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2 years ago

A reaction produces 92.50 g FeSO4. How many grams of CuSO4 are necessary for this to occur?

Chemistry

2 answers:

KIM [24]2 years ago

5 0

Answer : The mass of $CuSO_4$ needed will be, 97.2 grams.

Explanation : Given,

Mass of $FeSO_4$ = 92.50 g

Molar mass of $FeSO_4$ = 151.908 g/mole

Molar mass of $CuSO_4$ = 159.609 g/mole

First we have to calculate the moles of $FeSO_4$ .

$\text{Moles of }FeSO_4=\frac{\text{Mass of }FeSO_4}{\text{Molar mass of }FeSO_4}=\frac{92.50g}{151.908g/mole}=0.609moles$

Now we have to calculate the moles of $CuSO_4$ .

The balanced chemical reaction is,

$Fe+CuSO_4\rightarrow Cu+FeSO_4$

From the balanced reaction we conclude that

As, 1 mole of $FeSO_4$ obtained from 1 mole of $CuSO_4$

So, 0.609 moles of $FeSO_4$ obtained 0.609 moles of $CuSO_4$

Now we have to calculate the mass of $CuSO_4$ .

$\text{Mass of }CuSO_4=\text{Moles of }CuSO_4\times \text{Molar mass of }CuSO_4$

$\text{Mass of }CuSO_4=(0.609mole)\times (159.609g/mole)=97.2g$

Therefore, the mass of $CuSO_4$ needed will be, 97.2 grams.

xxMikexx [17]2 years ago

3 0

Fe+CuSO4⟶Cu+FeSO4

Given that

FeSO4 = 92.50 g

Number of moles = amount in g / molar mass

=92.50 g / 151.908 g/mol

=0.609 moles FeSO4

Now calculate the moles of CuSO4 as follows:

0.609 moles FeSO4 * 1 mole CuSO4 /1 mole FeSO4

= 0.609 moles CuSO4

Amount in g = number of moles * molar mass

= 0.609 moles CuSO4 * 159.609 g/mol

= 97.19 g CuSO4

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2 years ago

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A certain gas is present in a 12.0 L cylinder at 4.0 atm pressure. If the pressure is increased to 8.0 atm the volume of the gas

diamong [38]

There are 3 parts in this question:
1) To find the initial Boyle's constant $k_{i}$
2) To find the final Boyle's constant $k_{f}$
3) To verify whether gas is obeying Boyle's law or not

Given data:
The initial volume of the cylinder(in litres) = $V_{i}$ = 12.0 L
The initial pressure(in atmospheric pressure) = $P_{i}$ = 4.0 atm

The final pressure(in atmospheric pressure) = $P_{f}$ = 8.0 atm
The final volume of the cylinder(in litres) = $V_{f}$ = 6.0 L

First you need to know what Boyle's law is:
<span>Boyle's law states that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.
</span>
The Mathematical form of Boyle's law is:
$P = \frac{k}{V}$

Where,
P = Pressure
V = Volume of the gas
k = Boyle's constant

Now let's solve aforementioned parts one by one:
1.
The initial volume of the cylinder(in litres) = $V_{i}$ = 12.0 L
The initial pressure(in atmospheric pressure) = $P_{i}$ = 4.0 atm
The Boyle's constant = $k_{i}$ = ?

According to the Boyle's law,

$P_{i} = \frac{k_{i}}{V_{i}}$

=> $k_{i}$ = $P_{i}V_{i}$
Plug-in the values in the above equation, you would get:
$k_{i}$ = 4.0 * 12.0 = 48

Ans-1) $k_{i}$ = 48

2.
The final pressure(in atmospheric pressure) = $P_{f}$ = 8.0 atm
The final volume of the cylinder(in litres) = $V_{f}$ = 6.0 L
The Boyle's constant = $k_{f}$ = ?

According to the Boyle's law,

$P_{f} = \frac{k_{f}}{V_{f}}$

=> $k_{f}$ = $P_{f}V_{f}$
Plug-in the values in the above equation, you would get:
$k_{f}$ = 8.0 * 6.0 = 48

Ans-2) $k_{f}$ = 48

3.
In order to verify Boyle's law, the initial Boyle's constant should be EQUAL to the final Boyle's constant, meaning:

$k_{i}$ = $k_{f}$

Since,
$k_{i}$ = 48
$k_{f}$ = 48

Therefore,
48=48.

Ans-3) Hence proved: The gas IS obeying the Boyle's law.

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Answer:

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Calculate the mass of olive oil in a 750mL bottle if the density of olive oil is 0.92 g/mL. *Use the GRASP method.

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The mass of olive oil : 690 g

<h3>Further explanation </h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density

m = mass

v = volume

Given

volume of olive oil = V=750 ml

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Required

the mass of olive oil

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the formula :

$\tt m=\rho\times V$

Solution

$\tt m=0.92\times 750\\\\m=690~g$

Paraphrase

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