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2 years ago

Would Li CO2 and LiOH produce the same colors? Explain your thinking.

Chemistry

1 answer:

saul85 [17]2 years ago

8 0

Answer:

Lithium hydroxide is a base.

Carbon dioxide is the anhydride of the carbonic acid, H₂CO₃.

Therefore, the reaction awaited is a typical neutralization reaction with the formation of a salt and water.

2LiOH + CO₂ → Li₂CO₃ + H₂O

So, 2*20 = 40 moles of LiOH react with 20 moles of CO₂.

Molar Mass of LiOH = 23.95 g/mol

So, 40 * 23.95 = 958 g

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NEED HELP ASAP!!! I WILL GIVE 100 POINTS!!!

Yanka [14]

In the first paragraph state how climate change is the earth warming. Tell them how the warmer atmosphere has lead to the melting of snow and ice in the arctic and sever droughts in other areas of the world.

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In the final paragraph write about how weather phenomena impact global warming. State that El Niño has led to lower floods in some place whereas other areas have experienced an increase in floods. Finally tell the reader how the lower water volumes and discharge leads to a change in the global ocean circulation.

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4 years ago

Part 1. Determine the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm. Show your work.

zaharov [31]

If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.

An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given :

V = 2.4 L = 0.0024

P = 86126.25 Pa

T = 287 K

m = 0.622

R = 8.314

The ideal gas equation is given below.

n = PV/RT

n = 86126.25 x 0.0024 / 8.314 x 287

n = 0.622 / molar mass (n = Avogardos number)

Molar mass = 7.18 g

Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g

More about the ideal gas equation link is given below.

brainly.com/question/4147359

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2 years ago

Which phase is heat energy being released?

scZoUnD [109]

The answer is B
Vaporization

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3 years ago

Create your own Model of an Atom. You may choose any element you would like to be represented by your model of an atom. Then pho

mart [117]

Here this might help you added the website link:

https://phet.colorado.edu/sims/html/build-an-atom/latest/build-an-atom_en.html

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3 years ago

A solution contains 10.20 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL

AveGali [126]

The question is incomplete, here is the complete question:

A solution contains 10.20 g of unknown compound dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.21°C. The mass percent composition of the compound is 60.98% C , 11.94% H , and the rest is O.

What is the molecular formula of the compound?

Answer: The molecular formula for the given organic compound is $C_6H_{14}O_2$

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 50.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{50.0mL}\\\\\text{Mass of water}=(1g/mL\times 50.0mL)=50g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and the freezing point of solution

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=i\times K_f\times m$

Or,

$\text{Freezing point of pure solution}-\text{freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution (water) = 0°C

Freezing point of solution = -3.21°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal boiling point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute = 10.20 g

$M_{solute}$ = Molar mass of solute = ?

$W_{solvent}$ = Mass of solvent (water) = 50.0 g

Putting values in above equation, we get:

$(0-(-3.21))^oC=1\times 1.86^oC/m\times \frac{10.20\times 1000}{M_{solute}\times 50}\\\\M_{solute}=\frac{1\times 1.86\times 10.20\times 1000}{3.21\times 50}=118.2g$