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liraira [26]
2 years ago
15

Osmium (Os) is the densest element known (density =22.57 g/cm^3). Calculate the mass in pounds and in kilograms of an Os sphere

15 cm in diameter (about the size of a grapefruit) (volume of a sphere of radius r is 4/3Ïr^3).
Chemistry
1 answer:
Amiraneli [1.4K]2 years ago
7 0

<u>Answer:</u> The mass of osmium in kilograms and in pounds are 318.9141 kg and 703.085 pounds.

<u>Explanation:</u>

To calculate the volume of sphere, we use the formula:

V=\frac{4}{3}\pi r^3

where,

r = radius of sphere

We are given:

Radius of osmium = 15 cm

Volume of osmium = \frac{4}{3}\times 3.14\times (15)^3=14130cm^3

Density of osmium = 22.57g/cm^3

To calculate mass of a substance, we use the equation:

Density=\frac{Mass}{Volume}

Putting values in above equation, we get:

22.57g/cm^3=\frac{\text{Mass of osmium}}{14130cm^3}\\\\\text{Mass of osmium}=318914.g

To convert the given mass into kilo grams and pounds, we use the conversion factors:

1 kg = 1000 grams

So, 318914.1g=318914.1g\times \frac{1kg}{1000g}=318.9141kg

And,

1 pound = 453.592 g

So, 318914.1g=318914.1g\times \frac{1pound}{453.592g}=703.085pounds

Hence, the mass of osmium in kilograms and in pounds are 318.9141 kg and 703.085 pounds.

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1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur
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Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that P_{1} = 100\,kPa, V_{1} = 500\,mL and V_{2} = 1000\,mL, then the new pressure of the gas is:

P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)

P_{2} = 500\,kPa

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that V_{1} = 3.60\,L, P_{1} = 10\,kPa and P_{2} = 25\,kPa then the new volume of the gas is:

V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)

V_{2} = 1.44\,L

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that \frac{P_{2}}{P_{1}} = 3, then the volume ratio is:

\frac{V_{1}}{V_{2}} = 3

\frac{V_{2}}{V_{1}} = \frac{1}{3}

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

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The rate of disappearance of HBr in the gas phase reaction2HBr(g) → H2(g) + Br2(g)is 0.301 M s-1 at 150°C. The rate of appearanc
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Answer: 0.151

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2HBr(g)\rightarrow H_2(g)+Br_2(g

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}

Given: -\frac{d[HBr]}{dt}]=0.301

Putting in the values we get:

\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}

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choli [55]

Answer:

C = 18.29 g  

Explanation:

Given data:

Mass of beryllium needed = ?

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Solution:

Chemical equation:

3Be + N₂    →    Be₃N₂

now we will calculate the number of moles of nitrogen:

Number of moles = mass/molar mass

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Now we will compare the moles of nitrogen and Be from balance chemical equation.

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                       1          :       3

                  0.675       :      3/1×0.675 = 2.03 mol

Number of moles of Be needed are 2.03 mol.

Mass of Beryllium:

Mass = number of moles × molar mass

Mass = 2.03 mol ×   9.01 g/mol

Mass = 18.29 g  

4 0
3 years ago
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