ni

                    = 613 mg of Ni

3 0
3 years ago
Uranium has atomic number 92. Its most common isotope is 238U, but the form used in nuclear bombs and nuclear power plants is 23
Yakvenalex [24]

Explanation:

Atomic Number = Number of protons

Mass Number = Number of protons + Number of neutrons

Isotopes are simply atoms of an element with the same number of protons and different number of neutrons.

First Isotope -- 238U

Number of neutrons = Mass Number - Atomic Number

Number of neutrons =  238 - 92 = 146

Second Isotope -- 235U

Number of neutrons = Mass Number - Atomic Number

Number of neutrons =  235 - 92 = 143

3 0
3 years ago
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