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3 years ago

What comes after observations when scientific knowledge is gathered

Chemistry

1 answer:

Artist 52 [7]3 years ago

8 0

Going off of the scientific method, it would be Form a Question.

Hope this helped!

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Chromium (Cr) can combine with Chlorine (Cl2) to form Chromium chloride (CrCl3). Which equation is a correction representation o

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Balance the reaction so the same number of each type of atom is on each side. Which option seems balanced to you?

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3 years ago

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Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent

elixir [45]

Answer : The initial rate for a reaction will be $3.8\times 10^{-4}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The chemical equation will be:

$A+B+C\rightarrow P$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.1\times 10^{-5}=k(0.2)^a(0.2)^b(0.2)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.2)^a(0.2)^b(0.6)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.4)^a(0.2)^b(0.2)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.4)^a(0.4)^b(0.2)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.2)^a(0.2)^b(0.6)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.4)^a(0.2)^b(0.2)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.4)^a(0.4)^b(0.2)^c}{k(0.4)^a(0.2)^b(0.2)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.1\times 10^{-5}=k(0.2)^2(0.2)^0(0.2)^1$

$k=7.6\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.85 M of reagent A and 0.70 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.6\times 10^{-3})\times (0.85)^2(0.70)^0(0.70)^1$

$\text{Rate}=3.8\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.8\times 10^{-3}Ms^{-1}$

6 0

3 years ago

How many 325 mg aspirin tablets can be made from 875 g of aspirin?

oksano4ka [1.4K]

Answer: 35/13

Explanation:

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1 year ago

To eight significant figures, avogadro's constant is 6.0221367×1023mol−1. which of the following choices demonstrates correct ro

FrozenT [24]

6.0221367 rounded from the last digit would be 6.022137 but when using it calculations, you should just use 6.022 x 10^23.

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3 years ago

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How does an increase in temperature affect the equilibrium concentration of the underlined substance and K for each of the follo

Darya [45]

Explanation:

Due to the positive value of the change in temperature, this is an endothermic reaction.

Since the forward reaction is endothermic, increasing the temperature increases the equilibrium constant (k).

In an equilibrium system, the position of the equilibrium will move in a way to annul the change made to the system. An increase in temperature for an endothermic reaction would favour the reaction, leading to increase in amount of products and decrease in amount of reactants.

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3 years ago