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2 years ago

¿Cuál es el % m/m de una disolución en que hay disueltos 22 g de soluto en 44 g de disolvente?

1 answer:

5 0

The question is as follows: What is the% m / m of a solution in which 22 g of solute are dissolved in 44 g of solvent?

Answer: The% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.

Explanation:

Given: Mass of solute = 22 g

Mass of solvent = 44 g

The percentage m/m is calculated using the following formula.

$Mass percentage = \frac{mass of solute}{mass of solvent} \times 100$

Substitute the values into above formula as follows.

$Mass percentage = \frac{mass of solute}{mass of solvent} \times 100\\= \frac{22 g}{44 g} \times 100\\= 50 percent$

Thus, we can conclude that the% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.

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Because oxygen is more electronegative than hydrogen, the water molecule is

lapo4ka [179]

The water molecule has two hydrogen atoms joined to one oxygen atom by single covalent bonds. Because oxygen is more electronegative than hydrogen, the electrons of the covalent bond spend more time closer to the oxygen then hydrogen which creates a polar covalent bond. Hope this answers your question. Have a great day!

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3 years ago

1. If 80.0 ml of 3.00 M HCl is used to make a 100. ml of dilute acid, what is the molarity

Aleonysh [2.5K]

We are given:

M1 = 3 Molar V1 = 80 mL

M2 = x Molar V2 = 100 mL

Finding the molarity:

We know that:

M₁V₁ = M₂V₂

where V can be in any units

(3)(80) = (x)(100)

x = 240/100 [dividing both sides by 100]

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2 years ago

Which of the following is a product of aerobic respiration?

sasho [114]

Answer:

The product of aerobic respiration is Carbon dioxide.

Explanation:

The process of breaking down glucose to produce energy and waste products is called respiration. Livings beings need respiration process to generate energy so that they can survive.
The types of respiration are : Anaerobic and aerobic respiration.
Aerobic respiration takes place in presence of oxygen and produces large amount of energy.
The final product of aerobic respiration are carbon dioxide, water and 38 ATP of energy.

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3 years ago

Read 2 more answers

A mixture of0.161 moles of C is reacted with 0.117 moles of O2 in a sealed, 10.0 L-vessel at 500.0 K, producing a mixture of CO

labwork [276]

Answer:

number of moles of CO2 is 0.054

number of moles of CO is 0.107

number of moles of O2 remaining is 0.01 mole

mole fraction of CO is 0.63

Explanation:

Firstly, we write the equation of reaction;

3C(s) +2O2(g) → CO2(g) +2CO(g)

Now, we proceed.

From the written equation, we can deduce that

3 mol C = 2 mol O2 = 1 mol CO2 = 2 mol CO

No of mol of C reacted = 0.161 mol

limiting reactant according to the question is Carbon

a. no of mol of CO2 formed = 0.161*1/3 = 0.054 moles ( no of moles of CO2 formed is one-third of no of moles of carbon reacted. This is obtainable from their mole ratio 1:3)

b. no of mol of CO formed = 0.161*2/3 = 0.107 mol

c. no of mol of O2 remaining = 0.117 - (0.151*2/3) = 0.117-0.107 = 0.01 mole

d. mole fraction of CO = no of mol of CO/Total number of moles

= 0.107/(0.107+0.054+0.01)

= 0.625730994152 which is approximately 0.63

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2 years ago

calculate pressure exerted by 1.255 mol of CI2 in a volume of 5.005 L at a temperature 273.5 k using ideal gas equation

balu736 [363]

Answer:

The pressure is 5.62 atm.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

P= ?
V= 5.005 L
n= 1.255 mol
R= 0.082 $\frac{atm*L}{mol*K}$
T= 273.5 K

Replacing:

P* 5.005 L= 1.255 mol* 0.082 $\frac{atm*L}{mol*K}$ *273.5 K

Solving:

$P=\frac{1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K}{5.005 L}$

P= 5.62 atm

The pressure is 5.62 atm.

8 0

2 years ago