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3 years ago

How many moles of potassium hydroxide are needed to completely react with 2.94 moles of aluminum sulfate

Chemistry

1 answer:

ArbitrLikvidat [17]3 years ago

6 0

Answer:- Third choice is correct, 17.6 moles

Solution:- The given balanced equation is:

Al_2(SO_4)_3+6KOH\rightarrow 2Al(OH)_3+3K_2SO_4

We are asked to calculate the moles of potassium hydroxide needed to completely react with 2.94 moles of aluminium sulfate.

From the balanced equation, there is 1:6 mol ratio between aluminium sulfate and potassium hydroxide.

It is a simple mole to mole conversion problem. We solve it using dimensional set up as:

2.94molAl_2(SO_4)_3(\frac{6molKOH}{1molAl_2(SO_4)_3})

= 17.6 mol KOH

So, Third choice is correct, 17.6 moles of potassium hydroxide are required to react with 2.94 moles of aluminium sulfate.

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Why were microscopes essential in the discovery of cells

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Answer: hey nice to see you again, the answer is B. Microscopes allowed scientists to see the cells.

Explanation:

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2 years ago

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Using the same sample of gas (P1 = 565 torr , T1 = 27 ∘C ), we wish to change the pressure to 5650 torr with no accompanying cha

Stels [109]

Answer:

2726.85 °C

Explanation:

Given data:

Initial pressure = 565 torr

Initial temperature = 27°C

Final temperature = ?

Final pressure = 5650 torr

Solution:

Initial temperature = 27°C (27+273 = 300 K)

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

T₂ = P₂T₁ /P₁

T₂ = 5650 torr × 300 K / 565 torr

T₂ = 1695000 torr. K /565 torr

T₂ = 3000 K

Kelvin to degree Celsius:

3000 K - 273.15 = 2726.85 °C

4 0

2 years ago

A chemistry student needs 45.0mL of pentane for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the stud

Masja [62]

The mass of pentane the student should weigh out is

The density of pentane is 0.626 gcm-3

To calculate the mass of pentane following expression is used,

(Density is defined as the mass divide by volume)

Density = mass / volume

mass of pentane = Density of pentane * Volume of pentane

mass of pentane = 0.626 gcm-3 * 45.0 mL

= 28.17 g

Here the unit of mass of pentane is g,

However the unit of density is gcm-3 and unit of volume is mL i.e. cm3

Hence, Mass = gcm-3 * cm3

Mass = g

The mass of pentane the student should weigh out is 28.17g

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5 0

1 year ago

Help please I really really need it now

postnew [5]

1. A 2. B 5. B this is all I know hope it helps

5 0

2 years ago

Please help me, I really don't want to fail but I don't know how to do this

Brrunno [24]

Answer:

<u>12 moles</u>

Explanation:

$NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)$

__↑______↑

8.00 mol | 14.00 mol

________________

$NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)$

You can turn this into a system of variables which are solvable.

To do this, create variables for the coefficients of each compound in the reaction respectively.

$a(NH_{3}(g)) + b(O_{2}(g)) → \\c(NO_{2}) + d(H_{2}O(g))$

Because to be balanced, the count of atoms in each element of the compound correspond to the coefficient of the variable in that compound so that the count of the left (reactant) side is set equal to the right (product) side.

a corresponds to the coefficient of the first compound, b corresponds to the coefficient of the second compound, c corresponds to the coefficient of the third compound, and d corresponds to the coefficient of the fourth compound.

(Reactant = Product)

Reactant: 1a [N] Product: 1c.

Reactant: 3a [H] Product: 2d.

Reactant: 2b [O] Product: 2c + 1d.

Thus the system is:

1a = 1c

3a = 2d

2b = 2c + 1d.

Then just use the substitution methods to solve.

3 0

2 years ago