Arrange the following substances in order of increasing solubility of water. C6H14, C6H13Br, C6H13OH, C6H12(OH)2.

Answers

Answer 1

Answer:

Answer:

C6H14 < C6H13Br < C6H13OH < C6H12(OH)2

Explanation:

Hello,

In this case, since the solubility in water is related with the presence of polar bonds in the given molecules we can see that C6H12(OH)2 has the presence two O-H bonds which promote the highest solubility via hydrogen bonds as well as the C6H13OH but in a lower degree as only on O-H bond is present. Next since the bond C-Br in is slightly close to the polar bond C6H13Br rather than the C-C bonds only had by C6H14 we can infer that C6H13Br is more soluble in water than C6H14, therefore the required order is:

C6H14 < C6H13Br < C6H13OH < C6H12(OH)2

Whereas C6H12(OH)2 is the most soluble and C6H14 the least soluble in water.

Best regards.

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What is the molarity of the following solutions?a. 19.5 g NaHCO3 in 460.0 ml solution
b. 26.0 g H2SO4 in 200.0 mL solution
c. 15.0 g NaCl dissolved to make 420.0 mL solution

Answers

Answer:

a) NaHCO3 = 0.504 M

b) H2SO4 = 1.325 M

c) NaCl = 0.610 M

Explanation:

Step 1: Data given

Moles = mass / molar mass

Molarity = moles / volume

a. 19.5 g NaHCO3 in 460.0 ml solution

Step 1: Data given

Mass NaHCO3 = 19.5 grams

Volume = 460.0 mL = 0.460 L

Molar mass NaHCO3 = 84.0 g/mol

Step 2: Calculate moles NaHCO3

Moles NaHCO3 = 19.5 grams / 84.0 g/mol

Moles NaHCO3 = 0.232 moles

Step 3: Calculate molarity

Molarity = 0.232 moles / 0.460 L

Molarity = 0.504 M

b. 26.0 g H2SO4 in 200.0 mL solution

Step 1: Data given

Mass H2SO4 = 26.0 grams

Volume = 200.0 mL = 0.200 L

Molar mass H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = 26.0 grams / 98.08 g/mol

Moles H2SO4 = 0.265 moles

Step 3: Calculate molarity

Molarity = 0.265 moles / 0.200 L

Molarity =1.325 M

c. 15.0 g NaCl dissolved to make 420.0 mL solution

Step 1: Data given

Mass NaCl = 15.0 grams

Volume = 420.0 mL = 0.420 L

Molar mass NaCl = 58.44 g/mol

Step 2: Calculate moles NaCl

Moles NaCl = 15.0 grams / 58.44 g/mol

Moles NaCl = 0.256 moles

Step 3: Calculate molarity

Molarity = 0.256 moles / 0.420 L

Molarity =0.610 M

A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of the solution. The chemical reaction that takes place is represented by the following equation. Pb(NO3)2(s) + NaI(aq) → PbI2(s) + NaNO3(aq) How many moles of PbI2 can be produced?/Identify the limiting reactant/If the actual yield was 0.550 g of PbI2, what is the percent yield?

Answers

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield = 87.72%

Further explanation

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

moles of PbI₂
Limiting reactant
% yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

= 1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

the reaction A->B is second order in A and the rate constant is 0.039L/mol s. if it took 23 sec for the concentration of A to decrease to 0.30 M, what was the starting concentration of A

Answers

Answer:

0.41 M

Explanation:

A -> B

rate constant (k) = 0.039L/mol s

t = 23

Final concentration, [A] = 0.30M

Initial concentration, [A]o = x

1 / [A] = kt + 1 / [A]o

1 / [A]o = 1 / [A] - kt

1 / [A]o = 1 / 0.30 - 0.039 (23)

1 / [A] = 3.33 - 0.897 = 2.433

[A] = 0.41 M

What type of reaction is this? Hint: howmany reactants are there?
Decomposition
Combustion
Single Replacement

Answers

Answer:

Decomposition= one-reactant

Combustion= 2-reactants

Single displacement= 2-reactant

Explanation:

What is colloidal solutions

Answers

Explanation:

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid. ... Colloidal systems can occur in any of the three key states of matter gas, liquid or solid. However, a colloidal solution usually refers to a liquid concoction.

Answer:

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid.

Which of the following is not a reason why chemical bonds are important

Answers

What are the followings?

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