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4 years ago

A chemical reaction yields 3 moles of lithium hydroxide (LiOH). How many grams of

Chemistry

1 answer:

grin007 [14]4 years ago

8 0

72g

Explanation:

Given parameters:

Number of moles of LiOH = 3moles

Unknown:

Mass of LiOH = ?

Solution:

A mole of substance is a unit used to make quantitative measures in chemistry.

It is the amount of substance that contains the avogadro's number of particles.

The mole is related to mass using the expression below;

Mass of a substance = number of moles x molar mass

Molar mass of LiOH = 7 + 16 + 1 = 24g/mol

Mass of LiOH = 3 x 24 = 72g

learn more:

Number of moles brainly.com/question/1841136

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An element has five isotopes. Calculate the atomic mass of this element using the information below. Show all your work. Using t

Katena32 [7]

Answer: Sol:-

Data provided in the question is :-

Atomic mass of isotope -1 = 64 amu

Atomic mass of isotope -2 = 66 amu

Atomic mass of isotope -3 = 67 amu

Atomic mass of isotope -4 = 68 amu

Atomic mass of isotope - 5 = 70 amu

Percentage abundace of isotope - 1 = 48.89 %

Percentage abundance of isotope -2 = 27.81 %

Percentage abundance of isotope - 3 = 4.11%

Percentage abundance of isotope-4 = 18.57%

Percentage abundance of isotope - 5 = 0.62 %

Formula used :-

Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]

Calculation :-

Put all the value in the formula :-

Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu

= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu

= {(6528.04) / 100} amu

= 65.2804 amu

Average atomic mass of an element is = 65.2804 amu

Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc

atomic mass of zinc = 65.38 \approx 65.2804 amu

5 0

3 years ago

How many kJ are 3,340J?

forsale [732]

The answer I believe is 3.340kj.

8 0

3 years ago

At a certain temperature, the equilibrium pressures of NO2 and N2O4 are 1.4 bar and 0.46 bar, respectively. If the volume of the

Butoxors [25]

The partial pressure of NO₂ at equilibrium is 0.70 bar and the partial pressure of N₂O₄, at equilibrium is 0.23 bar.

Let x be the mole fraction of NO₂ and x' be the mole fraction of N₂O₄.

The total pressure according to Dalton's law of partial pressure is the sum of the partial pressures of each gas.

Let P be the total pressure of the gases, P' be the partial pressure of NO₂ = 1.4 bar and P" be the partial pressure of N₂O₄ = 0.46 bar.

So, P = P' + P"

= 1.4 bar + 0.46 bar

= 1.86 bar

Since P' = xP

x = P'/P

= 1.4 bar/1.86 bar

= 0.753

Also, P" = xP

x' = P"/P

= 0.46 bar/1.86 bar

= 0.247

Now, since the volume of the container is doubled at constant temperature, we use Boyle's law to determine the new pressure. P₁.

Boyle's law states that the pressure of a given mass of gas is inversely proportional to its volume provided the temperature remains constant. It is written mathematicaly as PV = constant

So, PV = P₁V₁

P₁ = (V/V₁)P

Since V/V₁ = 1/2

P₁ = (V/V₁)P

P₁ = P/2

P₁ = 1.86/2 bar

P₁ = 0.93 bar

So, the <u>new</u> partial pressure of NO₂, P₂ = xP₁

= 0.753 × 0.93 bar

= 0.70 bar

The <u>new</u> partial pressure of N₂O₄, P₃ = x'P₁

= 0.247 × 0.93 bar

= 0.23 bar

So, the partial pressure of NO₂ at equilibrium is 0.70 bar and the partial pressure of N₂O₄, at equilibrium is 0.23 bar.

Learn more about partial pressure here:

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3 years ago

In 1770, ___________ published the first map of the Gulf Steam.

larisa [96]

Answer: Ben Franklin

Explanation: Hope this helps! :)

8 0

3 years ago

Read 2 more answers

Which compound contains both sigma and pi bonds? HCCl3 H2CO H2S HBr.

Aliun [14]

The compound that contains both sigma and pi bonds has been $\rm \bold{H_2CO}$ . Thus, option B is correct.

The compounds have been resulted by the sharing of the valence electrons between atoms, for the completion of octet of each elements.

The bonds can be saturated with the presence of only sigma bond. The unsaturated bonds has presence of pi bonds as well. The bond with one pi and one sigma has been a double bond, while 1 sigma and 2 pi has been a triple bond.

The bonds present in the following structures has been:

$\rm HCCl_3=4\;\sigma\;bonds\\H_2CO=2\;\sigma,\;1\;\pi\;bond\\H_2S=2\;\sigma\;bonds\\HBr\;=1\;\sigma\;bond$

The compound with the presence of both sigma and pi bonds has been $\rm \bold{H_2CO}$ . Thus, option B is correct.

For more information about the sigma and pi bonds, refer to the link:

brainly.com/question/14018074

6 0

2 years ago