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Tatiana [17]
3 years ago
7

what+is+the+theoretical+yield+of+ammonia(in+grams)+of+19.25+grams+of+nitrogen+gas+and+11.35+grams+of+hydrogen+gas+are+allowed+to

+react
Chemistry
1 answer:
kondor19780726 [428]3 years ago
7 0
We are given the amount of Nitrogen gas and hydrogen gas reacted to form ammonia:

N2 = 19.25 grams
H2 = 11.35 grams

Set-up a balanced chemical equation:

N2 + 3H2 ==> 2NH3

The theoretical amount of ammonia that will be produced from the given amounts is:

First, we need to determine the limiting reactant to serve as our basis for calculation.

number of moles / stoichiometric ratio 

N2 = 19.25  g/ 28 g/mol / 1 = 0.6875
H2 = 11.35 g/ 2 g/mol /3 = 1.89 

The limiting reactant is N2.

0.6875 moles N2 * (2 NH3/ 1 N2) * 17 g/mol NH3

The amount of NH3 produced is 23.375 grams of ammonia. <span />
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Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the following reaction equation. 2H202(aq)
inn [45]

Answer : The temperature for non-catalyzed reaction needed will be 456 K

Explanation :

Activation energy : The energy required to initiate the reaction is known as activation energy.

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

Since, the rate for both the reaction are equal.

K_1=K_2

A\times e^{\frac{-Ea_1}{RT_1}}=A\times e^{\frac{-Ea_2}{RT_2}}

\frac{Ea_1}{T_1}=\frac{Ea_2}{T_2} ..........(1)

where,

Ea_1 = activation energy for non-catalyzed reaction = 75 kJ/mol

Ea_2 = activation energy for catalyzed reaction = 49 kJ/mol

T_1 =  temperature for non-catalyzed reaction = ?

T_2 = temperature for catalyzed reaction = 25^oC=273+25=298K

Now put all the given values in the above formula 1, we get:

\frac{Ea_1}{T_1}=\frac{Ea_2}{T_2}

\frac{75kJ/mol}{T_1}=\frac{49kJ/mol}{298K}

T_1=456K

Therefore, the temperature for non-catalyzed reaction needed will be 456 K

3 0
3 years ago
What the molar mass of carbon2hydrogen6
Alborosie
I’m pretty sure it’s 30.06904
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3 years ago
Calculate the vapor pressure of spherical water droplets of radius (a) 17 nm and (b) 2.0 μm surrounded by water vapor at 298 K.
Montano1993 [528]

Explanation:

Relation between pressure of water and its droplet is as follows.

           ln (\frac{p}{p_{o}}) = \frac{2 \gamma M}{r \rho RT}

where,   p = pressure of droplet

          p_{o} = water pressure in given temperature

          \gamma = 7.99 \times 10^{-3}

           M = Molecular Weight in Kg/Mol (0.018 for water)

            r = radius in meters

     \rho = density of water in Kg/m^{3} (1000 kg/m^{3})

           R = ideal gas constant (8.31)

           T = temperature in Kelvin

(a)   We will calculate the value of p as follows.

           p = e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}

              = e^{\frac{2 \times 0.07199 \times 0.018}{1.7 \times 10^{-8} \times 1000 \times 8.31 \times 298 K} \times 25.2

              = 26.8 torr

(b)  And, vapor pressure of spherical water droplets of radius 2.0 \mu m or 2 \times 10^{-6} m

             p = e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}

              = e^{\frac{2 \times 0.07199 \times 0.018}{2 \times 10^{-6} \times 1000 \times 8.31 \times 298 K} \times 25.2

              = 25.2 torr

7 0
4 years ago
Given the reaction that occurs in an electrochemical cell: Zn(s) + CuSO4(aq) --&gt; ZnSO4(aq) + Cu(s) During this reaction, the
liq [111]
<span>Zn⁰(s) + CuSO4(aq) --> Zn²⁺SO4²⁻(aq) + Cu(s)

</span>Zn⁰ --- in free elements oxidation number is 0.
Zn²⁺SO4²⁻ ---- ionic compound , it has ion Zn²⁺, so oxidation number here is +2.
<span>The oxidation number of Zn changes from 0 to +2.</span>
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What are other ways you can use wind energy to create electricity?
Phantasy [73]
You can use a fan on the wire. That's one way.
8 0
3 years ago
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