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REY [17]
2 years ago
10

Given the following equation: Cu + 2 AgNO3 ---> Cu(NO3)2 + 2

Chemistry
1 answer:
Temka [501]2 years ago
8 0

Given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

<h3>How to calculate mass of substances?</h3>

The mass of a substance can be calculated using the following steps:

Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

1 mole of Cu react with 2 moles of AgNO3

  • Molar mass of AgNO3 = 169.87 g/mol
  • Molar mass of Cu = 63.5g/mol

moles of AgNO3 = 262g/169.87g/mol = 1.54mol

1.54 moles of AgNO3 will react with 0.77 moles of Cu.

mass of Cu = 0.77 × 63.5 = 48.97g

Therefore, given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

Learn more about mass at: brainly.com/question/6876669

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Your dad is working on creating a brick border for the lake in your backyard each brick has a mass of 100 g and a volume of 20 c
wel

Answer:

5000kg/m³

Explanation:

density=mass/volume

d=m/v

d=100/20

=5g/cm³

g/cm³*1000=kg/m³

5g/m³*1000=5000kg/m³

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3 years ago
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What causes a top to stop spinning?
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The following thermochemical equation is for the reaction of nitrogen(g) with oxygen(g) to form nitrogen dioxide(g). N2(g) + 2O2
Rzqust [24]

<u>Answer:</u> The mass of nitrogen gas reacted to produce given amount of energy is 5.99 grams.

<u>Explanation:</u>

The given chemical reaction follows:

N_2(g)+2O_2(g)\rightarrow 2NO_2(g);\Delta H=66.4kJ

We know that:

Molar mass of nitrogen gas = 28 g/mol

We are given:

Enthalpy change of the reaction = 14.2 kJ

To calculate the mass of nitrogen gas reacted, we use unitary method:

When enthalpy change of the reaction is 66.4 kJ, the mass of nitrogen gas reacted is 28 grams.

So, when enthalpy change of the reaction is 14.2 kJ, the mass of nitrogen gas reacted will be = \frac{28}{66.4}\times 14.2=5.99g

Hence, the mass of nitrogen gas reacted to produce given amount of energy is 5.99 grams.

8 0
4 years ago
How are the following aspects of a reaction affected by the addition of a catalyst? 1) activation energy of the reverse reaction
snow_tiger [21]

These are four questions and four answers.

Answers:

1) activation energy of the reverse reaction

     b. Decreased

2) Rate of the forward reaction

    a. Increased

3) Rate of the reverse reaction

    a. Increased

4) Activation energy of the forward reaction

    b. decreased

Explanation:

<em>Activarion energy</em> is the energy required by the reactants to form the intermediate transition state and become products.

<em>Catalysts</em> are substances that change the path of the chemical reactions, lowering the activation energy, and thus speeding up the rate of the reactions, since the products can reach the new lower activation energy faster.

The equilibrium reactions are the chemical process in which two reactions, the <em>forward and the reverese reactions</em>, occur simultaneously and at the same rate.  The equlibrium reactions may be represented by:

  • A ⇄ B

Where A → B is the direct or forward reaction, and A ← B is the reverse reaction (note the inversed arrow, from right to left).

For the direct reaction A represents the reactants and B represents the products. On the other hand, B represents the reactants and A represents the reactants of the reverse reaction and A. This, is A is the reactant of the forward reaction and the product of the reverse reaction, while B is the reactant of the reverse reaction and the product of the forward reaction.

Since, <em>the addition of a catalyst</em> lowers the activation energy of the process, the new activation energy is lower for both the forward and the reverse reaction, meaning that:

1. <em>The activation energy of the reverse reaction is decreased</em> (option b. of the first question)

2.<em> The rate of the forward reaction is increased</em> (option a. of the second question)

3. <em>The rate of the reverse reaction is increased</em> (option a. of the third question).

4. <em>Activation energy of the forward reaction is decreased</em> (option b. of the fourth question).

In summary, the addition of a catalyst decreases the activation energy for both forward and reverse reactions, and increases the rate of both forward and reverse reactions.

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