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yulyashka [42]
3 years ago
8

3 H2 (g) + N2 (g) 2 NH3 (g)

Chemistry
1 answer:
ale4655 [162]3 years ago
4 0

Answer:

Mass = 0.697 g

Explanation:

Given data:

Volume of hydrogen = 1.36 L

Mass of ammonia produced = ?

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Solution:

Chemical equation:

3H₂ + N₂       →      2NH₃

First of all we will calculate the number of moles of hydrogen:

PV  = nRT

R = general gas constant = 0.0821 atm.L/mol.K

1atm ×1.36 L = n × 0.0821 atm.L/mol.K × 273.15 K

1.36 atm.L = n × 22.43 atm.L/mol

n = 1.36 atm.L / 22.43 atm.L/mol

n = 0.061 mol

Now we will compare the moles of hydrogen and ammonia:

                 H₂         :          NH₃

                  3          :            2

                0.061     :         2/3×0.061 = 0.041

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.041 mol × 17 g/mol

Mass = 0.697 g

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olga_2 [115]
D=m/v

so,

d=8.0g/4.0cm

d=2

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3 years ago
Which macromolecules break apart by hydrolysis?
ankoles [38]

Answer:

Dehydration synthesis reactions build molecules up and generally require energy, while hydrolysis reactions break molecules down and generally release energy. Carbohydrates, proteins, and nucleic acids are built up and broken down via these types of reactions, although the monomers involved are different in each case.

Explanation:

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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
A metal oxide with the formula mo contains 15.44% oxygen. in the box below, type the symbol for the element represented by m.
lana66690 [7]

<u>Answer:</u> The element represented by M is Strontium.

<u>Explanation:</u>

Let us consider the molar mass of metal be 'x'.

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It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:

\frac{15.44}{100}\times (x+16)=16g/mol\\\\(x+16)=\frac{16g/mol\times 100}{15.44}\\\\x=(103.626-16)g/mol\\\\x=87.62g/mol

The metal atom having molar mass as 87.62/mol is Strontium.

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8 0
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Answer:

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Explanation:

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