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Bingel [31]
3 years ago
11

Change the given word equation into balanced equation Hydrogen+Chloride=Hydrogen chloride​

Chemistry
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

h2 +2cl -> 2hcl

Explanation:

h2 + cl -> HCl

h2 +2cl -> 2hcl

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ycow [4]

Answer:

Heterogeneous

Explanation:

3 0
2 years ago
The atomic particles given off as a result of nuclear fusion are?
Vilka [71]

Answer:

I think <em><u>alpha</u></em> and <em><u>beta</u></em> is the answer.

4 0
3 years ago
Read 2 more answers
5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
2 years ago
What does a balanced equation allows you to determine
Allushta [10]

It allows you to determine the relation between the reactants and the products.

7 0
3 years ago
If the valency of oxygen is 2, derive the valencies of the other elements present in the following oxides. Na2O, ZnO , Al2O3 , M
11Alexandr11 [23.1K]

Answer:

1)Na2O

let the valency of Na is x

2(x)+(2)=0

2x+2=0

2x=-2

x=-1

2)ZnO

let the valency of Zn is x

x+2=0

x=-2

3)Al2O3

let the valency of Al is x

2(x)+3(2)=0

2x+6=0

2x=-6

x=-3

4)MgO

let the valency of Mg is x

x+2=0

x=-2

7 0
3 years ago
Read 2 more answers
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