PLEASE ANSWER

one mole of sodium chloride is added to 100 ml of water in beaker a. one mole of glucose, c6h12o6, is added to 100 ml of water i

Viefleur [7K]

Twice as much more will the freezing point of water be lowered in beaker a than in beaker b.

<h3>What determines freezing point?</h3>

A liquid's freezing point rises if the intermolecular interactions between its molecules are strong. The freezing point, however, drops if the molecules of inter - molecular are minimal. The process through which a substance transforms from a liquid into a solid is known as freezing.

<h3>How significant is freezing point?</h3>

Freezing points play a big role in occupational safety. A chemical may perhaps turn harmful if held below its freezing point. A critical safety benchmark for assessing the effects of worker exposure to cold environments is the freezing point.

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5 0

1 year ago

11. What is the mole ratio of C to B in the following generic chemical reaction?2A + 3B → C + 4D (correct numbers 1 point, corre

krok68 [10]

Is this for the honors unit 6 chem test?

4 0

3 years ago

After distilling your crude methyl benzoate, you set aside 5.12 grams of the purified ester. You then prepare the grignard reage

marysya [2.9K]

Answer:

Percent yield is 54%

Explanation:

The Grignard reaction is:

Mg + bromobenzene → phenylmagnesium bromide

Moles of Mg are:

2,3g×(1mol/24.3g) = 0.095 moles Mg.

Moles of bromobenzene are:

9,45mL×(1.495g/1mL)×(1mol/157.01g)= 0.0900 of bromobenzene.

The limiting reactant is bromobenzene. That means moles of phenylmagnesium bromide are 0.0900 moles.

To produce triphenyl carbinol you require:

2 phenylmagnesium bromide + methyl benzoate → triphenyl carbinol

Moles of methyl benzoate are:

5.12g×(1mol/136,15g) = 0.0376 moles of methyl benzoate.

The complete reaction of 0.0376 moles of methyl benzoate requires:

0.0376 moles of methyl benzoate× $\frac{2mol PhenylmagnesiumBromide}{1molMethylBenzoate}$ = 0.0752 moles of phenylmagnesium bromide. As you have 0.900 moles of phenylmagnesium bromide, limiting reagent is Methyl benzoate and moles of triphenyl carbinol are 0.0376. In grams:

0.0376 moles of triphenyl carbinol×(260.33g/mol) = 9,79g of triphenyl carbinol <em>-This is the teorethical yield- </em>

Percent yield is the ratio of actual yield to the theoretical yield. That is:

5,3g / 9,79g ×100 = <em>54%</em>

I hope it helps!

3 0

3 years ago

250.0 mL of 0.250 M calcium chloride is mixed with 440.0 mL of 0.155 M sodium hydroxide and a precipitation reaction occurs. Wha

Vladimir [108]

Answer:

Solid: 2.52 g

Concentrations: [CaCl₂] = 0.041 M, [NaCl] = 0.100 M

Explanation:

When calcium chloride (CaCl₂) reacts with sodium hydroxide (NaOH), a double replacement reaction occurs, forming NaCl and Ca(OH)₂. NaCl is a soluble salt, and Ca(OH)₂ is a little soluble base, thus, Ca(OH)₂ will be the precipiate.

The balanced reaction equation is:

CaCl₂(aq) + 2NaOH(aq) → 2NaCl(aq) + Ca(OH)₂(s)

The number of moles of the reactants mixed are their volume multiplied by their concentration:

nCaCl₂ = 0.250 L * 0.25 mol/L = 0.0625 mol

nNaOH = 0.440 L*0.155 mol/L = 0.0682 mol

One of the reactants is limiting, and the other is in excess. Let's suppose that CaCl₂ is limiting, then, by the stoichiometry:

1 mol of CaCl₂ ------------- 2 moles of NaOH

0.0625 mol ------------ x

By a simple direct three rule:

x = 0.125 mol of NaOH

Because there's less NaOH than the value found, NaOH must be the limiting reactant and CaCl₂ is in excess. Thus, by the stoichiometry:

1 mol of CaCl₂ ------------- 2 moles of NaOH

x ------------- 0.0682 mol

By a simple direct three rule:

2x = 0.0682

x = 0.0341 mol of CaCl₂ reacts

The number of moles of CaCl₂ that remains is: 0.0625 - 0.0341 = 0.0284 mol. The final volume is 250.0 mL + 440.0 mL = 690. mL = 0.69 L

[CaCl₂] = 0.0284/0.69 = 0.041 M

For the solube product:

2 moles of NaOH ------------ 2 moles of NaCl

0.0682 mol ------------ x

x = 0.0682 mol of NaCl formed

[NaCl] = 0.0682/0.69 = 0.100 M

For the precipitate:

2 moles of NaOH ----------- 1 mol of Ca(OH)₂

0.0682 mol ---------- x

x = 0.0341 mol of Ca(OH)₂ formed

The molar of Ca(OH)₂ is 74.0 g/mol, and the mass is the number of moles multiplied by the molar mass:

mCa(OH)₂ = 0.0341*74 = 2.52 g

8 0

3 years ago

If you had 240 L container at 479 k and 300 kpa, what would the volume be if you changed the conditions to STP

nydimaria [60]

Answer:

The answer to your question is V2 = 434.7 l

Explanation:

Data

Volume 1 = V1 = 240 l Volume 2 = ?

Temperature 1 = T1 = 479°K Temperature 2 = T2 = 293°K

Pressure 1 = P1 = 300 KPa Pressure 2 = P2 = 101.325 Kpa

Process

1.- Use the combined gas law to solve this problem

P1V1/T1 = P2V2/t2

-Solve for V2

V2 = P1V1T2 / T1P2

2.- Substitution

V2 = (300)(240)(293) / (479)(101.325)

3.- Simplification

V2 = 21096000 / 48534.675

4.- Result

V2 = 434.7 l

6 0

3 years ago