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lakkis [162]
3 years ago
9

How many mililiters is 0.55 L

Chemistry
1 answer:
svp [43]3 years ago
7 0

Answer:

It's 550 milliliters.

Explanation:

I hope this helps.

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Which activity is least likely to increase greenhouse gases in the atmosphere?
DanielleElmas [232]

Answer:using fertilizers on crops

Explanation:carbon dioxide in the atmosphere is absorbed by plants.plants use it to produce their food. An increase in atmospheric carbon dioxide is caused by cutting down of forest and burning of fossil fuels.this increase in the emissions of carbon dioxide is harmful to the earth. Gases like chlorofluorocarbons ,carbon dioxide etc can cause global warming. This gases causes the heat not to escape from the earth atmosphere,just as it is in a green house. When this happens heat is trapped in the earth it causes global warming. These gases are released by burning of fuel, CFCs from old refrigerators and air conditioner.

Increase in the heat in the earth can cause melting of ice in the polar region.this can lead to a raise in sea level and flooding.

8 0
3 years ago
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Which state of matter expands when heated and is easy to compress? gas solid liquid
butalik [34]

Answer:

gas

Explanation:

The gas molecules are far apart and as such, when heated they gain more kinetic energy thereby colliding with one another and with the wall of the container thereby exerting pressure and also they can be compressed by reducing the distance between between the molecules.

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3 years ago
How much heat is absorbed when 70 grams of
mart [117]

Answer:

Thus, the joules of heat absorbed when 70.0 grams of water is completely vaporised at its boiling point are 158,200.

8 0
3 years ago
Translate the following chemical equation into a sentence. CH4+2O2 CO2+2H2O
muminat
One mole methane combusts to form one mole CO2 and 2 moles H2O
3 0
4 years ago
The pH of 0.50 M benzoic acid is 2.24. Calculate the change in pH when 2.64 g of C6H5COONa is added to 38 mL of 0.50 M benzoic a
meriva

Answer:

1.93

Explanation:

Moles of C_{6}H_{5}COOH = 38/1000 × 0.50 = 0.019mol

Moles of C_{6}H_{5}COONa = Mass/Molar mass = 2.64/144.10 = 0.018321mol

Final pH = pKa + log([C_{6}H_{5}COONa]/[C_{6}H_{5}COOH]

= -log Ka  + log(mols of C_{6}H_{5}COONa]/mols of C_{6}H_{5}COOH

= -log(6.5 × 10^(-5)) + log (0.018321/0.019)=4.17

change in pH = final - initial pH

= 4.17 - 2.24

=1.93

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4 years ago
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