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sasho [114]
3 years ago
13

How much energy is required to raise the temperature of 10.6 grams of gaseous neon from

Chemistry
1 answer:
Alona [7]3 years ago
3 0

Answer:

Approximately 1.95 \times 10^{2}\; \rm J.

Explanation:

Look up the specific heat of gaseous neon:

c = 1.03 \; \rm J \cdot g^{-1} \cdot K^{-1}.

Calculate the required temperature change:

\Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K.

Let m denote the mass of a sample of specific heat C. Energy required to raise the temperature of this sample by \Delta T:

Q = c \cdot m \cdot \Delta T.

For the neon gas in this question:

  • c = 1.03\; \rm J \cdot g^{-1}\cdot K^{-1}.
  • m = 10.6\; \rm g.
  • \Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K.

Calculate the energy associated with this temperature change:

\begin{aligned}Q &= c \cdot m \cdot \Delta T \\ &= 1.03\; \rm J \cdot g^{-1}\cdot K^{-1} \times 10.6\; \rm g \times 17.9\; \rm K \\ &\approx 1.95 \times 10^{2}\; \rm J\end{aligned}.

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Which formula equation represents the burning of sulfur to produce sulfur dioxide?
Dmitriy789 [7]

Answer:

option A = S(s) + O₂(g)   →   SO₂ (s)

Explanation:

Chemical equation:

S(s) + O₂(g)   →   SO₂ (s)

when sulfur burned in the presence of oxygen it produce sulfur dioxide. The sulfur dioxide can further react with oxygen to produce sulfur trioxide and then react with water to form sulfuric acid.

Uses of sulfur dioxde:

It is used as a solvent and reagent in laboratory.

Sulfur dioxide is used to produce sulfuric acid.

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7 0
3 years ago
The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C
Vikentia [17]

Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})

1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})

T₂=371.77 K= 98.62 °C

5 0
3 years ago
Annabelle was explaining the carbon cycle to her friend. She said that all the carbon
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3 years ago
Provided is a diagram showing forces on a box.
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Answer:

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Explanation:

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