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Nitella [24]
3 years ago
13

How can I balance chemical equations by providing the correct coefficients? For example: [?]Pb(NO3)2 + [?]NaCrO4 = [?]PbCrO4 + [

?]NaNO3
Chemistry
1 answer:
melomori [17]3 years ago
3 0

Answer:

Pb(NO₃)₂ + Na₂CrO₄ —> PbCrO₄ + 2NaNO₃

The coefficients are: 1, 1, 1, 2

Explanation:

Pb(NO₃)₂ + Na₂CrO₄ —> PbCrO₄ + NaNO₃

The above equation can be balance as follow:

Pb(NO₃)₂ + Na₂CrO₄ —> PbCrO₄ + NaNO₃

There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by writing 2 before NaNO₃ as shown below:

Pb(NO₃)₂ + Na₂CrO₄ —> PbCrO₄ + 2NaNO₃

Now the equation is balanced.

The coefficients are: 1, 1, 1, 2

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lawyer [7]

Answer:

1.25 moles of HF

Explanation:

1 mole = 20.0 grams of HF

So, to calculate mass of 25.0 grams

                                       =25.0*1/20.0

                                       =1.25 moles

7 0
3 years ago
Which statement explains why a molecule of CH4 is nonpolar?
Lerok [7]

Answer:

The geometric shape of a CH4 molecule distributes the charges symmetrically.

Explanation:

The dipole moment is ruled by the compound's geometry and elements electronegativity difference.

Arranging the forces acording to the elements electronegativity difference withing the molecule geometry will yeld the dipole moment.  

If the forces cancel out, it is a nonpolar compound.

If the forces does not cancel out it is a polar compound.

Picture of such analysis for methane is attached.

3 0
4 years ago
Read 2 more answers
A chemist must prepare 0.200 L of aqueous silver nitrate working solution. He'll do this by pouring out some aqueous silver nitr
jeyben [28]

A chemist must prepare 0.200 L of 1.00 M aqueous silver nitrate working solution. He'll do this by pouring out 1.82 mol/L aqueous silver nitrate stock solution into a graduated cylinder and diluting it with distilled water. How many mL of the silver nitrate stock solution should the chemist pour out?

Answer: 0.110 L

Explanation:

According to the dilution law,

M_1V_1=M_2V_2

where,  

M_1 = molarity of stock silver nitrate solution = 1.82 M

V_1 = volume of stock silver nitrate solution = ?  

M_1 = molarity of diluted silver nitrate solution = 1.00 M

V_1 = volume of diluted silver nitrate solution = 0.200 L

Putting in the values we get:

1.82M\times V_1=1.00M\times 0.200L

V_1=0.110L

Therefore, volume of silver nitrate stock solution required is 0.110 L

3 0
3 years ago
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