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mel-nik [20]
3 years ago
12

Potassium chlorate (kclo3) decomposes in a reaction described by this chemical equation:2kclo3(s) → 2kcl(s) + 3o2(g)if 36.0 gram

s of potassium chlorate enter into this reaction, what is the total mass of the two products?
Chemistry
1 answer:
cricket20 [7]3 years ago
6 0
<span>If 36 gm of potassium chlorate enter into the reaction, the total mass of the two products will still be 36 gm because if there is only one reactant, the mass of the compounds after the reaction will be same that reactant based on the law of conservation of matter.</span>
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As you move to a higher elevation (e.g. in the mountains), the air pressure drops significantly. In turn,
soldi70 [24.7K]
The answer is C ( The evaporation rate of water increases at all temperatures causing an increase in volume.)                           
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2 years ago
A sample of gas in a cylinder as in the example in Part A has an initial volume of 48.0 L , and you have determined that it cont
masya89 [10]

Answer:

0.45 moles

Explanation:

The computation of the number of moles left in the cylinder is shown below:

As we know that

\frac{n1}{V1} = \frac{n2}{V2}

we can say that

n2 = n1 \times \frac{V2}{V1}

where,

n1 = 1.80 moles of gas

V2 = 12.0 L

And, the V1 = 48.0 L

Now placing these values to the above formula

So, the moles of gas in n2 left is

= 1.80 \times \frac{12.0\ L}{48.0\ L}

= 0.45 moles

We simply applied the above formulas so that the n2 moles of gas could arrive

5 0
3 years ago
PLEASE HELP
lozanna [386]

Answer:

The Answer is A

Explanation:

particles in gas much like any other particles vibrate. but gas is moving constantly so the answer would be A

5 0
2 years ago
The chemical reaction for the formation of syngas is: CH4 + H2O -&gt; CO + 3 H2 What is the rate for the formation of hydrogen,
grin007 [14]

Answer :  The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

CH_4+H_2O\rightarrow CO+3H_2

The expression for rate of reaction :

\text{Rate of disappearance of }CH_4=-\frac{d[CH_4]}{dt}

\text{Rate of disappearance of }H_2O=-\frac{d[H_2O]}{dt}

\text{Rate of formation of }CO=+\frac{d[CO]}{dt}

\text{Rate of formation of }H_2=+\frac{1}{3}\frac{d[H_2]}{dt}

The rate of reaction expression is:

\text{Rate of reaction}=-\frac{d[CH_4]}{dt}=-\frac{d[H_2O]}{dt}=+\frac{d[CO]}{dt}=+\frac{1}{3}\frac{d[H_2]}{dt}

As we are given that:

+\frac{d[CO]}{dt}=0.35M/s

Now we to determine the rate for the formation of hydrogen.

+\frac{1}{3}\frac{d[H_2]}{dt}=+\frac{d[CO]}{dt}

+\frac{1}{3}\frac{d[H_2]}{dt}=0.35M/s

\frac{d[H_2]}{dt}=3\times 0.35M/s

\frac{d[H_2]}{dt}=1.05M/s

Thus, the rate for the formation of hydrogen is, 1.05 M/s

6 0
3 years ago
a 0.5678 of KHP required 26.64cm³ of NaOH to complete neutralization.calculate the molarity of the NaOH solution​
vladimir2022 [97]

Answer:

Explanation:

0.5678 G        X GRAMS

KHC8H4O4 + NaOH = NaKC8H4O4 + H2O

1 MOL               1 MOL

0.5678G X 204G/MOL = 0.00278 MOL KHC8H4O4

0.00278 MOL KHC8H4O4 X 1 MOLE NaOH/1 MOLE  KHC8H4O4=0.00278 MOL NaOH

0.00278 MOL NaOH/26.26ml=0.106 molar

4 0
3 years ago
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