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3 years ago

How do photosynthesis and cellular respiration compare to each other

Chemistry

1 answer:

musickatia [10]3 years ago

4 0

Answer:

Photosynthesis takes place in the chloroplasts of cells. ... While photosynthesis requires energy and produces food, cellular respiration breaks down food and releases energy. Plants perform both photosynthesis and respiration, while animals can only perform respiration.

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A balloon is filled with 100.0 Lair at 1.300 atm pressure. what pressure is needed to change the volume to 25.00 L

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Answer:

4 atmospheric pressure is needed to get a volume of 25 litres

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2 years ago

The goal of this experiment is to separate a mixture of two unknown compounds into individual components and to identify the com

kolbaska11 [484]

Answer: Ethyl Ethanoate can be used as a developing solvent. It’s safer.

Explanation:Di ethyl ether should be carefully used because it’s highly flammable and intoxicating when inhaled and can cause explosions because of its high reactivity to air and light.

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2 years ago

Which of the following will produce a precipitate?

nekit [7.7K]

Reaction of option c produces precipitate.

Rhodium on reacting with potassium phosphate produces rhodium phosphate which remain in solution due to low lattice energy for rhodium phosphate.

Niobium on reacting with lithium carbonate produces niobium carbonate and it will remain in aqueous form.

Cobalt on reacting with zinc nitrate produces cobalt nitrate. This, Co(NO3 )2 is insoluble precipitate and settles at bottom whereas zinc ion will remain in solution as follows:

$Co^2^+ + Zn(NO¬_3 )¬_2 Co(NO¬_3 )¬_2 + Zn^2^+$

Potassium ion on reacting with sodium sulfide produces potassium sulfide which remain in solution

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2 years ago

A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate

inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

water content of sample is taken as;

w = (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 / 1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

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3 years ago

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