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3 years ago

Which has a greater density, a pound of feathers or a pound of gold

Chemistry

1 answer:

artcher [175]3 years ago

3 0

Answer:

Explanation:

Hello!!

Please remember that density=mass/volume.

In this case, we have the same mass for the feathers and the gold (1 pound).

So, if the mass is the same for both we have:

Df*Vg=Dg*Vf where f means feathers and g means gold.

In this case we see density is inversely related to volume. Both materials have the same mass. So which one have a bigger volume?

For this take a pound of feathers and submerge in a known volume of water. Note the displacement. Do the same for a pound of gold. Whichever has the smaller displacement, has the greater density.

In this case, Gold has the greatest density.

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Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.

weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

7 0

3 years ago

A 0.2688 g sample of a monoprotic acid neutralized 16.4 mL of 0.08133 M KOH solution. Calculate the molar mass of the acid

ivanzaharov [21]

Answer:

202 g/mol

Explanation:

Let's consider the neutralization between a generic monoprotic acid and KOH.

HA + KOH → KA + H₂O

The moles of KOH that reacted are:

0.0164 L × 0.08133 mol/L = 1.33 × 10⁻³ mol

The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 × 10⁻³ moles.

1.33 × 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:

0.2688 g/1.33 × 10⁻³ mol = 202 g/mol

3 0

3 years ago

Check the approach to obtain the correct conversion equation. Use the temperature in °F to find the temperature in °C.

Maru [420]

Answer:

( °F − 32) × 5/9 = °C

Explanation:

Also there is a mental calculation to convert from Fahrenheit to Celsius. The ratio 5/9 is approximately equal 0.55555….

Subtract 32º to adapt the equivalent in the Fahrenheit scale.

Divide the degrees Celsius by 2 (multiply by 0.5).

Take 1/10 of this number (0.5 * 1/10 = 0.05) and add it to the number obtained previously.

Example: Convert 98.6º F to Centigrade.

98.6 - 32 = 66.6

66.6 * 1/2 = 33.3

33.3 * 1/10 = 3.3

33.3 + 3.3 = 36.6 which is an approximation in degrees Centigrade

8 0

3 years ago

What volume is occupied by 0.109 molmol of helium gas at a pressure of 0.98 atmatm and a temperature of 307 K

KATRIN_1 [288]

Answer:

2.8 L

Explanation:

From the question given above, the following data were obtained:

Number of mole (n) = 0.109 mole

Pressure (P) = 0.98 atm

Temperature (T) = 307 K

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

The volume of the helium gas can be obtained by using the ideal gas equation as follow:

PV = nRT

0.98 × V = 0.109 × 0.0821 × 307

0.98 × V = 2.7473123

Divide both side by 0.98

V = 2.7473123 / 0.98

V = 2.8 L

Thus, the volume of the helium gas is 2.8 L.

6 0

2 years ago

In the winter , people often buy large bags of rock salt to sprinkle on their walkways . Why do people do this?

slamgirl [31]

The salt causes the water to freeze at a lower temperature. When a solute, aka salt, is introduced to the system, the freezing point is lowered. This makes the water freeze at a lower temperature.

6 0

3 years ago