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serg [7]
1 year ago
7

Vinegar is a solution of acetic acid (the solute) in water (the solvent) with a solution density of 1010 g/L. If vinegar is 0.80

M acetic acid, what is the % by mass concentration of acetic acid in vinegar?
Chemistry
1 answer:
Damm [24]1 year ago
6 0

Answer:

4.8 %

Explanation:

We are asked the concentration in % by mass, given the molarity of the solution and its density.

0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:

MW acetic acid = 60.0 g/mol

mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g

mass of solution = 1000 cm³ x 1.010 g/ cm³      (1l= 1000 cm³)

                            = 1010 g

% (by mass) = 48.00 g/ 1010 g  x 100 = 4.8 %

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A student pours exactly 26.9 mL of HCl acid of unknown molarity into a beaker. The student then adds 2 drops of the indicator an
Assoli18 [71]
a.
Acids react with bases and give salt and water and the products.

Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
            HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

To balance the reaction equation, both sides hould have same number of elements.

Left hand side,                                             Right hand side,
             
H atoms = 2                                               H atoms = 2
            Cl atoms = 1                                               Cl atoms = 1
            Na atoms = 1                                               Na atoms = 1 
           O atoms = 1                                                   O atoms = 1

Hence, the reaction equation is already balanced.

b. 
Molarity (M)= moles of solute (mol) / Volume of the solution (L)
 
          HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Molarity of NaOH = <span>0.13 M
</span>Volume of NaOH added = <span>43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10</span>⁻³ L
                                                 = 5.681 x 10⁻³ mol

Stoichiometric ratio between NaOH and HCl is 1 : 1

Hence, moles of HCl = moles of NaOH
                                    = 
5.681 x 10⁻³ mol

5.681 x 10⁻³ mol of HCl was in <span>26.9 mL.

Hence, molarity of HCl = </span>5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
                                     = 0.21 M
6 0
2 years ago
A container of hydrogen at 172 kPa was decreased to 85.0 kPa producing a new volume of 3L. What was the original volume?
Aneli [31]

Answer:

<h2>The answer is 1.48 L</h2>

Explanation:

In order to find the original volume we use the same for Boyle's law which is

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the original volume

V_1 =  \frac{P_2V_2}{P_1}  \\

From the question

P1 = 172 kPa = 172000 Pa

P2 = 85 kPa = 85000 Pa

V2 = 3 L

We have

V_1 =  \frac{85000 \times 3}{172000}  =  \frac{255000}{172000}  =  \frac{255}{172}  \\  = 1.482558139...

We have the final answer as

<h3>1.48 L</h3>

Hope this helps you

4 0
1 year ago
A cylinder container can hold 2.45 L of water. It’s radius is 4.00 cm. What is the volume of it in cubic centimeters?
tatuchka [14]

Answer:

2450 cm3

Explanation:

Volume of cylinder = V=πr2h

2.45L = 2450mL

1mL = 1 cm cubed

2450mL = 2450 cm cubed

7 0
2 years ago
How does the equilibrium change to counter the removal of A in this reaction? A + B ⇌ AB
atroni [7]

 To counter the removal of A  the  equilibrium  change by <u>s</u><em>hifting toward the left</em>


<em>          </em><u><em>explanation</em></u>

<u><em> </em></u>If the  reaction  is at equilibrium  and we   alter  the condition a new equilibrium  state   is created

<u><em>   </em></u>The  removal  of   A led to the shift of equilibrium  toward the left since  it led to  less  molecules  in reactant side  which favor the backward  reaction.(  equilibrium  shift to the left)

6 0
2 years ago
Read 2 more answers
4.82 g of an unknown metal is heated to 115.0∘C and then placed in 35 mL of water at 28.7∘C, which then heats up to 34.5∘C. What
nikitadnepr [17]

<u>Answer:</u> The specific heat of metal is 2.34 J/g°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 35 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of metal = 4.82 g

m_2 = mass of water = 35 g

T_{final} = final temperature = 34.5°C

T_1 = initial temperature of metal = 115°C

T_2 = initial temperature of water = 28.7°C

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)]

c_1=2.34J/g^oC

Hence, the specific heat of metal is 2.34 J/g°C

4 0
1 year ago
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