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Naya [18.7K]
2 years ago
14

Which of the following is not an example of potential energy? Group of answer choices electrical energy, chemical energy, gravit

ational energy, elastic energy.
PLS HELP FAST
Chemistry
1 answer:
Sholpan [36]2 years ago
6 0

Answer: The one listed below that's NOT an example of potential energy is mechanical energy. Mechanical energy is categorized as a kinetic energy with light, sound, and thermal/heat energy.

HOPE THIS HELPS

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One of the intermediates in the synthesis of glycine from ammonia, carbon dioxide, and methane is aminoacetonitrile, C2H4N2.
mart [117]

Answer:

  • <em>You could expect 3.48 grams of C₂H₄N₂</em>

Explanation:

You must start by stating the chemical equation for the reaction of ammonia, carbon dioxide, and methane to produce aminoaceto nitrile.

1. Word equation:

Ammonia + Carbon dioxide + Methane → Aminoacetonitrile + Water

2. Balanced chemical equation:

       8NH_3+5CO_2+3CH_4\rightarrow 4C_2H_4N_2 +10H_2O

3. Convert the mass of each reactant into number of moles:

<u>Formula:</u>

  • Number of moles = mass in grams/molar mass

<u>2.11g NH₃</u>

  • Number of moles = 2.11g / 17.03g/mol = 0.124 mol NH₃

<u>14.9g CO₂</u>

  • Number of moles = 14.9g/44.01g/mol = 0.339 mol CO₂

<u>1.75g CH₄</u>

  • Number of moles = 1.75g/16.04g/mol = 0.109 mol CH₄

4. Theoretical mol ratio

From the balanced chemical equation, using the coefficientes:

         8molNH_3:5molCO_2:3molCH_4:4molC_2H_4N_2:10molH_2O

5. Limiting reagent

The available amounts of the reactants are:

  • 0.124 mol NH₃
  • 0.339 mol CO₂
  • 0.109 mol CH₄

Fom the theoretical mole ration, to react with 0.124 mol of NH₃ you would need:

  • 0.124molNH₃ × (5molCO₂/8molNH₃) = 0.0775 mol CO₂

Since there are 0.339 moles available, this is in excess.

  • 0.124molNH₃ × (3molCH₄/8molNH₃) =  0.0465mol CO₂

Since there are 0.109 moles available, this is in excess too.

Hence, the limiting reagent is NH₃.

6. Yield

Use the theoretical ratio:

  • 0.124molNH₃ × (4molC₂H₄N₂ / 8molNH₃) = 0.0620 mol C₂H₄N₂

Convert to grams:

  • Mass = number of moles × molar mass
  • 0..0620 mol × 56.068g/mol = 3.48 g of C₂H₄N₂ ← answer

4 0
3 years ago
Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent
elixir [45]

Answer : The initial rate for a reaction will be 3.8\times 10^{-4}Ms^{-1}

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The chemical equation will be:

A+B+C\rightarrow P

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.1\times 10^{-5}=k(0.2)^a(0.2)^b(0.2)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.2)^a(0.2)^b(0.6)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.4)^a(0.2)^b(0.2)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.4)^a(0.4)^b(0.2)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.2)^a(0.2)^b(0.6)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.4)^a(0.2)^b(0.2)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.4)^a(0.4)^b(0.2)^c}{k(0.4)^a(0.2)^b(0.2)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.1\times 10^{-5}=k(0.2)^2(0.2)^0(0.2)^1

k=7.6\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.85 M of reagent A and 0.70 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.6\times 10^{-3})\times (0.85)^2(0.70)^0(0.70)^1

\text{Rate}=3.8\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.8\times 10^{-3}Ms^{-1}

6 0
3 years ago
The rate constant for a reaction is 4.65 L mol-1s-1. What is the overall order of the reaction? zero
il63 [147K]

Answer:

second order

Explanation:

units of reaction and their order.

Zero order --> M^1 s^-1 = M/s

First order --> M^0 s^-1 = 1/s

Second order --> M^-1 s^-1 = L/mol s

In the question rate constant k =  4.65 L mol-1s-1. = 4.65 L/mol s

Hence, the reaction is a second order reaction

8 0
2 years ago
Read 2 more answers
Wh
sergey [27]

The correct option would be 3.

Only thermal energy changes


Hope this helps you

Brainliest would be appreciated

-AaronWiseIsBae

5 0
2 years ago
Read 2 more answers
Some chemical changes can be ______________ through another _______________ change.
zalisa [80]
Some chemical changes can be reactive through another chemical change?
3 0
2 years ago
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