Anyone want to be friends????? Not necessary to answer

A solution is prepared by mixing 525 mL of ethanol with 597 mL of water. The molarity of ethanol in the resulting solution is 8.

Alenkinab [10]

Answer:

$\Delta V = 234.736\,mL$

Explanation:

The quantity of moles of ethanol in the solution is:

$n_{C_{2}H_{5}OH} = \left(\frac{597\,mL}{1000\,mL} \right)\cdot \left(8.35\,\frac{mol}{L} \right)$

$n_{C_{2}H_{5}OH} = 4.985\,mol$

The mass and volume of ethanol in the solution are, respectively:

$m_{C_{2}H_{5}OH} = (4.985\,mol)\cdot \left(46.07\,\frac{g}{mol} \right)$

$m_{C_{2}H_{5}OH} = 229.658\,g$

$V_{C_{2}H_{5}OH} = \frac{229.658\,g}{0.7893\,\frac{g}{mL} }$

$V_{C_{2}H_{5}OH} = 290.964\,mL$

The difference between the total volume of water and ethanol mixed to prepare the solution and the actual volume of solution is:

$\Delta V = (525\,mL+597\,mL) - (597\,mL + 290.964\,mL)$

$\Delta V = 234.736\,mL$

8 0

3 years ago

A student is told to use 20.0 grams of sodium chloride to make an aqueous solution that has a concentration of 10.0 grams of sod

Butoxors [25]

Volume of the solution = $20.0 g NaCl * \frac{1 L solution}{10.0 g NaCl}$

= 2 L solution x $\frac{1000 mL}{1 L} =2000 mL$

Volume of solute = 7.5 mL

Volume of water (solvent) = 2000 mL - 7.5 mL = 1992.5 mL water

7 0

3 years ago

A 260.0 −mL buffer solution initially contains 2.5×10−2 M of HCHO2 and 2.5×10−2 M of NaCHO2

FinnZ [79.3K]

The mass of HCHO2 and NaCHO2 to be added to the buffer solution are 0.23g and 0.44g respectively

Data;

Volume of solution = 260mL
conc. of HCHO2 = 2.5*10^-2M
conc. of NaCHO2 = 2.5*10^-2M

<h3>Mass of Reagent Added</h3>

To calculate the mass of reagent added, let's start with HCHO2

The mass of HCHO2 to be added is the number of moles of HCHO2 multiplied by it's molar mass.

$260*10^-^3 * 2.5*10^-^2 * 46 = 0.229 = 0.23g$

The mass of NaCHO2 to be added in the buffer solution is

$260*10^-^3*2.5*10^-^2*68 = 0.442 = 0.44g$

The mass of HCHO2 and NaCHO2 to be added to the buffer solution are 0.23g and 0.44g respectively

Learn more about buffer solution here;

brainly.com/question/22390063

3 0

2 years ago

An organism that hunts other organisms for food is a

Zepler [3.9K]

Answer:

B) a predator

6 0

3 years ago

Read 2 more answers

Percentage Yield

scoundrel [369]

Q.No. 1:

You have 20.0 g of CaCO₃. You decompose it by heat, and weigh the calcium oxide that remains. You have 10.3 grams. What is the % yield?

Answer:

%age Yield = 92.37 %

Solution:

The balance chemical equation for given decomposition reaction is;

CaCO₃ → CaO + CO₂

Step 1: Calculate Moles of CaCO₃:

Moles = Mass / M.Mass

Moles = 20.0 g / 100.08 g/mol

Moles = 0.199 moles of CaCO₃

Step 2: Calculate theoretical amount of CaO produced;

According to equation,

1 moles of CaCO₃ produced = 1 mole of CaO

So,

0.199 moles of CaCO₃ will produce = X moles of CaO

Solving for X,

X = 0.199 mol × 1 mol / 1 mol

X = 0.199 mol of CaO

Also,

Mass = Moles × M.Mass

Mass = 0.199 mol × 56.07 g/mol

Mass = 11.15 g of CaO

Step 3: Calculate %age Yield as;

%age Yield = Actual Yield / Theoretical Yield × 100

%age Yield = 10.3 g / 11.15 g × 100

%age Yield = 92.37 %

___________________________________________

Q.No. 2:

A student makes sodium chloride by buming 2.3 grams of sodium in chlorine gas. If the yield is 90%, how much sodium chloride is made?

Answer:

Actual Yield = 5.785 g of NaCl

Solution:

The balance chemical equation for given decomposition reaction is;

2 Na + Cl₂ → 2 NaCl

Step 1: Calculate Moles of Na:

Moles = Mass / M.Mass

Moles = 2.3 g / 23 g/mol

Moles = 0.10 moles of Na

Step 2: Calculate theoretical amount of NaCl produced;

According to equation,

2 moles of Na produced = 2 moles of NaCl

So,

0.10 moles of Na will produce = X moles of NaCl

Solving for X,

X = 0.10 mol × 2 mol / 2 mol

X = 0.10 mol of NaCl

Also,

Mass = Moles × M.Mass

Mass = 0.10 mol × 58.44 g/mol

Mass = 5.844 g of NaCl

Step 3: Calculate Actual Yield as;

%age Yield = Actual Yield / Theoretical Yield × 100

Or,

Actual Yield = %age Yield × Theoretical Yield ÷ 100

Actual Yield = 99 × 5.844 g ÷ 100

Actual Yield = 5.785 g of NaCl

___________________________________________

Q.No. 3:

For the chemical reaction Mg(s) + 2 HCl (aq) → H₂ (g) + MgCl₂ (aq) calculate the % yield if 100 grams of magnesium react with excess HCl to produce 310 grams of MgCl₂.

Answer:

%age Yield = 79.13 %

Solution:

The balance chemical equation for given decomposition reaction is;

Mg + 2 HCl → MgCl₂ + H₂

Step 1: Calculate Moles of Mg:

Moles = Mass / M.Mass

Moles = 100 g / 24.30 g/mol

Moles = 4.11 moles of Mg

Step 2: Calculate theoretical amount of MgCl₂ produced;

According to equation,

1 mole of Mg produced = 1 mole of MgCl₂

So,

4.11 moles of Mg will produce = X moles of MgCl₂

Solving for X,

X = 4.11 mol × 1 mol / 1 mol

X = 4.11 mol of MgCl₂

Also,

Mass = Moles × M.Mass

Mass = 4.11 mol × 95.21 g/mol

Mass = 391.73 g of MgCl₂

Step 3: Calculate %age Yield as;

%age Yield = Actual Yield / Theoretical Yield × 100

%age Yield = 310.0 g / 391.73 g × 100

%age Yield = 79.13 %

5 0

4 years ago