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laiz [17]
3 years ago
15

The solubility product for barium sulfate is 1.1 × 10–10. calculate the molar solubility of barium sulfate.

Chemistry
1 answer:
Triss [41]3 years ago
8 0
When it comes to equilibrium reactions in chemistry, there are a lot of equilibrium constants that can be used. In the case of solubility, the appropriate one to use is the equilibrium constant of solubility product denotes as Ksp. This is the concentration of products raised to their coefficients. For example,

 cC ⇔ aA + bB 

Ksp = {[A^a][B^b]}

Now, for the this problem, the reaction is

BaSO₄ ⇔ Ba²⁺ + SO₄²⁻

The reaction is already balanced. Since we don't know the value of Ba²⁺ and SO₄²⁻, let's denote this at x.

1.1 × 10⁻¹⁰ = [x][x] =[x²]
[x] = [Ba²⁺] = [SO₄²⁻] = [BaSO₄] = 1.049 × 10⁻⁵ M
You might be interested in
A graduated cylinder is filled with 50.7 mL of water. If a 10.0 g piece of zinc metal is added to the water, and the volume rise
Triss [41]

Answer:

<h3>The answer is 7.14 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 10 g

volume = final volume of water - initial volume of water

volume = 52.1 - 50.7 = 1.4 mL

We have

density =  \frac{10}{1.4}  \\  = 7.142857...

We have the final answer as

<h3>7.14 g/mL</h3>

Hope this helps you

5 0
3 years ago
Calculate the freezing point of a 12.25 m 12.25m aqueous solution of propanol. Freezing point constants can be found in the list
vichka [17]

Answer:

∆t(f) = 49.755C

Explanation:

Freezing point is defined as the point in which a liquid changes from liquid to solid state. Therefore, to calculate the freezing point of aqueous solution of propanol

∆t = k × I × m

Where k is cryscopic constant =

I = vant Hoff factor = 2

M = molar concentration = 12.25m

∆t (f) = 1.9 × 12.25 × 2 = 49.755C

5 0
4 years ago
A wooden block has a mass of 20 g and dimensions of 5 cm x 1 cm x 1 cm.
marta [7]

Answer:

V = 5 cm³

ρ = 4 g/cm³

Explanation:

Step 1: Calculate the volume (V)

We have a wooden cuboid of dimensions 5 cm × 1 cm × 1 cm. We can calculate its volume using the following expression.

V = 5 cm × 1 cm × 1 cm

V = 5 cm³

Step 2: Calculate the density (ρ)

The density is equal to the mass divided by the volume.

ρ = m / V

ρ = 20 g / 5 cm³

ρ = 4 g/cm³

4 0
3 years ago
When discussing acids and bases, any substance that donates a proton, by definition, is considered a(n)
blsea [12.9K]
Bronsted-Lowry acid is your answer
4 0
3 years ago
Any help would be appreciated. Confused.
masya89 [10]

Answer:

q(problem 1) = 25,050 joules;  q(problem 2) = 4.52 x 10⁶ joules

Explanation:

To understand these type problems one needs to go through a simple set of calculations relating to the 'HEATING CURVE OF WATER'. That is, consider the following problem ...

=> Calculate the total amount of heat needed to convert 10g ice at -10°C to steam at 110°C. Given are the following constants:

Heat of fusion (ΔHₓ) = 80 cal/gram

Heat of vaporization (ΔHv) = 540 cal/gram

specific heat of ice [c(i)] = 0.50 cal/gram·°C

specific heat of water [c(w)] = 1.00 cal/gram·°C

specific heat of steam [c(s)] = 0.48 cal/gram·°C

Now, the problem calculates the heat flow in each of five (5) phase transition regions based on the heating curve of water (see attached graph below this post) ...   Note two types of regions (1) regions of increasing slopes use q = mcΔT and (2) regions of zero slopes use q = m·ΔH.

q(warming ice) =  m·c(i)·ΔT = (10g)(0.50 cal/g°C)(10°C) = 50 cal

q(melting) = m·ΔHₓ = (10g)(80cal/g) 800 cal

q(warming water) = m·c(w)·ΔT = (10g)(1.00 cal/g°C)(100°C) = 1000 cal

q(evaporation of water) =  m·ΔHv = (10g)(540cal/g) = 5400 cal

q(heating steam) = m·c(s)·ΔT = (10g)(0.48 cal/g°C)(10°C) = 48 cal

Q(total) = ∑q = (50 + 800 + 1000 + 5400 + 48) = 7298 cals. => to convert to joules, multiply by 4.184 j/cal => q = 7298 cals x 4.184 j/cal = 30,534 joules = 30.5 Kj.

Now, for the problems in your post ... they represent fragments of the above problem. All you need to do is decide if the problem contains a temperature change (use q = m·c·ΔT) or does NOT contain a temperature change (use q = m·ΔH).    

Problem 1: Given Heat of Fusion of Water = 334 j/g, determine heat needed to melt 75g ice.

Since this is a phase transition (melting), NO temperature change occurs; use q = m·ΔHₓ = (75g)(334 j/g) = 25,050 joules.

Problem 2: Given Heat of Vaporization = 2260 j/g; determine the amount of heat needed to boil to vapor 2 Liters water ( = 2000 grams water ).

Since this is a phase transition (boiling = evaporation), NO temperature change occurs; use q = m·ΔHf = (2000g)(2260 j/g) = 4,520,000 joules = 4.52 x 10⁶ joules.

Problems containing a temperature change:

NOTE: A specific temperature change will be evident in the context of problems containing temperature change => use q = m·c·ΔT. Such is associated with the increasing slope regions of the heating curve.  Good luck on your efforts. Doc :-)

5 0
3 years ago
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