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weqwewe [10]
3 years ago
15

Hhhhhhheeeeeeelllllllppppppppppp

Chemistry
2 answers:
Ymorist [56]3 years ago
5 0

Answer:

c

Explanation:

zimovet [89]3 years ago
3 0

Answer:

the answer is C hope I helped

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1. How many molecules of S2 gas are in 756.2 L?
AfilCa [17]

Answer: There are 2.032 \times 10^{25} molecules S_{2} gas are in 756.2 L.

Explanation:

It is known that 1 mole of any gas equals 22.4 L at STP. Hence, number of  moles present in 756.2 L are calculated as follows.

Mole = \frac{Volume}{22.4 L}\\= \frac{756.2 L}{22.4 L}\\= 33.76 mol

According to mole concept, 1 mole of every substance contains 6.022 \times 10^{23} molecules.

Therefore, molecules of S present in 33.76 moles are calculated as follows.

1 mol = 6.022 \times 10^{23}\\33.76 mol = 33.76 \times 6.022 \times 10^{23}\\= 2.032 \times 10^{25}

Thus, we can conclude that there are 2.032 \times 10^{25} molecules S_{2} gas are in 756.2 L.

5 0
3 years ago
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
4 years ago
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