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worty [1.4K]
3 years ago
7

CAN SOMEBODY PLS HELP...

Chemistry
2 answers:
slavikrds [6]3 years ago
8 0
You could do something based on body heat or algae, or solar panels would be a good idea. maybe how much energy is used through solar panels
Aleonysh [2.5K]3 years ago
5 0

Answer:

Maybe you could do something with nuclear power plants and how we use those to create electrical energy for our phones, houses etc?

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However, when the temperature increases, the solubility of gaseous solutes will _______?
babunello [35]

Answer:

Decrease

Explanation:

Since the speed in which the gas molecules are faster as they are heated, they fly around in the container and logically, it is harder to insert a moving object into water than something more stationary or slower.

7 0
3 years ago
How many protons, neutrons, and electrons make up an ion of
yKpoI14uk [10]

Answer:

protons=15

electron=15

neutron=16

6 0
3 years ago
What is a safety guide line?
FromTheMoon [43]

Answer:

A safety guide lines are rules and regulations that are made to make sure place is safe and conducive for everyone.

5 0
3 years ago
How do you find the scientific notation of a Avogadro's number??
ruslelena [56]

Answer: ummmmmm hmmmmm

Explanation:

8 0
3 years ago
At the equilibrium point in the decomposition of phosphorus pentachloride to chlorine and phosphorus trichloride, the following
Hitman42 [59]

Answer: K_{eq} for the reaction is 5.55

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

                        PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)          

At eqm. conc.   (0.010) M     (0.15)  M   (0.37) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[Cl_2]\times [PCl_3]}{[PCl_5]}

Now put all the given values in this expression, we get :

K_c=\frac{(0.37)\times (0.15)}{(0.010)}

K_c=5.55

Thus the K_{eq} for the reaction is 5.55

5 0
3 years ago
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