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madam [21]
3 years ago
8

Answer the following questions for this equation: 2 H2 + O2 --> 2 H2O

Chemistry
1 answer:
laiz [17]3 years ago
5 0

Answer:

                       1 Mole O₂ / 2 Mole H₂O

Explanation:

                  The balance chemical equation for the synthesis of water is as follow,

                                             2 H₂ + O₂ → 2 H₂O

In this equation the moles are specified by the coefficients i.e.

2 moles of H₂ reacting with 1 mole of O₂

2 moles of H₂ producing 2 moles of H₂O

1 mole of O₂ producing 2 moles of H₂O

Hence the molar ratio for O₂ and H₂O can be written as,

                            1 Mole O₂ : 2 Mole H₂O

or,

                           1 Mole O₂ / 2 Mole H₂O

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Complete these metric conversions: 53 m - mm*
EleoNora [17]

Answer:

53 meters = 53000 millimeters

Explanation:

In this question we have to convert meters into millimeters .

By metric conversion,

Since, 1 meter = 1000 mm

Therefore, 53 meters = 53 × 1000

                                  = 53000 millimeters

53 meters = 53000 millimeters is the answer.

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3 years ago
5. How many molecules are in 0.550 moles of N2O5?
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What is the compound name to C5H6?
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3 years ago
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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
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pashok25 [27]
I got B. Because the answer is K2SO4, I got it right on my test so I know its right 

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