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prohojiy [21]
3 years ago
13

Which liquid would be the most difficult to raise or lower the temperature of

Chemistry
2 answers:
vaieri [72.5K]3 years ago
7 0
I think it’s A. One with low density
Pavel [41]3 years ago
5 0

Answer:i dont know

Explanation:

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Which do you think has more mass, 1 mole of fructose, 1 mole of sucrose or 1 mole of aspartame? Explain your reasoning.
pogonyaev

Answer:

i think they would be all the same

Explanation:

they sound like sugars

6 0
3 years ago
Read 2 more answers
PbSO4 has a Ksp = 1.3 * 10-8 (mol/L)2.
Oduvanchick [21]

i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:

\mathrm{PbSO_{4}}(aq) \rightleftharpoons \mathrm{Pb^{2+}}(aq)+\mathrm{SO_4^{2-}}(aq).

---

ii. Given the dissolution of some substance

xA{(s)} \rightleftharpoons yB{(aq)} + zC{(aq)},

the Ksp, or the solubility product constant, of the preceding equation takes the general form

K_{sp} = [B]^y [C]^z.

The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.

So, given our dissociation equation in question i., our Ksp expression would be written as:

K_{sp} = \mathrm{[Pb^{2+}] [SO_4^{2-}]}.

---

iii. Presumably, what we're being asked for here is the <em>molar </em>solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).

We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:

1.3 \times 10^{-8} = s \times s = s^2 \\s = \sqrt{1.3 \times 10^{-8}} = 1.14 \times 10^{-4} \text{ mol/L}.

So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.

8 0
3 years ago
NEED HELP QUICK!!!!!!!!!!!!!!
Zigmanuir [339]
The answer is Index fossils its the only answer it could be

3 0
3 years ago
A solution containing an equal number of hydrogen ions and hydroxide ions isA. basicB. alkalineC. acidicD. neutral.
Goryan [66]

Answer:

Neutral (Option D)

Explanation:

[H₃O⁺] = [OH⁻]  → Neutral solution   pH = 7

[H₃O⁺] < [OH⁻]  → Alkaline / Basic solution   pH > 7

[H₃O⁺] > [OH⁻] → Acidic solution   pH < 7

5 0
3 years ago
If 4.50 g of HCl are reacted with 15.00 g of Caco, according to the following balanced chemical equation, calculate the theoreti
Tom [10]

Answer: HCl is the limiting reactant and the theoretical yield is 2.72 g of CO2. If the actual yield was 2.50 g then, the percent yield is 92.0% when rounding off is done only for the final answer.  

Further Explanation:

In order to determine the theoretical yield and the percent yield of CO2, the following steps must be done:

  1. Determine the limiting reactant. This is the reactant that will determine the amount of CO2 that will actually form.
  2. Determine the theoretical yield for CO2 when the limiting reactant is used.
  3. Get the percent yield by getting the ratio of the actual yield stated in the problem and the calculated theoretical yield multiplied by 100.

Determining the Limiting Reactant

The Limiting Reactant (LR) will produce fewer moles of the products. To check  which of the reactants HCl or CaCO3 is the LR, we do dimensional analysis:

For HCl:

moles\ CO_{2}\ = (4.50\ g\ HCl)\(\frac{1\ mol\ HCl}{36.46094\ g})( \frac{1\ mol\ CO_{2} }{2\ mol\ HCl}) \\moles\ CO_{2}\ =\ 0.0617098

For CaCO3:

moles\ of\ CO_{2}\ = (15.00\ g\ CaCO_{3})\ (\frac{1\ mol\ CaCO_{3} }{100.0869\ g\ CaCO_{3} })\ (\frac{1\ mol\ CO_{2} }{1\ mol\ CaCO_{3} })\\moles\ of\ CO_{2}\ = \ 0.1499

Since HCl produces fewer moles of CO2, then it is the limiting reactant. We will use the given amount to determine the theoretical yield for CO2.

Determining the Theoretical Yield

From Step 1, we know that 0.0617098 moles of CO2 will be produced. We will just convert this to grams.

grams\ CO_{2}\ =\ (0.0617098\ mol\ CO_{2})  (\frac{44.01\ g\ CO_{2}}{1\ mol\ CO_{2}})\\grams\ CO_{2}\ =\ 2.71585

Since the answer only requires 3 significant figures, the final answer is 2.72 grams CO2.

Determining the Percent Yield

Dividing the actual yield by the theoretical yield will give us the percent yield, which is an indicator of how efficient the experiment or the method used was.

From the problem, the actual yield was 2.50 g, hence, the percent yield is:

percent\ yield\ of\ CO_{2}\ = (\frac{2.50\ g}{2.71585\ g}) (100)\\percent\ yield\ of\ CO_{2}\ = 92.05221

Rounding off to three significant figures, the percent yield is 92.0%. This suggests that the method used is somewhat efficient in producing CO2.

Learn More

  1. Learn More about Limiting Reactant brainly.com/question/7144022
  2. Learn More about Excess Reactant brainly.com/question/6091457  
  3. Learn More about Stoichiometry brainly.com/question/9743981

Keywords: stoichiometry, theoretical yield, actual yield

3 0
3 years ago
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