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skelet666 [1.2K]

2 years ago

6

Fred mixes a solution using two types of solutions: 0.60 liters containing 10% alcohol and 0.40 liters containing 35% alcohol. W

hat is the alcohol concentration of the mixed solution

1 answer:

insens350 [35]2 years ago

5 0

The alcohol concentration of the mixed solution is 20%

Simplification :

Based on the given condition, formulate :

35% ×0.40 + 0.6 ×10% ÷{ 0.4+0.6}

Calculate the product : $\frac{0.14+0.06}{0.4 + 0.6}$

Calculate the sum or difference : $\frac{0.2}{1}$

Any fraction with denominator 1 is equal to numerator : 0.2

Multiply a number to both numerator, denominator : 0.2 × $\frac{100}{100}$

Calculate the product or quotient : $\frac{20}{100}$

A fraction with denominator equals to 100 to a percentage 20%.

How do you find the concentration of a mixed solution?

In general when your are mixing two different concentrations together first calculate number of moles for each solution (n=CV ,V-in liter) then add them together it will be total moles,then concentration of mixture will be = total moles / total volume(liter).

Learn more about concentration of alcohol :

brainly.com/question/13220698

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3 years ago

What is the mass present in a 10.0L container of oxygen at a pressure of 105kPa and 20 degrees Celsius

omeli [17]

1.31 × 10⁴ grams.

<h3>Explanation</h3>

Assume that oxygen acts like an ideal gas. In other words, assume that the oxygen here satisfies the ideal gas law:

$P \cdot V = n \cdot R\cdot T$ ,

where

$P$ the pressure on the gas, $\bf P = 10^{5}\;\textbf{kPa}=10^{8}\;\textbf{Pa}$ ;
$V$ the volume of the gas, $V = 10.0 \;\text{L} = 10.0\times 10^{-3}\;\text{m}^{3}=10^{-2}\;\text{m}^{3}$ ;
$n$ the number of moles of the gas, which needs to be found;
$T$ the absolute temperature of the gas, $T=20\;\textdegree{}\text{C} = (20 + 273.15)\;\text{K} = 293.15\;\text{K}$ .
$R$ the ideal gas constant, $R = 8.314$ if P, V, and T are in their corresponding SI units: Pa, m³, and K.

Apply the ideal gas law to find $n$ :

$n = \dfrac{P\cdot V}{R\cdot T} = \dfrac{{\bf 10^{8}\;\textbf{Pa}}\times 10^{-2}\;\text{m}^{3}}{8.314 \;\text{Pa}\cdot\text{m}^{3}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\times 293.15\;\text{K}} = 410.3\;\text{mol}$ .

In other words, there are 410.3 moles of O₂ molecules in that container.

There are two oxygen atoms in each O₂ molecules. The mass of mole of O₂ molecules will be ${\bf 2}\times 16.00 = 32.00\;\text{g}$ . The mass of 410.3 moles of O₂ will be:

$410.3 \times 32.00 = 1.31\times10^{4}\;\text{g}$ .

What would be the mass of oxygen in the container if the pressure is approximately the same as STP at $10^{5}\;\textbf{Pa}$ or $10^{2}\;\text{kPa}$ instead?

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4 years ago

In standardizing the solution of aqueous sodium hydroxide, a chemist overshoots the end point and adds too much naoh(aq). would

irinina [24]

Answer: Concentration of NaOH calculated will be underestimated.

Explanation:

End point is an observational point , which tells us about the completion of reaction between the titrant (solution in burette) and titre(solution in conical flask) in titration experiment.

In this case , NaOH is titrant whose concentration is unknown.

$M_1=\text{molarity of titre}$ , $M_2=\text{molarity of NaOH}$

$V_1=\text{volume of titre}$ , $V_2=\text{volume of NaOH}$

$M_1V_1=M_2V_2$

$M_2=\frac{M_1V_1}{V_2}$ ....(1)

According to question a chemist overshoots the end point and adds to much of NaOH solution, which means increase in the value of $V_2$ .

Then the value of $M_2$ in equation (1), will get lowered , which means that the concentration of NaOH was lower than that of the actual value. Hence underestimated concentration of NaOH.

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