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DedPeter [7]
3 years ago
5

How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate is combined with 15.0 g of hydrochloric aci

d?
Chemistry
1 answer:
horsena [70]3 years ago
5 0

Answer:

22.7 g of CaCl₂ are produced in the reaction

Explanation:

This is the reaction:

CaCO₃ +  2HCl  →  CaCl₂  + CO₂  +  H₂O

Now, let's determine the limiting reactant.

Let's divide the mass between the molar mass, to find out moles of each reactant.

29 g / 100.08 g/m = 0.289 of carbonate

15 g / 36.45 g/m = 0.411 of acid

1 mol of carbonate must react with 2 moles of acid

0.289 moles of carbonate will react with the double of moles (0.578)

I only have 0.411 of HCl, so the acid is the limiting reactant.

Ratio is 2:1, so I will produce the half of moles, of salt.

0.411 / 2 = 0.205 moles of CaCl₂

Mol . molar mass = mass → 0.205 m . 110.98 g/m = 22.7 g

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Consider the titration of 10.00 mL of a monoprotic weak acid with 0.1234 M NaOH. If the equivalence point volume of NaOH was det
gogolik [260]

The initial concentration of the unknown acid is 0.1900 M.

Explanation:

Titration is a chemical method of analysis to know the concentration and volume of the unknown chemical or analyte.

The formula for the titration is:

Macid x Vacid = Mbase x V base

The volume must be in litres. The volume is given in ml it should be divided with 1000 to obtain values in litre.

Data given are:

volume of acid= 10 ml 0.01 L

Molarity of the acid = ?

volume of the NaOH or base = 15.4 ml or 0.0154 L (equivalence point of the base)

molarity of the base = 0.1234 M

Applying the formula and putting the values, we get

Macid x 0.01 = 0.1234 x 0.0154

Macid =  0.1900 M

The weak acid is having molarity of 0.1900 M against the strong base with molarity of 0.1234M.

4 0
2 years ago
Taking advantage of their large differences in pKa values, describe how a mixture of phenol and benzoic acid in diethyl ether so
Marat540 [252]

Answer:

By adding bicarbonate.

Explanation:

The pka of the phenol (C₆H₅OH) is 10 and the pka of the benzoic acid (C₆H₅COOH) is 4, which means that the benzoic acid is a stronger acid than phenol, so if we want to separate phenol from benzoic acid in diethyl ether we need to first use a weak base that will react with benzoic acid and not with the phenol:  

C₆H₅-COOH + HCO₃⁻  ⇄  C₆H₅-COO⁻  +  H₂CO₃

C₆H₅-OH + HCO₃⁻  ⇄  no reaction

The reaction of the benzoic acid with bicarbonate will produce the benzoate ion that will be soluble in the aqueous layer, while the phenol will remain dissolved in the organic layer, so we can separate the two of them by the separation of the two immiscible layers.      

Having the two layers separated, the benzoic acid can be recovered from the aqueous layer by adding HCl:

C₆H₅-COO⁻ + HCl  ⇄  C₆H₅-COOH + Cl⁻

<u>This acid will precipitate from the aqueous solution, and the solid can be isolated by filtration</u>.  

The phenol in the organic layer can be dissolved into an aqueous layer by the adding of a strong base like NaOH:

C₆H₅-OH + OH⁻  ⇄  C₆H₅-O⁻ + H₂O

The phenoxide ion soluble in the aqueous layer can be recovered later by the adding of HCl, which will form the original phenol:

C₆H₅-O⁻ + HCl  ⇄  C₆H₅-OH + Cl⁻  

<u>The precipitated phenol can be isolated by filtration. </u>

This way we can separate a mixture of phenol and benzoic acid in diethyl ether solution.  

I hope it helps you!

6 0
3 years ago
Consider 0.01 m aqueous solutions of each of the following. a) NaI; b) CaCl2; c) K3PO4; and d) C6H12O6 (glucose) Arrange the sol
stealth61 [152]

Answer:

The solutions are ordered by this way (from lowest to highest freezing point):  K₃PO₄ < CaCl₂ < NaI < glucose

Option d, b, a and c

Explanation:

Colligative property: Freezing point depression

The formula is: ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.

Option d, which is glucose is non electrolyte so the i = 1

a. NaI →  Na⁺  +  I⁻        i =2

b. CaCl₂ →  Ca²⁺  +  2Cl⁻      i =3

c. K₃PO₄ → 3K⁺ + PO₄⁻³     i=4

Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.

7 0
3 years ago
Reactions rates for reactions occurring in solution can be increased by increasing the concentration of the solution. With gases
LUCKY_DIMON [66]
"Increase Pressure " is the right answer. if you need help , let me know
4 0
3 years ago
(2 x 109) x (5 x 105).
Finger [1]

Answer:

114450

Explanation:

2*109 = 218

5*105 = 525

218*525 = 114450

8 0
2 years ago
Read 2 more answers
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