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11111nata11111 [884]
4 years ago
6

Silver nitrate solution reacts with calcium chloride solution according to the equation: 2 AgNO3 + CaCl2 → Ca(NO3)2 + 2 AgCl All

of the substances involved in this reaction are soluble in water except silver chloride, AgCl, which forms a solid (precipitate) at the bottom of the flask. Suppose we mix together a solution containing 6.97 g of AgNO3 and one containing 6.39 g of CaCl2. What mass, in grams, of AgCl is formed?
Chemistry
1 answer:
HACTEHA [7]4 years ago
5 0

Answer:

14.33 g

Explanation:

Solve this problem based on the stoichiometry of the reaction.

To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and  finally calculate the mass of AgCl.

                      2 AgNO₃   + CaCl₂    ⇒   Ca(NO₃)₂   + 2 AgCl

mass, g                  6.97        6.39                                     ?

MW ,g/mol         169.87      110.98                                  143.32

mol =m/MW           0.10         0.06                                    0.10

From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :

0.10 mol x 143.32 g/mol = 14.33 g

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Igoryamba

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internal fertilization

Explanation:

Internal fertilization enhances the fertilization of eggs by a specific male. Fewer offspring are produced through this method, but their survival rate is higher than that for external fertilization.

6 0
3 years ago
Yasir wants to determine if there is a relationship between room color and sleep. Based on his research, Yasir makes an educated
Fynjy0 [20]

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3 0
3 years ago
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Classify each of the following as a strong acid or a weak acid and indicate how each should be written in aqueous solution. Clas
JulijaS [17]

Answer:

HCN, weak acid

H⁺, Br⁻, strong acid

Explanation:

Hydrocyanic acid is a weak acid, according to the following equation.

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

Thus, it should be written in the undissociated form (HCN).

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Thus, it should be written in the ionic form (H⁺, Br⁻).

7 0
3 years ago
Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.
lianna [129]

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

4 0
3 years ago
A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of
Leto [7]

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B =  21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A:   M  = m1 *0.8662    O = m1 *0.1338

compound B:   M  = m2 *0.90666    O = m2 *0.09334

compound C:   M  = m3 *0.92832    O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO  / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO  / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO  / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338   m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48  ⇒ m2 = 227.74 g

0.92832m3 = 206.48  ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams  ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

6 0
3 years ago
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