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deff fn [24]
3 years ago
8

A sample of tap water contains 1.95 x 10^3 ppm caco3. assuming the density of the tap water is 1.00 g/ml, calculate the mass of

calcium carbonate in 2.0 l of water
Chemistry
1 answer:
Pavlova-9 [17]3 years ago
7 0
3.9 grams CaCO3  
The mass of 2.0 L of water with a density of 1.00 g/ml is 2000 grams.And 1 ppm of that is 2000 / 1000000 = 0.002 grams. So just multiply by the ppm of CaCO3, giving 0.002 g * 1.95x10^3 = 3.90 grams.  
Since the least accurate datum we have is 2 significant figures, the result should be rounded to 2 significant figures, giving 3.9 grams.
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What volume of Co2 (carbon (iv) oxide)
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Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL

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<u>Answer: </u>The volume of the solution is 85.7 mL

<u>Explanation:</u>

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (mL)}} .....(1)

We are given:

Molarity of solution = 0.600 M

Given mass of BaSO_4 = 12.00 g

We know, molar mass of BaSO_4=[(1\times 137.33)+(1\times 32.07)+(4\times 16)]=233.4g/mol

Putting values in equation 1, we get:

0.600=\frac{12.00\times 1000}{233.4\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{12.00\times 1000}{233.4\times 0.600}=85.68mL=85.7mL

The rule of significant number that is applied for the problems having multiplication and division:

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

Here, the least number of significant figures is 3 that is determined by the number, 0.600. Thus, the answer must have these many significant figures only.

Hence, the volume of the solution is 85.7 mL

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