What are the energy transformations that occur in a cellphone that is plugged in? Is it an open or a closed system?

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muminat

1. Mg + CuSO4 —> MgSO4 + Cu

=> Single displacement reaction

2. 2HCI + Mg(OH)2 —> MgCI2 + H2O

=> Acid - base reaction

3.AgNO3 + NaCI —> AgCI + NaNO3

=> Double displacement reaction

8 0

2 years ago

Help please thanks you

scoundrel [369]

Explanation:

Ok so water is H2O and cabon dioxide in the air is CO2, so the water goes through the carbon dioxide and makes acid rain H2SO4.

Now we have our limestone which is CaCO3.

What happens is that the acid breaks apart our limestone into Ca2+ and CO3 2-. This then reforms into Calcium bicarbonate Ca(CO3)2.

Calcium bicarbonate is soluble in water and is hence washed away by the rain eroding the limestone.

5 0

2 years ago

After wood burns, the ash weighs much less than the original wood. explain why the law of conservation of mass is not violated i

Annette [7]

When wood burns it makes 1 ash and 2 CO and so law of conversation matter wouldn't be violated as the ash + the smoke each would have a fraction of the original matter from the wood

8 0

2 years ago

Read 2 more answers

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$