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3 years ago

Which compound has the highest percent composition by mass of strontium?

2 answers:

5 0

The answer is (3). The number of Sr is the same so if the compound has the smallest gram formula mass, it has the highest percent composition by mass of strontium. So the answer is (3).

lesya [120]3 years ago

4 0

Answer: option (3) SrO, with 84.56%

Explanation:

You have to calculate the mass of strontium and the mass of the compound, for each of the four compounds, and compute the percent:

percent of strontium = (mass of strontium / mass of the chemical formula) × 100 %.

These are the atomic masses of the elements that you have to use:

Atomic mass of Sr: 87.62 g/mol

Atomic mass of Cl: 35.453 g/mol

Atomic mass of I: 126.904 g/mol

Atomic mass of O: 15.999 g/mol

Atomic mass of S: 32.065 g/mol

(1) SrCl₂:

i) mass of Sr: 87.62 g/mol

ii) mass of Cl₂: 2×35.453 g/mol = 70.906 g/mol

iii) mass of SrCl₂: 87.62 g/mol + 70.906 g/mol = 158.526 g/mol

iv) Sr % = [87.62 / 158.526 ] × 100 = 55.27%

(2) SrI₂

Following the same procedure:

i) Sr % = [87.62 / (87.62 + 2× 126.904) ] × 100 = 25.66%

(3) SrO

i) Sr % = [87.62 / (87.62 + 15.999) ] × 100 = 84.56%

(4) SrS

i) Sr % = [87.62 / (87.62 + 32.065) ] × 100 = 73.21%

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What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m

djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL (1cm³ = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³ mol / 0.4327 L = 0.0075 M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x 432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0

3 years ago

A cylinder with a movable piston originally has a volume of 2805 mL and is filled with nitrogen to a pressure of 4.00

Sergio039 [100]

This problem is providing the initial volume and pressure of nitrogen in a piston-cylinder system and asks for the final pressure it will have when the volume increases. At the end, the answer turns out to be 2.90 atm.

<h3>Boyle's law</h3>

In chemistry, gas laws are used so as to understand the volume-pressure-temperature-moles behavior in ideal gases and relate different pairs of variables.

In this case, we focus on the Boyle's law as an inversely proportional relationship between both pressure and volume at constant both temperature and moles:

$P_1V_1=P_2V_2$