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Blizzard [7]
3 years ago
15

What is the molar mass of the EF?

Chemistry
2 answers:
inysia [295]3 years ago
8 0

Answer:

47.778

Explanation:

Kisachek [45]3 years ago
7 0

Answer:

97.11 should be correct

explanation

EF = C4H5ON2

molar mass= 4 (12.01) + 5 (1.9) +16 + 2 (14.01) = 97.11 glmol

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Which type of ion usually undergoes reduction?
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Answer:

anion

Explanation:

anion undergoes reduction.

cation undergoes oxidation.

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3 years ago
At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s
Shalnov [3]

Answer : The time taken for the reaction is, 28 s.

Explanation :

Expression for rate law for first order kinetics is given by :

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant  = 0.0632

t = time taken for the process  = ?

[A_o] = initial amount or concentration of the reactant  = 1.28 M

[A] = amount or concentration left time 't' = 1.28\times \frac{17}{100}=0.2176M

Now put all the given values in above equation, we get:

0.0632=\frac{2.303}{t}\log\frac{1.28}{0.2176}

t=28s

Therefore, the time taken for the reaction is, 28 s.

8 0
3 years ago
What is radiation exactly
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6 0
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Read 2 more answers
The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
3 years ago
Identify areas of Job growth?
kumpel [21]

Answer:

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Explanation:

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3 years ago
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