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Gre4nikov [31]
3 years ago
5

How do you balance the equation in number one?

Chemistry
1 answer:
AleksAgata [21]3 years ago
3 0
____NaNO3 + ___PbO --> ___Pb(NO3)2 + ___Na[2]O

To balace the eqaution, you need to have the same number of atoms for each element on both the reactant (left) and product (right) side. 

To start off, you wanna know the number of atoms in each element on both sides, so take it apart:

[reactants]        [product]
Na- 1               Na- 2

N- 1               N- 2(it's 2 because the the subscript [2] is outside of the parenthesis)

O- 4              O- 7 (same reason as above) 

Pb- 1            Pb- 1 

Na is not balanced out, so add a coefficient to make it the same on both sides.In this case, multiply by 2: 

2NaNO3

Now Na is balanced, but the N and O are also effected by this, so they also have to be multiplied by 2 and they become: 

Na- 2             Na- 2 

N- 2              N- 2 (it balanced out) 

O- 7 (coefficient times subscript, plus lone O)  O- 7 (balanced out) 

Pb was already balanced so no need to mess with it, just put a 1 where needed (it doesn't change anything). 

Now to put it back together, it will look like this: 

2NaNO3 + 1PbO --> 1Pb(NO3)2 + 1Na[2]O
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How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben
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Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(C) 17 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the \frac{weight}{volume} in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be \frac{1000mL}{200mL}=\frac{1g}{xg}

200mL × 1g = 1000 mL × x(g)

x(g) = \frac{200mL*1g}{1000mL}

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ \frac{10mL}{250mL}=\frac{0.2g}{y(g)}

y(g) = \frac{250mL*0.2g}{10mL}

y(g) = 5g of benzalkonium chloride.

Now, at 17% \frac{weight}{volume} concentrate contains 17g/100ml:

∴  the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= \frac{17g}{5g} = \frac{100mL}{z(mL)}

z(mL) = \frac{100mL*5g}{17g}

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

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