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3 years ago

Rory answered that absolute zero is 0°K. Is this answer correct? Explain why or why not.

Chemistry

1 answer:

Ierofanga [76]3 years ago

5 0

Answer:

Yes, this answer is correct.

Explanation:

Absolute zero is -273.15 degrees celsius

T= Tc+273.15

T is the kelvin temperature

Tc is the temperature in degrees celsius

substitute;

T= -273.15 + 273.15

T=0 K

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When 1.5 grams of magnesium burns, how many grams of magnesium oxide will be formed?

SashulF [63]

The balanced chemical reaction for this would be written as:

2Mg + O2 = 2MgO

We use this reaction and the amount of the reactant given to calculate for the amount of magnesium oxide that is produced. We do as follows:

1.5 g Mg (1 mol / 24.31 g) ( 2 mol MgO / 2 mol Mg ) (40.30 g /1 mol ) = 2.49 g MgO produced

6 0

2 years ago

HELP ASAP ITS TIMED!!!

Tpy6a [65]

Answer:

C would be the right answer

5 0

3 years ago

Read 2 more answers

What is the molar mass of Ba3(PO4)2?

Yakvenalex [24]

Answer:

<h3>explanation:</h3>

Problem: The Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g.

5 0

2 years ago

A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa

Elina [12.6K]

<u>Answer:</u> The identity of the unknown acid is butanoic acid or ascorbic acid.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

$0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol$

The chemical equation for the reaction of NaOH and monoprotic acid follows:

$NaOH+HX\rightarrow NaX+H_2O$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

$2NaOH+H_2X\rightarrow 2NaX+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = $\frac{0.0225}{2}=0.01125moles$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

<u>For butanoic acid:</u>

Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

$\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol$

<u>For L-tartaric acid:</u>

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

$\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol$

<u>For ascorbic acid:</u>

Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

$\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol$

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

5 0

3 years ago

How many grams of barium chloride (BaCl2) are there in 0.187 mol?<br> 0.187 mol

igor_vitrenko [27]

Answer:

38.9 grams of $BaCl_2$

Explanation:

0.187 mol BaCl2 x $\frac{208 g}{1 mol}$

0.187 m x 208 g/m

0.187 x 208 g

38.896 g --> 38.9 g BaCl2

8 0

2 years ago