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siniylev [52]
3 years ago
10

If the Moon rises at 7 P.M. on a particular day, then approximately what time will it rise six days later?

Chemistry
1 answer:
OLga [1]3 years ago
5 0

Answer:

below

Explanation:

28th 10;24 am

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Which process is endothermic?a)b)c)d)The evaporation of water from the skin.The burning of candle wax.The oxidation of iron in a
BlackZzzverrR [31]

Answer:

The answer is option a. The evaporation of water from the skin

Explanation:

An endothermic reaction is any synthetic reaction that retains heat from its environment. The ingested energy gives the enactment energy to the reaction to happen.

No new bonds formed, no bonds broken in this example. water molecules go from liquid to vapour (gas ), intermolecular bond exist between polar (-OH) molecules and energy is required to break these bonds during the change from liquid to gas. So, it is endothermic .

Remaining three above examples are exothermic. combustion of gas releases energy(exothermic). burninig of the candle exothermic because the energy released from the oxygen into carbon dioxide and water.oxidation of iron also exothermic

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How much NaNO3 can I dissolve in 400 grams of water at 283 K?
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Which of the following is a chemical property? a
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The answer is A (The ability to react with oxygen)
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Stars, such as our sun, use fusion to combine hydrogen atoms into helium atoms, and in the process, create energy. As massive st
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4 0
3 years ago
1. A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0°C to 35.5°C. If the
Maurinko [17]

Answer:

Mass of aluminium in sample = 3.591 g ≅ 3.6 grams

Explanation:

Given that,  A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0°C to 35.5°C.

the specific heat of aluminum is 0.900 J/g- °C

The relation between heat absorbed and change in temperature is given by,   Q = msΔT.

where Q = heat absorbed

            m = mass of the substance

            s = specific heat of substance

          ΔT  = change in temperature

Now, in our case, Q = 50.1 J ; s = 0.900 J/g- °C; ΔT= 35.5-20 = 15.5°C

⇒ m =  \frac{Q}{s(T_{2} -T_{1}) }

⇒ m = \frac{50.1}{0.900(15.5)} = 3.591 g ≅ 3.6 g

⇒ m ≅ 3.6 g

5 0
3 years ago
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