Which is an indicator of poor air quality? salt bacteria fertilizer smog

Meera added blue copper sulphate crystals to some water in a beaker.

professor190 [17]

Answer:

The solid crystals disappeared

Explanation:

When a soluble solid solute is added to water, the solid solute disappears after a little while. The disappearance of this solute indicates that the solute has been dissolved in water.

In this case, blue copper sulphate crystals are added to water, the blue crystals disappear leaving only a blue solution. The disappearance of these blue copper sulphate crystals indicates that the substance has dissolved in water.

7 0

3 years ago

Why is co2 collected by downward delivery in preparation

ozzi

Answer:

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5 0

3 years ago

How would you explain the Earth’s atmosphere?

CaHeK987 [17]

The atmosphere is a thin layer of gases that surrounds the Earth. It seals the planet and protects us from the vacuum of space. It protects us from electromagnetic radiation given off by the Sun and small objects flying through space such as meteoroids.

So it's like a blanket protecting us but it has air hole/patches where sunlight can get through

4 0

3 years ago

5. calculate the percentage of water of crystallization in blue vitriol (copper sulphate pentahydrate), cuso4.5h2o

ki77a [65]

The percentage of water of crystallization in blue vitriol is 36.07%.

M(H₂O) = 2Ar(H) + Ar(O) x g/mol

M(H₂O) = 2 + 16 x g/mol

M(H₂O) = 18 g/mol; molar mass of water

M(CuSO₄·5H₂O) = Ar(Cu) + Ar(S) + 4Ar(O) + 5Mr(H₂O) x g/mol

M(CuSO₄·5H₂O) = 63.5 + 32 + 64 + 90 x g/mol

M(CuSO₄·5H₂O) = 249.5 g/mol; molar mass of copper sulphate pentahydrate

The percentage of water: 5M(H₂O) / M(CuSO₄·5H₂O) x 100%

The percentage of water: 90 g/mol / 249.5 g/mol x 100%

The percentage of water = 36.07%

More about blue vitriol: brainly.com/question/8895853

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5 0

1 year ago

Read 2 more answers

For the following systems at equilibrium C: CaCO3(s) ⇌ CaO(s)+CO2(g) ΔH=+178 kJ/mol D: PCl3(g)+Cl2(g) ⇌ PCl5(g) ΔH=−88 kJ/mol cl

Rama09 [41]

Explanation:

C: $CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$ ΔH=+178 kJ/mol

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

$A+\text{heat}\rightleftharpoons B$

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Treat heat as a reactant and on increasing a reactant at equilibrium, shifts the reaction in the forward direction.

Increase temperature → increase in heat → forward direction

Decrease temperature → decease in heat → backward direction

System C - Increase temperature : Reaction will move forward

System C - Decrease temperature : Reaction will move backward

D: $PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$ ΔH=−88 kJ/mol

The total enthalpy of the reaction comes out to be negative .

The temperature of the surrounding will increase.

For an exothermic reaction, heat is released during a chemical reaction and is written on the product side.

$A\rightleftharpoons B+\text{ heat}$

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Treat heat as a product and on increasing a product at equilibrium, shifts the reaction in the backward direction.

Increase temperature → increase in heat → backward direction

Decrease temperature → decease in heat → forward direction

System D - Increase temperature : Reaction will move backward

System D - Decrease temperature : Reaction will move forward

7 0

3 years ago