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VladimirAG [237]
4 years ago
11

A chemistry student must write down in her lab notebook the concentration of a solution of potassium chloride. The concentration

of a solution equals the mass of what's dissolved divided by the total volume of the solution.Here's how the student prepared the solution:a. The label on the graduated cylinder says: empty weight: 5.55gb. She put some solid potassium chloride into the graduated cylinder and weighed it. With the potassium chloride added, the cylinder weighed 92.7 g.c. She added water to the graduated cylinder and dissolved the potassium chloride completely. Then she read the total volume of the solution from the markings on the graduated cylinder. The total volume of the solution was 23.9 ml.What concentration should the student write down in her lab notebook? Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
almond37 [142]4 years ago
4 0

Answer:

d = 3.65 g/mL

Explanation:

This problem is solved by using the equation:

d= m/V

But we have to be careful with the number of significant figures and number of decimals to report our result.

There are two steps to calculate the density:

1. We should perform a substraction to determine the mass of potassium chloride.

mass KCl =( Weight cylinder + KCl ) - (Weight empty cylinder)

92.7 g - 5.55 g = 87.15 g = 87.2 ( rounded to 1 decimal place)

The rule for addition and substraction is that we round the result  to the number of decimal place  with the least  number of decimals ( 92.7 has one decimal, 5.55 has two)

2. We can now calculate the density by dividing the mass into the volume, but retaining the number of significant figures to the number with the smallest number of significant figures, the rule for multiplication and division.

d= m/V = 87.2 g / 23.9 mL = 3.648 g/mL

87.2 has three significant figures and so does 23.9, so we have to round to 3 significant figures.

The rule here is that if the left most digital to be dropped is greater o equal to 5, we round to the nearest higher digit, so 8 is greater than 5 and we rounded up 3.648 to 3.65.

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Answer:

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7 0
3 years ago
PLZ HELP *NO LINKS*
Radda [10]

Answer: (1). There are  0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP.

(2). The density of gaseous arsine is 3.45 g/L.

Explanation:

1). At STP the pressure is 1 atm and temperature is 273.15 K. So, using the ideal gas equation number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\1 atm \times 0.372 L = n \times 0.0821 L atm/mol K \times 273.15 K\\n = 0.0165 mol

2). As number of moles are also equal to mass of a substance divided by its molar mass.

So, number of moles of Arsine (AsH_{3}) (molar mass = 77.95 g/mol) is as follows.

No. of moles = \frac{mass}{molar mass}\\0.0165 mol = \frac{mass}{77.95 g/mol}\\mass = 1.286 g

Density is the mass of substance divided by its volume. Hence, density of arsine is calculated as follows.

Density = \frac{mass}{volume}\\= \frac{1.286 g}{0.372 L}\\= 3.45 g/L

Thus, we can conclude that 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP and the density of gaseous arsine is 3.45 g/L.

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3 years ago
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3 years ago
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lozanna [386]

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Paladinen [302]

The question is incomplete, here is the complete question:

Gaseous compound Q contains only xenon and oxygen. When a 0.100 g sample of Q is placed in a 50.0-mL steel vessel at 0°C, the pressure is 0.230 atm. What is the likely formula of the compound?

A. XeO

B. XeO_4

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D. Xe_2O_3

E. Xe_3O_2

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<u>Explanation:</u>

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w = Weight of the gas = 0.100 g

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R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gas = 0^oC=273K

Putting value in above equation, we get:

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The compound having mass as calculated is XeO_4

Hence, the chemical formula of the compound is XeO_4

5 0
4 years ago
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