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SashulF [63]
3 years ago
8

why was the mass lost when the reaction was done in the normal setup but stayed the same when it was done in the gas collection

setup
Chemistry
1 answer:
german3 years ago
6 0
<h2>Answer : Law of conservation of mass</h2><h3>Explanation :</h3>

The law of conservation of mass states that in any reaction mass is neither created nor lost it has to remain constant in a system.

In this case, when the reaction setup was done in normal way the mass was  lost in surrounding was not considered nor being calculated; whereas when the reaction was studied in a closed system where the gas was collected after the reaction the mass changes was noted down which helped to prove the point of law of conservation of mass and energy.

One can consider an example of soda can where the carbonated drink contains pressurized carbon dioxide gas. when opened the gas bubbles gets lost into the surroundings and we don't measure the mass changes. Instead if the soda can was opened in such a way where the gas evolved was measured then the mass changed would remain the same.

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5 0
3 years ago
The observed rotation of 2.0 gg of a compound in 50 mLmL of solution in a polarimeter tube 25-cmcm long is +47.2 ∘∘ . What is th
Anastaziya [24]

Answer:

[∝] = +472

Explanation:

Specific rotation in a solution is defined as:

[∝] = ∝ / c×l

Where:

[∝] is specific rotation, ∝ is observed rotation (In degrees), c is concentration in g/mL and l is path length (In dm).

∝: +47.2°

c: 2.0g / 50mL = 0.04g/mL

l: 25cm × (1dm /10cm) = 2.5dm

Replacing:

[∝] = +47.2° / 0.04g/mL×2.5dm = <em>+472</em>

I hope it helps!

4 0
3 years ago
A gas is collected at 30 Celsius and has a pressure of 200 kPa. What pressure would it exert if the temperature changed to 40 Ce
garik1379 [7]

Answer:

206.6\ \text{kPa}

Explanation:

P_1 = Initial pressure = 200 kPa

P_2 = Final pressure

T_1 = Initial temperature = 30^{\circ}\text{C}

T_2 = Final temperature = 40^{\circ}\text{C}

We have the relation

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}\\\Rightarrow P_2=\dfrac{T_2}{T_1}P_1\\\Rightarrow P_2=\dfrac{40+273.15}{30+273.15}\times 200\\\Rightarrow P_2=206.6\ \text{kPa}

The pressure that would be exerted after the temperature change is 206.6\ \text{kPa}.

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