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IrinaK [193]
3 years ago
9

Given the following equation,

Chemistry
1 answer:
VARVARA [1.3K]3 years ago
7 0

Answer:

5 moles

Explanation:

Given data:

Number of moles of HCl = 5 mol

Number of moles of H₂O produced = ?

Solution:

Chemical equation:

HCl + NaOH     →   NaCl + H₂O

Now we will compare the moles of HCl with H₂O.

                          HCl            :           H₂O

                             1               :             1

                             5               :            5

5 moles of water will be produced.

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In the first 15.0 s of the reaction, 1.9×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.480 L . What is the
N76 [4]

The question is incomplete, here is the complete question:

Consider the following reaction:  2N_2O(g)\rightarrow 2N_2(g)+O_2(g)

In the first 15.0 s of the reaction, 1.9×10⁻² mol of O₂ is produced in a reaction vessel with a volume of 0.480 L . What is the average rate of the reaction over this time interval?

<u>Answer:</u> The average rate of appearance of oxygen gas is 2.64\times 10^{-3}M/s

<u>Explanation:</u>

We are given:

Moles of oxygen gas = 1.9\times 10^{-2}moles

Volume of solution = 0.480 L

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

So, \text{Molarity of }O_2=\frac{1.9\times 10^{-2}mol}{0.480L}=0.0396M

The given chemical reaction follows:

2N_2O(g)\rightarrow 2N_2(g)+O_2(g)

The average rate of the reaction for appearance of O_2 is given as:

\text{Average rate of appearance of }O_2=\frac{\Delta [O_2]}{\Delta t}

Or,

\text{Average rate of appearance of }O_2=\frac{C_2-C_1}{t_2-t_1}

where,

C_2 = final concentration of oxygen gas = 0.0396 M

C_1 = initial concentration of oxygen gas = 0 M

t_2 = final time = 15.0 s

t_1 = initial time = 0 s

Putting values in above equation, we get:

\text{Average rate of appearance of }O_2=\frac{0.0396-0}{15-0}\\\\\text{Average rate of appearance of }O_2=2.64\times 10^{-3}M/s

Hence, the average rate of appearance of oxygen gas is 2.64\times 10^{-3}M/s

8 0
4 years ago
The balanced equation below shows the products that are formed when pentane (C5H12) is combusted. C5H12 + 8O2 mc029-1.jpg 10CO2
Maru [420]

Answer: The mole ratio of oxygen to pentane for the combustion of pentane is 8 : 1.

Explanation: The given reaction is a type of combustion reaction.

Combustion reaction is a reaction in which a hydrocarbon reacts with oxygen gas to produce water and carbon dioxide gas.

For the reaction of pentane with oxygen, the balanced equation would be:

C_5H_{12}+8O_2\rightarrow 5CO_2+6H_2O

In the reaction, 1 mole of C_5H_{12} reacts with 8 moles of O_2 gas.

Thus giving us the mole ratio as

Oxygen : Pentane = 8 : 1

4 0
3 years ago
Read 2 more answers
How many electron energy shells are occupied in an unstable atom of silicon
drek231 [11]

Answer:

4

Explanation:

it will share its valence electrons with other elements to acquire noble gas confriguration of the nearest inert element.

8 0
3 years ago
Please explain your answer! <br><br> Thanks!
Yuri [45]
Hello Gary!
*Sorry if I'm late*

Your answer is going to be 65.
Element A (which is actually Zinc) has the atomic number of 65. (I remember having got memorize this also).
Pretty much the explanation is that the number of patrons is equivalent to the element's atomic number!

Hope this helps!
Have a nice day :D
8 0
3 years ago
Read 2 more answers
Considering the limiting reactant, what is the mass of zinc sulfide produced from 0.250 g of zinc and 0.750 g of sulfur? Zn(s)+S
DIA [1.3K]

Answer:

The mass of zinc sulfide produced is  M_{ZnS} =  0.76 \ g

Explanation:

From the question we are told that

   The mass of zinc is  m_z =  0.750 \ g

    The mass of sulfur is  m_s =  0.250 \ g

The molar mass   of  Zn_{(s)}  is a constant with value  65.39 g /mol

The molar mass of S_{(s)}  is a constant with value  32.01 g/mol

The molar mass of  ZnS_{(s)} is a constant with value 97.46  g/mol

The reaction is  

        Zn_{(s)} + S_{(s)}  ------> ZnS_{(s)}

   So from the reaction

       1 mole of  Zn_{(s)} react with 1 mole of  S_{(s)} to produce 1 mole of ZnS_{(s)}

This implies that

65.39 g /mol of  Zn_{(s)} react with 32.01 g/mol of  S_{(s)} to produce   97.46  g/mol  of ZnS_{(s)}

From the values given we can deduce that the limiting reactant is sulfur cause  of the smaller mass

 So  

    0.250 g of  Zn_{(s)} react with 0.250 of  S_{(s)} to produce x \  g of  ZnS_{(s)}

So

      x =  \frac{97.46 * 0.250}{32.01}

       x =  0.76 \ g

Thus the mass of the mass of zinc sulfide produced is

    M_{ZnS} =  0.76 \ g

 

     

7 0
4 years ago
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