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polet [3.4K]
3 years ago
6

Calcule el porcentaje de soluto (%vv) en una disolución preparada con 45,2 mL de ácido nítrico en 205 mL de agua.

Chemistry
1 answer:
ryzh [129]3 years ago
5 0

Answer:

18.1% V/V

Explanation:

El porcentaje volumen/volumen (%v/v) es una unidad de concentración en química definida como la relación entre volumen de soluto y volumen total de la solución (En las mismas unidades ambos volúmenes) por cien. La fórmula es:

% V/V = Volumen soluto / Volumen solución × 100

Analizando el problema, el soluto es ácido nítrico (Menor cantidad) y el solvente es el agua.

Recordando, el volumen de la solución es igual al volumen del soluto + el volumen del solvente. Así:

% V/V = 45.2mL soluto / (45.2mL + 205mL) × 100 =

<em>18.1% V/V</em> es la concentración de la solución.

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the hydrogen atom of one water molecule and the lone pair of electrons on an oxygen atom of a neighboring water molecule.

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2 years ago
the atomic number tells you the number of _______ in one atom of an element. It also tells you the number of ______ in a neutral
uysha [10]

Answer:

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Explanation:

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5 0
3 years ago
A 0.944 M solution of glucose, C6H12O6, in water has a density of 1.0624 g/mL at 20∘C. What is the concentration of this solutio
____ [38]

Answer:

Mole fraction: 0,0157.

Molality: 0,889m

Mass%: 16%

Explanation:

<em>The units are mole fraction, molality and mass%</em>

<em />

Units of mole fraction are moles of glucose per total moles.

Moles of glucose assuming 1L are:

0,944 moles.

Moles of water in 1L are:

1L × (1,0624kg/L) × (1000g / 1kg) × (1mol / 18,02g) = <em>59,0 moles of water</em>

<em />

Mole fraction is: 0,944 moles / (59,0 mol + 0,944mol) = <em>0,0157</em>

Molality is mole of solute (0,944) per kg of solution (1,0624kg):

0,944mol / 1,0624kg = <em>0,889m</em>

<em></em>

In mass percent total mass is 1062,4g and mass of 0,944 moles of glucose is:

0,944mol×(180,156g/1mol) = 170g of glucose. Mass%:

170g / 1062,4g ×100 = <em>16%</em>

<em></em>

I hope it helps!

6 0
3 years ago
1.) Calculation: If 9.02 x 1024 particles of vinegar (HC2H3O2)HC2H3O2) are added to 16.5 moles of eggshell (CaCO3) and 6.35 mole
blsea [12.9K]

The theoretical yield of acetate is 2607 g. The actual yield of acetate is 1066.8 g. The percentage yield of acetate is 41%.

If 1 mole of vinegar contains 6.02  x 10^23 particles

x moles of vinegar contains 9.02 x 10^24 particles

x = 1 mole x 9.02 x 10^24 /6.02 x 10^23

x = 15 moles of vinegar

The reaction is as follows;

2HC2H3O2 + CaCO3 -----> Ca(C2H3O2)2 + H2O + CO2

Since 2 moles of vinegar reacts with 1 mole of carbonate

x moles of vinegar reacts with 16.5 moles of carbonate

x =  2 moles x 16.5 moles/ 1 mole

x = 33 moles of vinegar

We can see that the vinegar is the reactant in excess hence the carbonate is the limiting reactant.

Theoretical yield = 16.5 moles x 158 g/mol = 2607 g

Actual yield = 6.35 moles  x 158 g/mol = 1066.8 g

Percent yield = 1066.8 g/2607 g × 100/1

= 41%

Learn more: brainly.com/question/13440572?

7 0
3 years ago
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