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3 years ago

Help please fast will mark as Brainliest

Chemistry

2 answers:

goblinko [34]3 years ago

6 0

Answer:

Zn + I2 > Zn+² + 2 I-

Explanation:

First identify the which is oxidised and reduced in the reaction.

Zn > Zn2+ is oxidised increase in ox. #

I2 > 2I- is reduced decrease in ox. #

Do half reactions and add electrons to each positive side to balance the charges

Zn > Zn2+ + 2e-

I2 + 2e- > 2 I-

Combine each product side and reactant. Cancel the electrons out since they are in the same amount.

wlad13 [49]3 years ago

3 0

Answer:

The answer is 2i on right hand side.

Explanation:

We should star by checking the equation from right.

First we check how many Zn r there in left hand side. Which is 1. Let us check how many Znr there in right hand side, there is 1.So Zn is balanced, and don't worry about Znplus2 on right hand side it is just the ions not how many zinc r there.

Now let us check how many I are there left hand side. Which is 2. Now how many I are there in right hand side? Only 1.

So we put 2 behind I.

Now there r 2 I on both sides.

Its simple actually.

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25.0 mL of an HBr solution were titrated with 29.15 mL of a 0.205 M LiOH solution to reach the equivalence point. What is the mo

trapecia [35]

The molarity of HBr solution is 0.239 M

calculation

write the equation for reaction

that is LiOH +HBr → LiBr +H20

find the moles of LiOH used

moles =molarity x volume

=0.205 x 29.15 = 5.976 moles

by use of mole ratio between LiOH :HBr which is 1:1 the moles of HBr is therefore= 5.976 moles

Molarity of HCL= moles /volume

= 5.976/25 = 0.239 M

7 0

3 years ago

If the solvent in a solution is water, the solution is

BlackZzzverrR [31]

Answer:

I don't know

Explanation:

Yet but I will find it for you

7 0

3 years ago

What volume of solution gives the desired moles?

motikmotik

Answer:

1. 0.073L

2. 0.028L

3. 0.014L

Explanation:

The volume for the different solutions are obtained as shown below:

1. Mole = 0.53mol

Molarity = 7.25M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.53/7.25

Volume = 0.073L

2. 0.035mol from a 1.25M

Mole = 0. 035mol

Molarity = 1.25M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.035/1.25

Volume = 0.028L

3. Mole = 0.0013mol

Molarity = 0.090M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.0013/0.090

Volume = 0.014L

8 0

3 years ago

The initial pressure of a mixture of C6H6 and an excess of H2 in a rigid vessel is 1.21 atm. A catalyst is introduced. After the

KengaRu [80]

Answer:

mole fraction of C6H6 = 0.613 atm

Explanation:

The equation for this reaction is :

$C_6H_6 _{(g)} + 3H_2_{(g)} \to C_6H_{12}_{(g)}$

Initial P₁ P₂ 0

Final 0 P₂ -P₁/2 P₁

After completion of the reaction;

P₁ + P₂ = 1.21 atm ----- (1)

P₂ - P₁/2 + P₁ = 0.839 atm

P₂ + P₁/2 = 0.839 atm ----- (2)

Subtracting (2) from (1); we have:

P₁/2 = 0.371

P₁ = 0.742 atm

From(1)

P₁ + P₂ = 1.21 atm

0.742 atm + P₂ = 1.21 atm

P₂ = 1.21 atm - 0.742 atm

P₂ = 0.468 atm

Thus, the partial pressure of C6H6 = 0.742 atm

∴

Partial pressure = Total pressure × mole fraction of C6H6

mole fraction of C6H6 = Partial pressure / Total pressure

mole fraction of C6H6 = 0.742 atm / 1.21 atm

mole fraction of C6H6 = 0.613 atm

5 0

3 years ago

A 2.350×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then b

sveta [45]

Answer:

Part A: m = 0.02356 mol/kg = 0.02356 m

Part B: Xsolute = 4.243×10⁻⁴

Part C: % m/m = 0.1376%

Part D: ppm = 1,376 ppm

Explanation:

1. Data:

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

2. Formulae

M = n solute / V sol (L)
m = n solute / Kg solvent
X solute = n solute / N total
% m/m = (mass of solute / mass of solution) × 100
ppm = (mass of solute / mass of solution) × 1,000,000
density = mass in grams / volume in mL

3. Solution

Part A: Calculate the molality of the salt solution.

m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

Part B: Calculate the mole fraction of salt in this solution.

X solute = n solute / N total

i) n solute = 2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

Part C: Calculate the concentration of the salt solution in percent by mass.

% m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm

5 0

3 years ago