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pickupchik [31]
3 years ago
8

If it takes 26.0 mL of 0.0250 M potassium dichromate to titrate 25.0 mL of a solution containing Fe2 , what is the molar concent

ration of Fe2
Chemistry
1 answer:
irina [24]3 years ago
8 0

Answer:

Explanation:

moles of potassium dichromate = .0250 x .026 = 65 x 10⁻⁵ moles

1 mole of potassium dichromate reacts with 6 moles of Fe⁺²

65 x 10⁻⁵ moles of potassium dichromate will react with

6 x 65 x 10⁻⁵ moles of Fe⁺²

= 390 x 10⁻⁵ moles

390 x 10⁻⁵ moles are contained in 25 mL of solution

molarity of solution = 390 x 10⁻⁵ / 25 x 10⁻³

= 15.6 x 10⁻² M  .

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Answer: Diffusion is the movement of particles from a high to low particle concentration, while osmosis is the movement of water from a high to a low water concentration.

Explanation:

6 0
3 years ago
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How many hydrogen atoms are in 2 mol of H2O?
Volgvan

Explanation:

2mol means 2H2O in 2 moles of water 4 H and 2Oxygen

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What happens when an atom has more electrons than can fit in one energy
Verizon [17]

Answer: A. The extra electrons start to fill higher sublevels in the energy level.

Explanation:

8 0
3 years ago
A sample of the sugar d-ribose (C5H10O5) of mass 0.727 g was placed in a calorimeter and then ignited in the presence of excess
alexandr1967 [171]

Answer:

The internal energy of combustion of d-ribose = -2127 kJ/mol

The enthalpy of formation of d-ribose = -1269.65 kJ/mol

Explanation:

Step 1: Data given

Mass of d-ribose = 0.727 grams

The temperature rose by 0.910 K

In a separate experiment in the same calorimeter, the combustion of 0.825 g of benzoic acid, for which the internal energy of combustion is −3251 kJ mol−1, gave a temperature rise of 1.940 K.

Molar mass of benzoic acid = 122.12 g/mol

Step 2: Calculate ΔU  for benzoic acid

The calorimeter is a constant-volume instrument so:

ΔU = q

ΔU = (0.825 g/ 122.12 g/mol) * (−3251 kJ /mol)

ΔU = -21.96 kJ

Step 3: Calculate ΔU  for d-ribose

c = |q| / ΔT

⇒ with ΔT = 1.940 K

c = 21.96 kJ / 1.940 K

c = 11.32 kJ /K

For d-ribose: ΔU = -cΔT

ΔU  = -11.32 kJ/K * 0.910 K

ΔU = - 10.3 kJ

Step 4: Calculate moles of d-ribose

moles ribose = 0.727 grams / 150.13 g/mol

moles ribose = 0.00484 moles

Step 5: Calculate the internal energy of combustion for d-ribose

ΔrU = ΔU / n

ΔrU  = -10.3 kJ / 0.004842 moles

ΔrU = -2127 kJ/mol

Step 6: Calculate The enthalpy of formation of d-ribose

The combustion of ribose is:

C5H10O5(s) + 5O2(g) → 5CO2(g) + 5H20(l)

Since there is no change in the number of moles of gas,  ΔrH = ΔU  

For the combustion of ribose, we consider the following reactions:

5CO2(g) + 5H2O(l) → C5H10O5(s) +5O2(g)      ΔH = -2127 kJ/mol

C(s) + O2(g) → CO2(g)      ΔH = -393.5 kJ/mol

H2(g) + 1/2 O2(g) → H2O(l)    ΔH = -285.83 kJ/mol

ΔH = 2127 kJ/mol + 5(-393.5 kJ/mol) + 5(-285.83 kJ/mol)

ΔH = 2127 kJ/mol - -1967.5 kJ/mol - 1429.15 kJ/mol

ΔH =  -1269.65 kJ/mol

The internal energy of combustion of d-ribose = -2127 kJ/mol

The enthalpy of formation of d-ribose = -1269.65 kJ/mol

3 0
3 years ago
1.86 g of ethanol reacts with 10.0 g of oxygen. What is the total volume of gas present (in L) after the reaction is complete, a
vladimir1956 [14]

Answer:

See explanation

Explanation:

The equation of the reaction is;

C 2 H 6 O(l)  +  3 O 2 (g)  → 2 CO 2 ( g )  +  3 H 2 O(l )

Next we have to determine the limiting reactant, this reactant gives the least number of moles of product.

Number of moles of C 2 H 6 O = mass/molar mass = 1.86g/ 46.07 g/mol = 0.04 mols

From the equation;

1 mol of ethanol yields 2 mols of CO2

0.04 moles of ethanol yields 0.04 * 2/1 = 0.08 mols of CO2

For water;

1 mol of ethanol yields 3 mols of water

0.04 moles of ethanol yields 3 * 0.04/1 = 0.12 mols of water

Also;

Number of moles of oxygen= 10g/32g/mol = 0.31 moles

3mols of O2 yields 2 moles of CO2

0.31 moles of O2 yields 0.31 * 2/3 = 0.21 moles of CO2

For water;

3 moles of O2 yields 3 moles of water

0.31 moles of O2 yields 0.31 * 3/3 = 0.31 moles of water

Hence ethanol is the limiting reactant.

From  PV=nRT

Volume of CO2 is;

V = nRT/P

V = 0.08 * 0.082 *298/1 = 1.95 L

Volume of water;

V = nRT/P

V= 0.12 * 0.082 * 298/1

V= 2.93 L

Total volume of gases after reaction = 1.95 L + 2.93 L = 4.88 L

6 0
3 years ago
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