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VMariaS [17]
3 years ago
14

Convert 7.8 moles of carbon tetrafluoride into grams.

Chemistry
1 answer:
Vesnalui [34]3 years ago
6 0

Answer:

686.43363984 is the answer when 7.8 moles is converted.

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notsponge [240]

Answer: Ca

Explanation:

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3 years ago
How many molecules does Ba3(PO4)2 contain
Natalija [7]

Answer:

Molecular mass of Ba3(PO4)2 = 3 × 137.5 + 2 [31 + 4 × 16] = 602.5

Explanation:

hope this helps

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6 0
2 years ago
How many grams (m) of glucose are in 165 ml of a 5.50% (m/v) glucose solution? express your answer numerically in grams?
Leya [2.2K]
     <span> mass of glucose = 0.055 *165 = 9.075 g

vol of methyl alc = 0.185 * 1.87 = 0.346 L = 346 ml

% NaCl ( m/v ) = mass NaCl * 100/ vol of soln

or Vol of Soln = mass NaCl / % NaCl (m/v)

= 32.1 * 100 / 6 = 535 ml the total vol of soln</span>
6 0
3 years ago
I need to know what the answer to number 4 letter a.
QveST [7]
I believe it is C. I hope
4 0
3 years ago
Read 2 more answers
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
3 years ago
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