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dsp73
3 years ago
9

Calcula el %m/v de alcohol en una mezcla utilizada para la desinfección de manos formada por: 15 ml de agua (densidad=1g/ml), 10

5 g de etanol (densidad: 0,798 g/ml) y 4,5 gramos de jabón líquido (densidad= 1,5 g/ml)
Chemistry
1 answer:
mr Goodwill [35]3 years ago
3 0

Answer:

%m/v =70%

Explanation:

El %m/v es una unidad de concentración que se define como cien veces la división entre la masa de una sustancia (En gramos) y el volumen total en el que esta sustancia se encuentra (en mL).

En el problema, debemos hallar la masa de etanol (Alcohol) y el volumen total de la solución.

<em>Masa alcohol: </em>

Ya te la dan en el problema: 105g

<em>Volumen solución:</em>

Volumen agua: 15mL

Volumen etanol: 105g × (1mL / 0.798g) = 131mL

Volumen Jabón líquido: 4.5g × (1mL / 1.5g) = 3mL

Volumen: 15mL + 131mL + 3mL

149mL

Así, el %m/v de alcohol en la solución es:

%m/v = (105g / 149mL) × 100

<h3>%m/v =70%</h3>

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A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
Balancee por tanteo las siguientes ecuaciones químicas. Escriba el nombre a reactantes y productos. H2O5 + H2O ---&gt; HNO3 Na2O
Elan Coil [88]

Answer:

a. N₂O₅ + H₂O ⇒ 2 HNO₃ (pentóxido de dinitrógeno + agua ⇒ ácido nítrico)

b. Na₂O + H₂O ⇒ 2 NaOH (óxido de sodio + agua ⇒ hidróxido de sodio)

Explanation:

Tenemos que balancear, por el método de tanteo, las siguientes ecuaciones químicas.

a. En la primera reacción, el pentóxido de dinitrógeno reacciona con agua para formar ácido nítrico. Es una reacción de síntesis o combinación.

N₂O₅ + H₂O ⇒ HNO₃

Podremos obtener la ecuación balanceada si multiplicamos HNO₃ por 2.

N₂O₅ + H₂O ⇒ 2 HNO₃

b. En la segunda reacción, óxido de sodio reacciona con agua para formar hidróxido de sodio. Es una reacción de síntesis o combinación.

Na₂O + H₂O ⇒ NaOH

Podremos obtener la ecuación balanceada si multiplicamos NaOH por 2.

Na₂O + H₂O ⇒ 2 NaOH

3 0
2 years ago
If 20.0 g LIOH react with excess KCl, giving a 17.0
adell [148]

The yield of lithium chloride is 1.92 grams.

Option D.

<h3><u>Explanation:</u></h3>

In this reaction, we can see that 1 mole of lithium hydroxide reacts with 1 mole of potassium chloride to produce 1 mole of lithium chloride and 1 mole of potassium hydroxide.

Molecular weight of lithium hydroxide is 24.

Molecular weight of lithium chloride is 42.5.

So 24 grams of lithium hydroxide produces 42.5 grams of lithium chloride.

So, 20 grams of lithium hydroxide produces 20 * 24 / 42.5  grams =11. 29 grams of lithium chloride.

But this is when the yield is 100%.

But yield is 17%.

So the yield is 1.92 grams of lithium chloride.

8 0
3 years ago
Which career field is an applied science?<br><br> Which hypothesis is testable?
VashaNatasha [74]

Chemists, biologist, pharmacist, medical researcher ect…

4 0
3 years ago
Write the balanced equation for the ionization of the weak base pyridine, c5h5n , in water, h2o. Phases are optional.
Lelu [443]

The balanced equation for the ionization of the weak base pyridine,C5H5N in water, H2O

C_5H_5N ( aq.) + H2O ( l) ---------> C5H5NH+ (aq.) + OH- (aq.)

<h3>What is the balanced equation for the ionization?</h3>

Generally, Pyridine is characterized by a ring structure, in this characteristic ring structure N is sp2 hybridized, hence creating a lone pair present on N so s - character is more, as well as lone pair, is present.

Therefore, Considering The following functions of the equation:weak base pyridine,C5H5N in water, H2O

We write the balanced equation for the ionization as

C_5H_5N ( aq.) + H2O ( l) ---------> C5H5NH+ (aq.) + OH- (aq.)

Read more about  Chemical Reaction

brainly.com/question/11231920

6 0
2 years ago
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