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Lisa [10]
3 years ago
7

Are elements considered to be substances?

Chemistry
1 answer:
Schach [20]3 years ago
7 0

Answer:

C. Yes, because they have a definite composition.

Explanation:

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>
You might be interested in
If 155 grams of potassium (K) reacts with 122 grams of potassium nitrate (KNO3), what is the limiting reagent?
GenaCL600 [577]
K:

m=155g
M=39g/mol

n = 155g / 39g/mol ≈ 3,97mol

KNO₃:

m=122g
M=101g/mol

n = 122g/101g/mol = 1,21mol

2K          +            10KNO₃  ⇒  6K₂O + N₂
2mol        :            10mol
3,97mol   :           1,21mol
                             limiting reagent

KNO₃ is limiting reagent

5 0
3 years ago
Why is water classified as a renewable resource?
mrs_skeptik [129]
The answer is a because the water we use it comes back again
6 0
3 years ago
Read 2 more answers
Why might a scientist repeat an experiment if he/she did not make a mistake in the first one? An experiment should be repeated t
makvit [3.9K]

The answer is C. To ensure the results are accurate.

5 0
3 years ago
Read 2 more answers
If a body with a mass of 4 kg is moved by a force of 20 N, what’s is the rate of acceleration ?
kaheart [24]

Answer:

The rate of acceleration is 5.

Explanation:

In order to calculate acceleration we need to divide the force by the mass.

Acceleration = net force/mass

In this case, it would be 20/4. Simplify that and we get 5.

6 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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