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2 years ago

When 1. 0 l of 0. 00010 m naoh and 1. 0 l of 0. 0014 m mgso4 are mixed, would a precipitate be formed? show work

Chemistry

1 answer:

koban [17]2 years ago

5 0

When 1. 0 l of 0. 00010 m NaOH and 1. 0 l of 0. 0014 m mgso4 are mixed, there will be no precipitate formed.

<h3>What is a precipitate?</h3>

The precipitate is the solid concentration of a substance that is collected over a solution.

First, we determine the concentration of magnesium and hydroxide

(Mg2+) = 7.00 × 10⁻⁴

(OH−) = 5.00 × 10⁻⁵

Now, we calculate the solubility quotient

Qc = (Mg2+) (OH−) ²

Qc = 7.00 × 10⁻⁴ x (5.00 × 10⁻⁵)²

Qc = 1.75 x 10⁻¹²

The solubility product of the magnesium hydroxide is 1.80 x 10⁻¹¹ which is more than the solubility quotient. Thus, there will be no precipitate form.

Thus, there will be no precipitate formed because the solubility quotient we calculated is less than the solubility product.

To learn more about precipitate, refer to the below link:

brainly.com/question/16950193

#SPJ4

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How many molecules is 0.360 mole of water?

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Answer: 2.17 x 10^23 molecules

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3 years ago

What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)

Genrish500 [490]

Answer:

34.28 L ( 1.5*22.4 L)

Explanation:

Calculation of the moles of aluminum as:-

Mass = 55 g

Molar mass of aluminum = 26.981539 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{55\ g}{26.981539\ g/mol}$

$Moles= 2.0384\ mol$

According to the reaction:-

$4Al+3O_2\rightarrow 2Al_2O_3$

4 moles of aluminum react with 3 moles of oxygen gas

1 mole of aluminum react with $\frac{3}{4}$ moles of oxygen gas

2.0384 moles of aluminum react with $\frac{3}{4}\times 2.0384$ moles of oxygen gas

Moles of oxygen gas = 1.5288 moles

At STP,

Pressure = 1 atm

Temperature = 273.15 K

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K

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3 years ago

An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at

lana [24]

Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

6 0

3 years ago