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yawa3891 [41]
3 years ago
11

How many grams of CaCl, are in 250 mL of 2.0 M CaCl ?

Chemistry
1 answer:
stellarik [79]3 years ago
6 0

Answer: 55 g CaCl2 hope this helps

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A 1.2516 gram sample of a mixture of caco3 and na2so4 was analyzed by dissolving the sample and completely precipitating the ca
Dennis_Churaev [7]

Answer:

0.009725 moles of H2C2O4

0.009725 moles CaCO3

Mass percentage =  77.77%

Explanation:

<u>Step 1</u>: The balanced equation

2MnO4- +5C2H2O4+6H+ →2Mn2+ +10CO2+8H2O

We can see that for 2 moles of Mno4- consumed , there is 5 moles of C2H2O4 needed and 6 moles H+ to produce 2 moles Mn2+, 10 moles of CO2 and 8 moles of H2O

<u>Step 2</u>: Calculate moles of MnO4-

Molarity = Moles/volume

Moles of Mno4- = Molarity of MnO4- * Volume of Mno4-

Moles of Mno4- = 0.1092M * 35.62 *10^-3 L

Moles of MnO4- = 0.00389 moles

<u>Step 3</u>: Calculate moles of H2C2O4

Since there is needed 5 moles of C2H2O4 to consume 2 moles of MnO4-

then for 0.00389 moles of MnO4-, there is 5/2 *0.00389 = <u>0.009725 moles of H2C2O4</u>

<u />

<u>Step 4:</u> Calculate moles of CaCO3

moles of H2C2O4 = moles CaCO3, therefore, 0.009725 moles H2C2O4 = 0.009725 moles CaCO3

<u>Step 5</u>: Calculate mass of CaCO3

Molar mass of CaCO3 = 100.09 g/mole

Mass of CaCO3 = moles of CaCO3 * Molar mass of CaCO3

Mass of CaCO3 = 0.009725 moles * 100.09 g/mole = 0.9734 g

<u>Step 6</u>: Calculate percentage by weight of CaCO3

Mass of CaCO3 = 0.9734g

Mass of original sample = 1.2516g

Mass percentage = 0.9734/1.2516 *100% = 77.77%

6 0
3 years ago
How many moles are in 3.0g of A1203?
vfiekz [6]

Answer:

0.065 moles

Explanation:

4 0
3 years ago
When heated, lithium reacts with nitrogen to form lithium nitride: 6Li(s) + N2(g) → 2Li3N(s) What is the theoretical yield of Li
anyanavicka [17]

Answer:

The % yield of the reaction = 27.5 %

Explanation:

Step 1: Data given

Mass of Li = 12.7 grams

Mass of N2 = 34.7 grams

Actual yield of Li3N = 5.85 grams

Molar mass of  Lithium = 6.94 g/mol

Molar mass of N2 = 28 g/mol

Molar mass of LI3N = 34.83 g/mol

Step 2: The balanced equation:

6Li(s) + N2(g) → 2Li3N(s)

Step 3: Calculate moles of Lithium

Moles Li = mass Li / Molar mass Li

Moles Li = 12.7 grams / 6.94 g/mol

Moles Li = 1.83 moles

Step 4: Calculate moles of N2

Moles N2 = 34.7 g/ 28 g/mol

Moles N2 = 1.24 moles

Step 5: Limiting reactant

For 6 moles Li consumed, we need 1 mole of N2 to produce 2 moles of Li3N

Lithium is the limiting reactant. It will completely be consumed (1.83 moles).

N2 is in excess. There will be consumed 1.83 / 6 = 0.305 moles

There will remain 1.24 - 0.305 = 0.935 moles

Step 6: Calculate moles of Li3N

For 6 moles Li consumed, we need 1 mole of N2 to produce 2 moles of Li3N

For 1.83 moles Li, we'll have 1.83/3 = 0.61 moles of Li3N

Step 7: Calculate mass of Li3N

Mass Li3N =moles LI3N * Molar Mass LI3N

Mass Li3N = 0.610 moles * 34.83 g/mol

Mass Li3N = 21.2463 grams = Theoretical yield

Step 8: Calculate % yield

% yield = actual yield / theoretical yield

% yield = (5.85 / 21.2463)*100% = 27.5%

The % yield of the reaction = 27.5 %

8 0
3 years ago
Describe how this instrument works.<br><br> Operation:
blsea [12.9K]

Answ

Explanation:

that is like a satelite it helps detect signals from like outer space and things

4 0
2 years ago
Read 2 more answers
The poisonous gas hydrogen sulfide, H2S, can be neutralized with a base such as sodium hydroxide,
mariarad [96]

Answer:

It's not correct. For balancing, we need to put the coefficients in the molecule, not in the athom. Because if you do this, you're creating another molecule, instead of a balacing, for which the reaction may not happen - but anyway, it would be another reaction.

The correct balacing is:

2NaOH + 1H2S → 1Na2S + 2H2O

Explanation:

Look: Na2OH does not even exist. OH has only one free link, so he can't - in normal conditions - make another one with any athom. That's why we should write 2NaOH instead of Na2OH. The first means "2 mols of NaOH".

4 0
3 years ago
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