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3 years ago

Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. K2SO4 NH4I CoCl3

1 answer:

8 0

The substance that releases the greatest amount of ions will have the greatest attractive forces within its solution, resulting in a reduced freezing point.
K₂SO₄ yields 3 ions
NH₄I yields 2 ions
CoCl₃ yields 4 ions

Freezing points:
CoCl₃ < K₂SO₄ < NH₄I

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Calculate the molar solubility of silver(i) bromate with ksp = 5. 5×10-5. also, convert the molar solubility to the solubility.

Sindrei [870]

The molar solubility is 7.4× $10^{-3}$ M and the solubility is 7.4× $10^{-3}$ g/L .

Calculation ,

The dissociation of silver bromide is given as ,

$AgBr$ → $Ag ^{+}$ + $Br^{-}$

- S S

Ksp = [ $Ag ^{+}$ ] [ $Br^{-}$ ] = [S] [ S ] = $S^{2}$

S = √ Ksp = √ 5. 5× $10^{-5}$ = 7.4× $10^{-3}$

The solubility =7.4× $10^{-3}$ g/L

The molar solubility is the solubility of one mole of the substance.

Since , one mole of $AgBr$ is dissociates and form one mole of each $Ag ^{+}$ and $Br^{-}$ ion . So, solubility is equal to molar solubility but unit is different.

Molar solubility = 7.4× $10^{-3}$ mol/L = 7.4× $10^{-3}$ M

To learn more about molar solubility ,

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7 0

1 year ago

What is the conjugate acid in the following equation hbr + H2O yields h30 positive + BR negative

kirill115 [55]

Answer:

HBr + H2O = H3O+ + Br-

So our conjugate acid is the H3O+ to H2O

Explanation:

A conjugate acid of a base results when the base accepts a proton.

Consider ammonia reacting with water to form an equilibrium with ammonium ions and hydroxide ions:

NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)

Ammonium, NH4+, acts as a conjugate acid to ammonia, NH3.

3 0

3 years ago

You are planning an investigation that explores properties of matter.

Elena-2011 [213]

Answer:

Placing a powder into a beaker that contains liquid, resulting the beaker to get hotter.

Explanation:

Physical property is something that you can observe that does not affect the mixture/solution/substance, and that includes temperature

7 0

3 years ago

In the laboratory, a student dilutes 13.5 mL of a 11.6 M hydroiodic acid solution to a total volume of 100.0 mL. What is the con

igomit [66]

Answer: The concentration of the diluted solution is 1.566M.

Explanation:

The dilution equation is presented as this: $M_{s} V_{s} =M_{d} V_{d}$ .

·M= molarity (labeled as M)

·V= volume (labeled as L)

·s= stock solution (what you started with)

·d= diluted solution (what you have after)

Now that we know what each part of the formula symbolizes, we can plug in our data.

$11.6M*13.5mL=M_{d} *100mL$

We cannot leave it like this because the volumes must be in Liters, not milliliters. To convert this, we divide the milliliters by 1000.

$13.5mL/1000=0.0135L$ $100mL/1000=0.1L$

Now that we have the conversions, let's plug them into the equation.

$11.6M*0.0135L=M_{d} *0.1L$

The only thing that we need to do now is actually solving the answer.

$M_{d} =\frac{11.6M*0.0135L}{0.1L}$ $M_{d} =1.566M$

From the work shown above, the answer is 1.566M.

I hope this helps!! Pls mark brainliest :)

7 0

2 years ago

What fuel source is Jan using if she exercises at 85% of her maximum aerobic capacity?

Marina86 [1]

Answer:

Carbohydrates

Explanation:

Increased exercise intensity means the overall need for energy increases. As we increase exercise intensity we increase our glucose uptake and oxidation which far exceeds uptake, indicating that muscle stores of glycogen are being used. At moderate intensities (65%) there is an increased need for muscle glycogen and muscle triglycerides which is fat. At higher levels of intensities (85%) there is an even greater need for energy, and this is met almost solely by an increased uptake of glucose from the blood and from muscle glycogen.

In the case of fats as an energy fuel source at high intensities, increasing levels of intensity increases fat oxidation but once we get into higher levels of intensity, we return to levels of fat oxidation similar to very low intensities.

4 0

3 years ago