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2 years ago

Convert 7.8 liters of carbon tetra fluoride cfa to grams

1 answer:

4 0

To answer the question, we assume that the given compound is an ideal gas that at STP, one mole of the substance will occupy 22.4 L. From the given volume, we determine the number of moles of substance.
7.8 L / (22.4 L /mole) = 0.3482 moles of cfa
Then, we multiply this number of moles by the molar weight of cfa which is equal to 88 g/mol.
Multiplying,
weight = (0.3482 moles of cfa) x (88 g/mol) = 30.64 grams

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The gas described in parts a and b has a mass of 1.66 g. the sample is most likely which monatomic gas?

nordsb [41]

Mass of the gas m = 1.66
The calculated temperature T = 273 + 20 = 293
We have to calculate molar mass to determine the gas
Molar Mass = mRT / PV
M = (1.66 x 8.314 x 293) / (101.3 x 1000 x 0.001)
M = 4043.76 / 101.3 = 39.92 g/mol
So this gas has to be Argon Ar based on the molar mass.

7 0

1 year ago

Read 2 more answers

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the a

UkoKoshka [18]

Answer:

$m_{Ga}=0.550gGa$

Explanation:

Hello!

In this case, since the applied current for the 50.0 mins provides the following charge to the system:

$q=0.760\frac{C}{s}*50.0min*\frac{60s}{1min}=2,280C$

As 1 mole of electrons carries a charge of 1 faraday, or 96,485 coulombs, we can compute the moles of electrons involved during the reduction:

$n_{e^-}=2,280C*\frac{1mole^-}{96,485C}=0.0236mole^-$

Then the reduction of Ga³⁺ to Ga involves the transference of three electrons, we are able to compute the moles and therefore the mass of deposited gallium:

$m_{Ga}=0.0236mole^-*\frac{1molGa}{3mole^-}*\frac{69.72gGa}{1molGa} \\\\m_{Ga}=0.550gGa$

Best regards!

5 0

2 years ago

One hour of bicycle racing can require 500-900 kcal of energy, depending on the speed of the race, the terrain, and the weight o

diamong [38]

Answer:

0.467 kilograms of protein or carbohydrates

Explanation:

First, there is a need to understand that 1 kJ = 0.239 kCal

Hence, 17 kJ = 17 x 0.239 = 4.063 kCal

The race requires 650.0 kCal/hr and has to last for 175 minutes.

175 minutes = 175/60 = 2.917 hrs

The total kCal requires for the race = 650 x 2.917 = 1,895.833 kCal

1 g of protein or carbohydrate food produces 17.0 kJ or 4.063 kCal of energy. Hence, the total g of protein or carbohydrate that will produce 1,895.833 kCal of energy would be;

1,895.833/4.063 = 466.609 g

1 g = 0.001 kg

466.609 g = 466.609 x 0.001 = 0.467 kg

Hence, 0.467 kilograms of proteins or carbohydrates must be consumed.

4 0

2 years ago

a drop of water weighing 0.48 g condenses on the surface of a 55-g block of aluminum that is initially at 25C. if the heat durin

Gwar [14]

Mass of water vapor is 0.48 gms.

Weight of the aluminium block is 55 gms.

Heat of vaporization of water at 25 degree Celsius is 44.0 kJ/mol.

The amount of heat given by the condensation is:

q = Heat of vaporization × Mass of vapor / Molar mass of steam

= 44.0 kJ/mol ×0.48 g / 18 g/mol

= 1.173 kJ

Now the final temperature of the metal block is calculated by the formula:

q = m × c × ΔT

q = m × c × (T₂ - T₁)

Here, q is the amount of heat, m is the mass of the metal, c is the specif heat of the metal, T₁ is initial temperature, and T₂ is the final temperature.

Now, substituting the values we get,

1.173 kJ = 55 g × 0.903 J/g. ° C * (T₂ - 25°C)

1.173 × 10³ J = 55 g × 0.903 J/g. degree C × (T₂ - 25°C)

1.173 × 10³ = 49.665 ° C * (T₂ - 25° C)

T₂ = 49° C.

Thus, the final temperature of the metal block is 49° C.

8 0

2 years ago

Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g co2 and 0.6551 g

dlinn [17]

The compound contains Carbon, Hydroxide and Oxide
1 mole of carbon iv oxide contains 44 g, out of which 12 g are carbon.
Therefore, 1.6004 g of CO2 will contain;
1.6004 ×12/44 = 0.4365 g of carbon
1 mole of water contains 18 g of which 2 g is hydrogen,
Therefore, 0.6551 g of H2O will hace ;
0.6551 × 2/18 = 0.0728 g of hydrogen.
The total mass of the compound is 0.8009 g,
Thus the mass of oxygen = 0.8009 -(0.4365 +0.0728)
= 0.2916 g
To get the empirical formula we first get the number of moles of each element;\
Carbon = 0.4365/12= 0.036375 moles
Hydrogen = 0.0728/1 = 0.0728 moles
Oxygen = 0.2916/16 = 0.018225 moles
Then, to get the smallest ratio we divide each with the smallest value;
Carbon : Hydrogen : Oxygen
= (0.036375/0.018225) : (0.0728/0.018225) : ( 0.018225/0.018225)
= 1.996 : 3.995 : 1
≈ 2 : 4 : 1
Therefore, the empirical formula is C2H4O

3 0

2 years ago