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ad-work [718]
3 years ago
8

A coyote and an oak tree both have DNA?

Chemistry
2 answers:
denis-greek [22]3 years ago
7 0

Answer:

true, they both have different types of DNA.

PtichkaEL [24]3 years ago
6 0

Answer:

Um yes... they have different DNA's.. ;-;

Explanation:

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the answer is C. 3. measure the mass of the liquid in both beakers

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What is the subscript of Carbon? 6 CaCO3
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subscript is 3

Explanation:

the  subscript is the number that is slightly lower than a # which in this case it's O

this indicates that there's 3 oxygen atoms

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Temperature is the measurement of how hot or cold something is. Agree Disagree
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What type of bond does carbon and sulphur form when they react. explain​
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For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The
Alik [6]

Answer : The rate for trial 5 will be 4.25\times 10^{-2}Ms^{-1}

Explanation :

Rate law 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.

For the given chemical equation:

A+B+C\rightarrow D+E

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:

6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^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:

6.0\times 10^{-4}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-2}M^{-2}s^{-1}

Thus, the value of the rate constant 'k' for this reaction is 7.5\times 10^{-2}M^{-2}s^{-1}

Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.

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

\text{Rate}=(7.5\times 10^{-2})\times (0.90)^2(0.60)^0(0.70)^1

\text{Rate}=4.25\times 10^{-2}Ms^{-1}

Therefore, the rate for trial 5 will be 4.25\times 10^{-2}Ms^{-1}

3 0
3 years ago
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