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slavikrds [6]
3 years ago
14

Which of the following terms refers to areas of gas on the suns surface that are cooler than the gases around them?

Chemistry
2 answers:
solniwko [45]3 years ago
8 0
The answer is B sunspots
arlik [135]3 years ago
3 0
B: sunspots :) hope I helped
You might be interested in
What is the pressure of 2.20 mol of a gas stored in a 4.5 L container at a temperature of 170 K? (Use PV=nRT and R = 8.314 L∙kPa
algol13

Answer:

I don't know how can i do

Explanation:

please give me hint

7 0
2 years ago
An 80.0-gram sample of water at 10.0°C absorbs 1680 Joules of heat energy. What is the final temperature of the water? a 50.0°C
ICE Princess25 [194]

Answer:

b)15.0°C

Explanation:

Specific Heat of Water=4.2 J/g°C

This means, that 1 g of Water will take 4.2 J of energy to increase its temperature by 1°C.

∴80 g Water will take 80×4.2 J of energy to increase its temperature by 1°C.

80×4.2 J=336 J

Total Energy Provided=1680 J

The temperature increase=\frac{\textrm{Total energy required}}{\textrm{energy required to increase temperature by one degree}}

Temperature increase=\frac{1680}{336}

=5°C

Initial Temperature =10°C

Final Temperature=Initial + Increase in Temperature

=10+5=15°C

7 0
3 years ago
A 7.47 g sample of calcium carbonate [CaCO3 (s)] absorbs 85 J of heat, upon which the temperature of the sample increases from 2
Alex_Xolod [135]

Answer:

Specific heat of calcium carbonate(C) = 0.82 (Approx)

Explanation:

Given:

Energy absorbs (q) = 85 J

Change in temperature (Δt) = 34.9 - 21 = 13.9°C  

Mass of calcium carbonate = 7.47 g

Find:

Specific heat of calcium carbonate(C)

Computation:

Specific heat of calcium carbonate(C) = q / m(Δt)

Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)

Specific heat of calcium carbonate(C) = 85 / 103.833

Specific heat of calcium carbonate(C) = 0.8186

Specific heat of calcium carbonate(C) = 0.82 (Approx)

7 0
2 years ago
How much energy (kJ) is required to change 0.18 mole of ice (s) at 0 C to water (l) at 0 C?
Dmitriy789 [7]

Answer:25,06 kJ of energy must be added to a 75 g block of ice.

ΔHfusion(H₂O) = 6,01 kJ/mol.

T(H₂O) = 0°C.

m(H₂O) = 75 g.

n(H₂O) = m(H₂O) ÷ M(H₂O).

n(H₂O) = 75 g ÷ 18 g/mol.

n(H₂O) = 4,17 mol.

Q = ΔHfusion(H₂O) · n(H₂O)

Q = 6,01 kJ/mol · 4,17 mol

Q = 25,06 kJ.

Explanation:

6 0
2 years ago
If a mass of a cube were 48.0g and its volume 24.0 cm3 what would its density be
EastWind [94]

Answer:

<h3>The answer is 2 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass = 48 g

volume = 24 cm³

We have

density =  \frac{48}{24}  \\

We have the final answer as

<h3>2 g/cm³</h3>

Hope this helps you

5 0
3 years ago
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