Part B

A sample that contains only SrCO3 and BaCO3 weighs 0.846 g. When it is dissolved in excess acid, 0.234 g carbon dioxide is liber

In-s [12.5K]

Answer:28.605

Explanation:First, the molar mass of of SrCO3, BaCO3 and CO2 has to be calculated, (using the molar mass of each element Sr = 87.62, Ba = 137.327, C=12.011, O= 16.00)

The molar masses are;

SrCO3 = 87.62 + 12.011 + (3*16) = 147.631g/mol

BaCO3 = 79.904 + 12.011 + (3*16) = 197.34 g/mol

CO2 = 12.011 + (2*16) = 44.011 g/mol

To obtain one of the equations to solve the problem;

The sample is made of SrCO3 and BaCO3 and has a mass of 0.846 g. Representing the mass of SrCO3 as ma and that of BaCO3 as mb. The first equation can be written as:

ma + mb = 0.846g (1)

To obtain another equation in order to be able to determine the different percentages of the compounds (SrCO3 and BaCO3) that make of the sample, a relationship can be obtained by determining the relationship between the number of moles of CO2 formed as the mass of the SrCO3 and BaCO3;

The number of moles of CO2 formed = (mass of CO2)/(molar mass) =0.234/44.011 =0.00532moles

CO2 contains 1 mole of carbon (C) so therefore 0.00532 moles of CO2 contains 0.00532 moles of C

The sample produced 0.00532 moles of CO2, therefore the number of moles SrCO3 and BaCO3 that produced this amount can be calculated using the formula;

= (mass )/(molar mass)

No of moles of SrCO3 and BaCO3 will be ma/147.631 and mb/197.34 moles respectively

The total amount of C molecules produced by SrCO3 and BaCO3 will be 0.00532 moles of C

The second equation can be written as

ma/147.631 + mb/197.34= 0.00532 (2)

Solving Equation (1) and (2) simultaneously;

ma = 0.604g; mb = 0.242g

Therefore the percentage of BaCO3 = (mass of BaCO3 )/(mass of sample )*100

= 0.242/(0.846 )*100

= 28.605%

5 0

4 years ago

??? What’s the answer???

inn [45]

Answer:

The first one is physical because the color is changing from clear to red. Red food coloring isn't a chemical lol.

The second is chemical because two chemicals come together to make the color yellow.

4 0

3 years ago

Read 2 more answers

a balloon will get smaller when it's filled with air, when air is heated, or cooled, or when it rises?

lora16 [44]

Assuming the balloon is untied aka the air inside isn't intact.

It will get smaller when it rises.

5 0

3 years ago

Read 2 more answers

Suppose you need to produce 25.0 g of pure copper by electrolysis of a solution of CuSO, (aq), by the reduction, Cu2+ (aq) + 2 e

Lelechka [254]

Answer:

2.5*10^4 secs

Explanation:

Using the formular for obtaining the quantity of charge passed Q= It

We remember that two electrons are transferred in the electrochemical reaction as shown in the image attached.

3 0

4 years ago

A reaction was performed in which 1.400 g of camphor was reduced by an excess of sodium borohydride to make 1.061 g of isoborneo

Mrrafil [7]

Answer:

The theoretical yield is 1.4191 grams of isoborneol. The % yield is 74.77 %

Explanation:

Step 1: Data given

Camphor = C10H16O

sodium borohydride = NaBH4

Isoborneol = C10H18O

Mass of camphor = 1.400 grams

Mass of sodium borohydride = 1.061 grams

Molar mass of camphor = 152.23 g/mol

Molar mass of sodium borohydride = 37.83 g/mol

Molar mass of isoborneol = 154.25 g/mol

sodium borohydride is in excess, this means camphor is the limiting reactant

Step 2: The balanced equation

2C10H16O + NaBH4 → 2C10H18O + NaB

Step 3: Calculate moles of camphor

Moles camphor = mass camphor / molar mass camphor

Moles camphor = 1.400 grams / 152.23 g/mol

Moles camphor = 0.00920 moles

Step 4: Calculate moles of isoborneol

For 2 moles camphor we need 1 mol of NaBH4 to produce 2 moles of isoborneol

For 0.00920 moles camphor we'll have 0.00920 moles isoborneol

Step 5: Calculate mass isoborneol

Mass isoborneol = moles isoborneol * molar mass isoborneol

Mass isoborneol = 0.00920 * 154.25 g/mol

Mass isoborneol = 1.4191 grams (= theoretical yield)

% yield = (actual yield / theoretical yield) *100%

% yield = (1.061 / 1.4191) *100 %

% yield = 74.77 %

The theoretical yield is 1.4191 grams of isoborneol. The % yield is 74.77 %

6 0

3 years ago