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Anuta_ua [19.1K]
3 years ago
15

Ions:

Chemistry
1 answer:
kozerog [31]3 years ago
5 0
Chlorine: gain, -1 charge.

iron: lose, +2 charge.

sulfur: 6, gain, -2 charge.

copper: lose, +1 charge.
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Given the equation representing a solution equilibrium: BaSO4(s) <—> Ba2+ (aq) + SO4 2- (aq)
goldenfox [79]

Answer:

The equilibrium shifts to the left, and the concentration of Ba2+(aq) decreases

Explanation:

Whenever a solution of an ionic substance comes into contact with another ionic compound with which it shares a common ion, the solubility of the ionic substance in solution decreases significantly.

In this case, both BaSO4 and Na2SO4 both possess the SO4^2- anion. Hence SO4^2- anion is the common ion. Given the equilibrium;

BaSO4(s) <—> Ba2+ (aq) + SO4 2- (aq), addition of Na2SO4 will decrease the solubility of BaSO4 due to the presence of a common SO4^2- anion compared to pure water.

This implies that the equilibrium will shift to the left, (more undissoctiated BaSO4) hence decreasing the Ba^2+(aq) concentration.

3 0
3 years ago
A certain liquid X has a normal boiling point of 133.60°C and a boiling point elevation constant Kb= 2.46°C kg mol^-1.Calculate
Afina-wow [57]

Answer:

136.63 °C

Explanation:

ΔTb=Tb solution - Tb pure

Where; Tb pure = 133.60°C

molar mass of solute = 121.14 g/mol

number of moles of solute; 52.2g/121.14 g/mol = 0.431 moles

molality = 0.431 moles/350 * 10^-3 = 1.23 molal

Then;

ΔTb = Kb * m * i

Kb = 2.46°C kg mol^-1

m = 1.23 molal

i = 1

ΔTb = 2.46 * 1.23 * 1

ΔTb = 3.03 °C

Hence;

Tb solution = ΔTb + Tb pure

Tb solution = 3.03 °C + 133.60°C

Tb solution = 136.63 °C

6 0
3 years ago
If 75.0 g of a liquid has a volume of 62.4 mL ,calculate the liquid’s density
Galina-37 [17]
<span>To calculate the density of a liquid, you have to first know that density is the amount of substance per unit of volume. In this specific question, density will be found with units of g/mL. Now, the density can be found by dividing the amount of liquid, 75.0g, by the volume, 62.4mL. Doing this we get: 75.0g/62.4mL= 1.2 g/mL as the density of the liquid.</span>
4 0
3 years ago
What is the pH of a neutral solution at 25°C?<br> 7<br> 1
Kaylis [27]

Answer:

I pretty sure it's the pH level 7

8 0
4 years ago
Lead can be prepared from galena [lead(II) sulfide] by first roasting the galena in oxygen gas to form lead(II) oxide and sulfur
Vesna [10]

Answer:

a) Step 1:

2PbS(g)+3O_2(g)\overset{roasting}\rightarrow 2PbO(s)+2SO_2(g)

Step 2:

2PbO(s)+PbS(s)\overset{\Delta }\rightarrow 3Pb(l)+SO_2(g)

b) The overall balanced reaction for given process is ;

3PbS(s)+3O_2(g)\rightarrow 3Pb(l)+3SO_2(g)

Explanation:

a)

Galena = PbS

Lead(II) oxide = PbO

Sulfur dioxide = SO_2

Step 1:

Roasting the galena in oxygen gas to form lead(II) oxide and sulfur dioxide.

Balanced equation of step 1:

2PbS(g)+3O_2(g)\overset{roasting}\rightarrow 2PbO(s)+2SO_2(g)..[1]

Step 2:

Heating the metal oxide with more galena forms the molten metal and more sulfur dioxide.

Balanced equation of step 2:

2PbO(s)+PbS(s)\overset{\Delta }\rightarrow 3Pb(l)+SO_2(g)..[2]

b)

For over all reaction add [1] and [2]. The overall balanced reaction for given process is ;

3PbS(s)+3O_2(g)\rightarrow 3Pb(l)+3SO_2(g)

5 0
4 years ago
Read 2 more answers
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