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Sav [38]
3 years ago
14

How many moles of NaCl will be produced from 83.0g of Na, assuming Cl2 is available in excess?

Chemistry
2 answers:
SashulF [63]3 years ago
5 0

Answer:

2.57

Explanation:

Sindrei [870]3 years ago
3 0

Answer: The number of moles of NaCl produced will be, 3.61 moles.

Explanation : Given,

Mass of Na = 83.0 g

Molar mass of Na = 23 g/mol

First we have to calculate the moles of Na

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

\text{Moles of }Na=\frac{83.0g}{23g/mol}=3.61mol

Now we have to calculate the moles of NaCl

The balanced chemical equation is:

2Na+Cl_2\rightarrow 2NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 3.61 mole of Na react to give 3.61 mole of NaCl

Therefore, the number of moles of NaCl produced will be, 3.61 moles.

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The action or process of making a copy of something.
3 0
3 years ago
Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,
7nadin3 [17]

Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 )   N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)  ; Kc = 1.54e - 31

2)   N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

     N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)   Kc = 1/2.16e - 24  

    now adding 1 and reversed equation (2)

       N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)

      N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)

   we get ,

                  N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)  Kc = 1.54e-31 × 1/2.61e - 24

       equilibrium constant of equation (3) is -

            Kc = 1.54e - 31 / 2.61e - 24

3 0
3 years ago
Which choice below would affect the rate of reaction in the opposite way from the other four?
bulgar [2K]
Adding a catalyst as this would speed up the reaction and the rest would slow it down
4 0
3 years ago
A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
A trench is a. Cold and shallow
nexus9112 [7]

Answer:

D

Explanation:

i think

4 0
3 years ago
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