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Juli2301 [7.4K]
1 year ago
5

A scientist makes an acid solution by adding drops of acid to 1.2 l of water. the final volume of the acid solution is 1.202 l.

assuming the volume of each drop is 0.05​ ml, how many drops were added to the​ water? about what percent of the acid solution is​ acid?
Chemistry
1 answer:
Nataliya [291]1 year ago
5 0

Answer:- 40 drops and the percentage of acid in acid solution is 0.166%.

Solution:- Acid is added drops wise to 1.2 L of water to make acid solution. The final volume of the acid solution is 1.202 L. From here we could calculate the volume of the acid added to water by subtracting water volume from final acid solution volume as:

volume of acid added = 1.202 L - 1.2 L = 0.002 L

It's given that the volume of each drop is 0.05 mL. let's convert the volume of acid added from L to mL.

0.002L(\frac{1000mL}{1L})

= 2mL

Now let's calculate the number of drops of acid added to water as:

2mL(\frac{1drop}{0.05mL})

= 40 drops

0.002 L of acid are present in 1.202 L of solution. It asks to calculate about what percent of acid solution is acid means how many L of acid are present in 100 L of acid solution. It's done as:

percentage of acid in acid solution = (\frac{0.002}{1.202})100

= 0.166%

So, 40 drops of acid are required and there is 0.166% of acid in acid solution.


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Determine the PH at the point in the titration of 40.0ml of 0.200M HC4H7o2 with 0.100 M Sr(OH)2 after 100ml of the strong base h
Dmitriy789 [7]

Answer:

Check the explanation

Explanation:

Mols HC4H7O2 = (volume in L)*(molarity) = (40.0 mL)*(0.200 M)

= (40.0 mL)*(1 L)/(1000 mL)*(0.200 M)

= 8.00*10-3 mol.

Mols Sr(OH)2 corresponding to 10.0 mL of 0.100 M solution =

(volume in L)*(molarity)

= (10.0 mL)*(0.100 M)

= (10.0 mL)*(1 L)/(1000 mL)*(0.100 M)

= 1.00*10-3 mol.

Consider the ionization of Sr(OH)2 as below.

Sr(OH)2 (aq) ----------> Sr2+ (aq) + 2 OH- (aq)

As per the stoichiometric equation,

1 mol Sr(OH)2 = 2 mols OH-.

Therefore,

0.0010 mol Sr(OH)2 = [0.0010 mol Sr(OH)2]*(2 mols OH-)/[1 mole Sr(OH)2]

= 0.0020 mol

= 2.00*10-3 mol

Set up the ICE charts as below.

HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)

Before (mol)        8.00*10-3         2.00*10-3                           -                -

Change (mol)      -2.00*10-3       -2.00*10-3                           -        +2.00*10-3

After (mol)           6.00*10-3                0                                  -          2.00*10-3

The change in a pure substance, e.g., H2O is not considered in an acid-base reaction.

Volume of the solution = (40.0 + 10.0) mL = 50.0 mL = (50.0 mL)*(1 L)/(1000 mL) = 0.05 L.

The initial concentrations are obtained by dividing the numbers of moles by the volume, 0.05 L.

Set up the ICE charts as below.

HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)

Initial (M)             0.160               0.0400                             -                -

Change (M)        -0.0400            -0.0400                            -           +0.0400

Equilibrium (M) 0.120                     0                                  -            0.0400

The acid-ionization constant is written as

Ka = [H3O+][C4H7O2-]/[HC4H7O2] = 1.5*10-5

Plug in the known values and get

Ka = [H3O+]*(0.0400)/(0.120) = 1.5*10-5

======> [H3O+] = (1.5*10-5)*(0.120)/(0.0400) (ignore units)

======> [H3O+] = 4.5*10-5

The proton concentration of the solution is 4.5*10-5 M.

pH = -log (4.5*10-5 M)

= 4.346

≈ 4.35 (ans).

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How many liters of a 0.0550 M NaF solution contain 0.163 moles of NaF?
Kobotan [32]

Answer : The volume of solution will be 2.96 liters.

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

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

In this question, the solute is NaF.

Now put all the given values in this formula, we get:

0.0550M=\frac{0.163mole}{\text{Volume of solution (in L)}}

\text{Volume of solution (in L)}=\frac{0.163mole}{0.0550M}

\text{Volume of solution (in L)}=2.96L

Therefore, the volume of solution will be 2.96 liters.

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II. Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.86 g of magnesium ribbo
Leto [7]

Answer:

Excess=3.53g

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2Mg(s)+O_2(g) \rightarrow 2MgO(s)

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg}  =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg

Thus, the mass in excess is:

Excess=3.86g-0.334g\\\\Excess=3.53g

Regards!

5 0
1 year ago
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