Ask question

Ask question

All categories

scoundrel [369]

4 years ago

5

if a molecule has gained an electron through a chemical reaction, it has been ____ , and if it has lost an electron, it has been

____

2 answers:

Alina [70]4 years ago

6 0

Reduced and Oxidized

soldier1979 [14.2K]4 years ago

5 0

Answer:

FIRST REDUCED THEN OXIDIZED

Explanation:

You might be interested in

Which of the following elements is inert, meaning is non-reactive?

swat32

Xenon. It is located in the last group of the periodic table.

3 0

4 years ago

Read 2 more answers

first to answer correctly gets brainleist! (you dont needa hear that vid thingy tahts just for ppl who cant read or need orial c

Soloha48 [4]

I think the answer is c

7 0

3 years ago

Read 2 more answers

How strong is sulfur

lorasvet [3.4K]

Atomic number= 16

<span>Atomic mass = </span>32.064

<span><span /></span>

8 0

4 years ago

How much heat does your body lose when 3.45 g of sweat evaporates from your skin at 25 âc? (assume that the sweat is only water.

Eddi Din [679]

<span>Water has a Mol.mass of 18g/mol. (2 x 1:H) + (1 x O:16) = 18. Mass / mol.mass = moles. 3.45g / 18g/mol. = 0.192 moles of sweat. 0.192mol. x 44.0kJ/mol = 8.43kJ of heat released from the body.</span>

4 0

4 years ago

An aqueous KNO3 solution is made using 72.5 g of KNO3 diluted to a total solution volume of 2.00 L. Calculate the molarity, mola

Sonja [21]

Answer:

The molarity is 0.359 $\frac{moles}{L}$

The molality is 0.354 $\frac{moles}{kg}$

The mass percent of the solution es 3.45%

Explanation:

Molarity is a unit of concentration that indicates the amount of moles of solute that appear dissolved in each liter of the mixture. It is determined by:

$Molarity (M)=\frac{number of moles of solute}{volume}$

Being:

K: 39 g/mole
N: 14 g/mole
O: 16 g/mole

The molar mass of KNO₃ is:

KNO₃= 39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole

You can apply the following rule of three: if 101 grams of KNO₃ are present in 1 mole, 72.5 grams in how many moles are present?

$moles of KNO_{3}=\frac{72.5 grams*1 mole}{101 grams}$

moles of KNO₃= 0.718

So you have:

moles of KNO₃= 0.718
volume= 2 L

Applying this quantity in the definition of molarity:

$molarity=\frac{0.718 moles}{2 L}$

Molarity= 0.359 $\frac{moles}{L}$

<u><em>The molarity is 0.359</em></u> $\frac{moles}{L}$ <u><em></em></u>

Molality is a way of measuring the concentration of solute in solvent and indicates the amount of moles of solute in each kilogram of solvent.

Then the molality is calculated by:

$Molality=\frac{moles of solute}{mass of solvent in kilograms}$

Density is defined as the property that matter, whether solid, liquid or gas, has to compress in a given space. That is, it is the amount of mass per unit volume. So, if the density of 1.05 g / mL for the solution indicates that in 1 mL of solution there are 1.05 grams of solution, in 2000 mL (where 2L = 2000 mL, because 1 L = 1000mL) how much mass is there?

$mass=\frac{2000 mL*1.05 grams}{1 mL}$

mass= 2100 grams

Since mass solution = mass water + mass KNO₃

then mass water = mass solution - mass KNO₃

Being mass solution 2100 grams and mass KNO₃ 72.5 grams, and replacing you get: mass water= 2100 grams - 72.5 grams

mass water= 2,027.5 grams

Then, being:

moles of KNO₃= 0.718
mass of solvent in kilograms= 2.0275 kg (being 2,027.5 grams= 2.0275 kilograms because 1,000 grams= 1 kilogram)

Replacing in the definition of molality:

$molality=\frac{0.718 moles}{2.0275 kg}$

molality= 0.354 $\frac{moles}{kg}$

<u><em>The molality is 0.354 </em></u> $\frac{moles}{kg}$ <u><em></em></u>

The mass percent of a solution is the number of grams of solute per 100 grams of solution. Then the mass percent is the mass of the element or solute divided by the mass of the compound or solute and the result of which is multiplied by 100 to give a percentage.

$mass percent=\frac{mass of solute}{mass of solution} *100$

So, in this case:

$mass percent=\frac{72.5 grams}{2100 grams} *100$

mass percent= 3.45 % KNO₃ by mass

<u><em>The mass percent of the solution es 3.45%</em></u>

3 0

3 years ago