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2 years ago

If the original sample of Zn had a mass of 1.4 g, what mass of ZnCl2 can be produced in the reaction?

1 answer:

4 0

Zinc react with chlorine to form zinc chloride
Zn + cl2 ---> ZnCl2
calculate the number of moles of Zn
that is mass/molar mass of Zn(65.38g/mol)
moles= 4g/65.38 g/mol= 0.021moles
the reacting ratio of zinc to ZnCl2 is 1:1 from the reaction above
therefore the moles of Zncl2 is also 0.021 moles
mass= number of moles x molar mass
The molar mass of ZNCl2 = 65.38 + (35.2 x2) = 136.38 g/mol
Therefore the mass of Zncl2 = 0. 021 moles x 136.38 g/mol=2.86 grams

You might be interested in

Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.

maria [59]

The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%

5 0

2 years ago

Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t

Dafna11 [192]

Answer:

The plane with aluminium can lift more mass of passangers than the plane of steel.

Explanation:

The total mass the airplane canc lift is:

$m_{tot}=m_{fuselage}+m_{passangers}$

For aluminium:

$m_{tot}=m_{fus-Al}+m_{pas-Al}$

$m_{fus-Al}=\delta _{Al}*V_{fuselage}$

and

$V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]$

where:

L is lenght
D is diameter
e is thickness

$m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}$

For steel (same procedure):

$m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel$

Knowing that the total mass the airplane can lift is constant and that aluminum has a lower density than the steel, we can afirm that the plane with aluminium can lift more mass of passangers.

Also you can estimate an average weight of passanger to estimate a number of passangers it can lift.

5 0

2 years ago

You have two compounds that you have spotted on the TLC plate. One compound is more polar than the other. You ran the TLC plate

goldenfox [79]

Answer:

we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf

Explanation:

when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf

The speed of the polar spot depends largely on the level of polarity, an increase in the polarity will see both spots of Neat hexane run when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate

3 0

2 years ago

If 3.491 grams of the precipitate was formed, how many moles of strontium bromide were reacted

AVprozaik [17]

The balanced chemical equation that represents the reaction is as follows:
SrBr2(aq) + 2AgNO3(aq) → Sr(NO3)2(aq) + 2AgBr(s)

From the periodic table:
mass of silver = 108 grams
mass of bromine = 80 grams

molar mass of silver bromide = 108 + 80 = 188 grams

number of moles = mass / molar mass
number of moles of produced precipitate = 3.491/188 = 0.018 moles

From the balanced equation:
1 mole of strontium bromide produces 2 moles of silver bromide. Therefore, to calculate the number of moles of strontium bromide that produces 0.018 moles of silver bromide, you will just do a cross multiplication as follows:
amount of strontium bromide = (0.018x1) / 2 = 9.28 x 10^-3 moles

4 0

2 years ago

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

2 years ago