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Naddik [55]
3 years ago
9

Why is it not advisable to wear black clothes in the afternoon​

Chemistry
1 answer:
Nimfa-mama [501]3 years ago
3 0
Uhhh, well if the sun is out there is something called blackbody radiation. Wearing black clothes could absorb the sunlight just like solar panels on a house. ( black plates placed on houses) Plus you could get a heat stroke but its pretty rare.
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During the hot summer days, there is a lot more water in the air. This is due to which change of state?
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With the evaporation
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When a car engine burns gasoline, the results of the reaction are similar to when cells burn glucose. both reactions release car
-Dominant- [34]

the answer is heat. while a car is in idol, the tailpipe gets very hot, (motorcycle, car, etc.) this also produces h20 which you can see dripping out of the tailpipe.

3 0
3 years ago
How do the lithosphere and the asthenosphere work together?
miskamm [114]
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2 years ago
Calculate the mass of H2OH2O produced by metabolism of 2.4 kgkg of fat, assuming the fat consists entirely of tristearin (C57H11
otez555 [7]

Answer:

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

Explanation:

2C_{57}H_{110}O_6+163O_2\rightarrow 114CO_2+110H_2O

Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g

1 kg = 1000 g

Molar mass of fat = M

M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]

Moles of fat = \frac{2400 g}{890 g/mol}=2.6966 mol

According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give  ;

\frac{110}{2}\times 2.6966 mol=148.31 mol of water

Mass of 148.31 moles of water ;

148.31 mol × 18 g/mol = 2,669.58 g

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

8 0
3 years ago
A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0
dimaraw [331]

Explanation:

It is known that for high concentration of M^{2+}, reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.

Now, E^{o}_{cell} = 0 and the general reaction equation is as follows.

         M^{2+} + M \rightarrow M + M^{2+}

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         E = 0 - \frac{0.0591}{2} log \frac{50 \times 10^{-3}}{1}

            = -\frac{0.0591}{2} log (5 \times 10^{-2})

            = 0.038 V

Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.

5 0
3 years ago
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