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3 years ago

What is the mass of one mole of calcium carbonate

Chemistry

2 answers:

vlada-n [284]3 years ago

6 0

100.08 grams
Use train track method!

valkas [14]3 years ago

3 0

The mass of one mole of calcium carbonate should be 100.087g. Hope this helps!

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Question 8 of 8> 0 Attempt 2 A chemist needs to determine the concentration of a sulfuric acid solution by titration with a s

Nesterboy [21]

Answer:

12.29 M

Explanation:

The reaction that takes place is:

H₂SO₄ + 2NaOH → 2Na⁺ + SO₄⁻² + 2H₂O

Now let's calculate the moles of H₂SO₄ that were titrated:

$0.1284 M * 0.1915L*\frac{1molAcid}{2molNaOH}$ = 0.01229 mol H₂SO₄.

Thus, the concentration of the diluted solution is:

0.01229 mol H₂SO₄ / 0.010 L = 1.229 M

Finally, the concentration of the original acid solution is:

$1.229 M*\frac{250mL}{25mL}$ = 12.29 M

5 0

3 years ago

Calculate the number of moles in 8 g of water.

polet [3.4K]

Answer:

refer the above attachment

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2 years ago

Given the formula representing a compound: what is a chemical name of this compound? 2-pentene 2-pentyne 3-pentene 3-pentyne

Aleks [24]

The question is incomplete. Complete question is attached below.
..........................................................................................................................

Correct Answer: Option 1) 2-pentene

Reason:
Following are the IUPAC rules for naming the compound
1) Select the longest carbon chain. In present case longest carbon chain has 5 carbon atom. Hence, it is a pentane derivative.
2) In case of alkene, replace 'e' of alkane by 'ene'
3) Give lowest number to function group. In present case, it is double bond.

Applying above rules, the IUPAC name of compound is 2-pentene

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3 years ago

1. When 8 moles of lithium metal react with excess oxygen gas, how many miles of lithium oxide can be produced?

mash [69]

2 moles of lithium oxide

5 0

4 years ago

Read 2 more answers

The reactant concentration in a second-order reaction was 0.560 m after 115 s and 3.30×10−2 m after 735 s . what is the rate con

Aloiza [94]

Use the formula for second order reaction:

$\frac{1}{C} = \frac{1}{C_0} + kt$

C = concentration at time t
C0 = initial conc.
k = rate constant
t = time

1st equation : $\frac{1}{0.56} = \frac{1}{C_0} + 115k$

2nd Equation: $\frac{1}{0.033} = \frac{1}{C_0} + 735k$

Find $\frac{1}{C_0}$ from 1st equation and put it in 2nd equation:

$\frac{1}{0.033} = \frac{1}{0.56} - 115k + 735k$

$\frac{1}{0.033} - \frac{1}{0.56} = 620k$

k = 0.046

3 0

3 years ago