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Galina-37 [17]
3 years ago
8

The burning of a sample of propane

Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
7 0

Answer:

70.0°C

Explanation:

We are given;

  • Amount of heat generated by propane as 104.6 kJ or 104600 Joules
  • Mass of water is 500 g
  • Initial temperature as 20.0 ° C

We are required to determine the final temperature of water;

Taking the initial temperature is x°C

We know that the specific heat of water is 4.18 J/g°C

Quantity of heat = Mass × specific heat × change in temperature

In this case;

Change in temp =(x-20)° C

Therefore;

104600 J = 500 g × 4.18 J/g°C × (x-20)

104600 J = 2090x -41800

146400 = 2090 x

  x = 70.0479

     =70.0 °C

Thus, the final temperature of water is 70.0°C

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Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT
marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

(c) L(T, α) = 60·α·T - 9000·α + 60

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L =  L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

(c) L = L₀ + ΔL  = L₀ +  α × L₀ × ΔT = L₀ +  α × L₀ × (T - T₀)

Therefore;

L = 60 +  α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

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3 years ago
The sun's inner core is the hottest part of the sun<br><br> true or false
harkovskaia [24]
False heat rises upwards therefore the hottest part of the sun would be upwards
3 0
3 years ago
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A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68mg of the compound yields 16.01 mg CO2 and 4.37 mg H2O
Zielflug [23.3K]
<span>Molecular formulas tell you how many atoms of each element are in a compound, and empirical formulastell you the simplest or most reduced ratio of elements in a compound. ... Also, many compounds with different molecular formula have the same<span>empirical formula</span></span>
6 0
3 years ago
Which example is the site of reduction when the dry cell is operating
Simora [160]

Answer:

A battery contains electrochemical cells that can store chemical energy to be converted to electrical energy. A dry-cell battery stores energy in an immobilized electrolyte paste, which minimizes the need for water. Common examples of dry-cell batteries include zinc-carbon batteries and alkaline batteries.

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4 0
3 years ago
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A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical
Katarina [22]

Answer:

empirical formula: H_2SO_4

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02  :  1.02/1.02  :  4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: H_2SO_4.

7 0
3 years ago
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