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Dmitrij [34]
3 years ago
10

Copper (II) sulfate pentahydrate may be heated to drive off the water of hydration. If 5 g of water are produced, what was the o

riginal weight of the copper (II) sulfate pentahydrate (CuSO4 • 5H2O)? CuSO4 • 5H2O → CuSO4 + H2O. The answer given was 13.9g. Can someone explain how this answer is found?

Chemistry
1 answer:
____ [38]3 years ago
6 0

Answer:

13.9g of CuSO4•5H2O

Explanation:

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

CuSO4•5H2O → CuSO4 + 5H2O

Next, we shall determine the mass of CuSO4•5H2O heated and the mass of H2O produced from the balanced equation.

This is illustrated below

Molar mass of CuSO4• 5H2O = 63.5 + 32 + (16x4) + 5(2x1 + 16)

= 63.5 + 32 + 64 + 5(18) = 249.5g/mol

Mass of CuSO4•5H2O from the balanced equation = 1 x 249.5 = 249.5g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 5 x 18 = 90g

From the balanced equation above,

249.5g of CuSO4•5H2O produced 90g of H2O.

Now, we can determine the mass of CuSO4•5H2O needed to produce 5g of H2O. This can be achieved as shown below:

From the balanced equation above,

249.5g of CuSO4•5H2O produced 90g of H2O.

Therefore, Xg of CuSO4•5H2O will produce 5g of H2O i.e

Xg of CuSO4•5H2O = (249.5 x 5)/90

Xg of CuSO4•5H2O = 13.9g

Therefore, 13.9g of CuSO4•5H2O is needed to produce 5g of H2O.

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Chloroform flows through a 4.26 inch inside-diameter pipe at the rate of 3.60 gallons per minute. What is the average velocity o
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Answer:

1) 0,081 ft/s

2) 0,746 lb/s

Explanation:

The relation between flow and velocity of a fluid is given by:

Q=Av

where:

  • Q, flow [ft3/s]
  • A, cross section of the pipe [ft2]
  • v, velocity of the fluid [ft/s]

1)

To convert our data to appropiate units, we use the following convertion factors:

1 ft=12 inches

1 ft3=7,48 gallons

1 minute=60 seconds

So,

Q=\frac{3,60 gallons}{1 min}*\frac{1min}{60 s}*\frac{1ft3}{7,48gallons}=0,00802 \frac{ft3}{s}

As the pipe has a circular section, we use A=πd^2/4:

d=4,26 inch *\frac{1ft}{12 inch}=0,355ft\\  A=\pi \frac{0,355^{2} }{4}=0,0989ft2

Finally:

Q=vA......................v=Q/A

v=\frac{0,00802ft^{3} /s}{0,0989ft^{2} }=0,081ft/s

2)

The following formula is used to calculate the specific gravity of a material:

SG = ρ / ρW  

where:

  • SG = specific gravity,
  • ρ = density of the material [lb/ft3]
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then:

ρ = SG*ρW   = 1,49* 62,4 lb/ft3 = 93 lb/ft3

To calculate the mass flow, we just use the density of the chloroform in lb/ft3 to relate mass and volume:

0,00802 \frac{ft3}{s}*\frac{93lb}{1ft3}=0,746lb/s

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Answer:

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Explanation:

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where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R  is the general gas constant,

T is the temperature of the gas in K.

  • If n and T are constant, and have different values of P and V:

<em>(P₁V₁) = (P₂V₂)</em>

<em></em>

Knowing that:

V₁  =  4.0 L, P₁ = 2.0 atm,

V₂  =  ??? L, P₂ = 6.0 atm.

  • Applying in the above equation

(P ₁V₁) = (P₂V₂)

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