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hichkok12 [17]
3 years ago
5

You need to make an aqueous solution of 0.207 M calcium acetate for an experiment in lab, using a 300 mL volumetric flask. How m

uch solid calcium acetate should you add?
Chemistry
1 answer:
jeka943 years ago
5 0

Answer:

We have to add 9.82 grams of calcium acetate

Explanation:

Step 1: Data given

Molarity of the calcium acetate solution = 0.207 M

Volume = 300 mL = 0.300 L

Molar mass calcium acetate = 158.17 g/mol

Step 2: Calculate moles calcium acetate

Moles calcium acetate = molarity * volume

Moles calcium acetate = 0.207 M * 0.300 L

Moles calcium acetate = 0.0621 moles

Step 3: Calculate mass calcium acetate

Mass calcium acetate = moles * molar mass

Mass calcium acetate = 0.0621 moles * 158.17 g/mol

Mass calcium acetate = 9.82 grams

We have to add 9.82 grams of calcium acetate

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What is the mass of oxygen in 300 grams of carbonic acid (H2CO3)
DENIUS [597]

mass of carbonic acid = 300g

molar mass of H2CO3 = 2H + C + 3 O

= 2 x 1.008+ 12.01 + 3 x  16

= 62.03g/mol


moles of H2CO3 = mass/Molar mass

= 300/62.03

= 4.8364 moles


1 mole H2CO3 has 3 moles Oxygen


4.8364 moles H2CO3 contains  

=   3 x 4.8364  moles Oxygen  =   14.509 moles Oxygen


moles = mass/Molar mass


mass of oxygen = moles x Molar mass of Oxygen

= 14.509 x 16

= 232.15g Oxygen

mass of oxygen in 300g of carbonic acid(H2CO3) = 232.15g

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What is that the theoretical yield of aluminum oxide I if 3.20 mol of aluminum metal is exposed to 2.70 mole of oxygen
photoshop1234 [79]

Answer:

163.2g

Explanation:

First let us generate a balanced equation for the reaction. This is shown below:

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From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.

From the equation,

4moles of Al produced 2moles of Al2O3.

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Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:

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