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Anna [14]
3 years ago
8

In an experiment, equal amounts of water and soil were first heated and then left to cool. The graph shows the change in tempera

ture for each with time.
A graph with the title Change in Temperature is shown. The y axis title is Temperature in degree C and the x axis title is Time in seconds. An inverted U shaped curve labeled A is plotted. A flatter curved line labeled B is also shown plotted on the graph.

Which statement about the curves A and B is true?

Curve A represents how soil takes longer to heat and longer to cool.
Curve B represents how water takes longer to heat and longer to cool.
Curve A represents water as it heats rapidly and cools slowly.
Curve B represents soil as it heats slowly and cools rapidly.

If you don't understand, here's the image.

Chemistry
2 answers:
poizon [28]3 years ago
5 0

Answer:

ITS A

Explanation:

im taking same quiz lol and its correct <3 have a nice day UwU

AveGali [126]3 years ago
3 0

Answer: A

Explanation:

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1 year ago
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For the chemical equation SO 2 ( g ) + NO 2 ( g ) − ⇀ ↽ − SO 3 ( g ) + NO ( g ) the equilibrium constant at a certain temperatur
MakcuM [25]

Answer:

Moles of NO₂ = 0.158

Explanation:

                         SO 2 ( g ) + NO 2 ( g ) ⇄  SO 3 ( g ) + NO ( g )

              According to the law of mass equation

                                     

                                       K_{c} = \frac{[SO_{3} ][NO]}{[SO_{2}][NO_{2}  ]}

                              ⇒   3.10 = \frac{(1.00)(1.00)}{(2.30) [NO_{2} ]}    At equilibrium [SO₃] = [NO]

                              ⇒ [NO₂] = \frac{1}{6.3}

                              ⇒ [NO₂] = 0.158

So. number of moles of NO₂ at equilibrium added = 0.158

8 0
3 years ago
The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?
n200080 [17]

<u>Answer:</u> The moles of oxygen and carbon dioxide in air is 3.63\times 10^{19}mol and 7.18\times 10^{16}mol respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of atmosphere = 5.00\times 10^{18}kg=5.00\times 10^{21}g

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = \frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol

Moles of carbon dioxide in air = \frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol

Hence, the moles of oxygen and carbon dioxide in air is 3.63\times 10^{19}mol and 7.18\times 10^{16}mol respectively

6 0
3 years ago
At 350 k, kc = 0.142 for the reaction 2 brcl(g) ⇀↽ br2(g) + cl2(g) an equilibrium mixture at this temperature contains equal con
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