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Aleonysh [2.5K]
2 years ago
5

A chemist conducts a titration on 50 mL of a solution of hydrobromic acid (HBr) of unknown strength. She finds the pH reaches th

e equivalency point with the addition of 15 mL of 0.30 M sodium hydroxide (NaOH). What is the concentration of the HBr solution?
Chemistry
1 answer:
Varvara68 [4.7K]2 years ago
6 0

The concentration of the HBr would be 0.09 M

<h3>What are stoichiometric problems?</h3>

First, let us look at the equation of the reaction:

HBr + NaOH --- > NaBr + H_2O

The mole ratio of HBr to NaOH is 1:1.

Mole of 15 mL, 0.30 M NaOH = 0.30 x 15/1000 = 0.0045 moles

Equivalent mole of HBr - 0.0045 moles

Molarity of HBr = mole/volume = 0.0045/50/1000 = 0.09 M

More on stoichiometric problems can be found here: brainly.com/question/15047541

#SPJ1

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What is formed when a BASE dissociates in water, according to the Brønsted-Lowry definition?
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2 years ago
What is the minimum pressure in kPa that must be applied at 25 °C to obtain pure water by reverse osmosis from water that is 0.1
Yuri [45]

Answer:

The minimum pressure should be 901.79 kPa

Explanation:

<u>Step 1: </u>Data given

Temperature = 25°C

Molarity of sodium chloride = 0.163 M

Molarity of magnesium sulfate = 0.019 M

<u>Step 2:</u> Calculate osmotic pressure

The formula for the osmotic pressure =

Π=MRT.

⇒ with M = the total molarity of all of the particles in the solution.

 ⇒ R = gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 25 °C = 298 K

NaCl→ Na+ + Cl-

MgSO4 → Mg^2+ + SO4^2-

M = 2(0.163) + 2(0.019 M)

M = 0.364 M

Π = (0.364 M)(0.08206 atm-L/mol-K)(25 + 273 K)

Π = 8.90 atm

(8.90 atm)(101.325 kPa/atm) = 901.79 kPa

The minimum pressure should be 901.79 kPa

6 0
4 years ago
Given a sample of poly[ethylene-stat-(vinyl acetate)] A.Calculate the mean repeat unit molar mass for a sample of poly[ethylene-
AlladinOne [14]

Answer:

a) The mean repeat unit molar mass for PEVA is 30.72 g/mol

b) degree of polymerization of the copolymer is 1300

Explanation:

Given that;

the wt% of copolymer consist of 12.9 wt% of vinyl acetate and 87.1 wt% Ethylene.

Basis: 100 g of PEVA consist of 12.9 of vinyl acetate and 87.1g of Ethylene.

now we calculate the mole fraction of vinyl acetate Ethylene in the copolymer;

the molecular weights of vinyl acetate and ethylene are 86.09 g/mol and 28.05 g/mol respectively

so

moles of vinyl acetate = wt. of vinyl acetate / molecular weights of vinyl acetate

moles of vinyl acetate = 12.9 g / 86.09 g/mol

moles of vinyl acetate = 0.1498 mol

moles of Ethylene = wt. of Ethylene / molecular weights of Ethylene

moles of Ethylene = 87.1 g / 28.05  d/mol

moles of Ethylene  = 3.1052 mol

Total moles = 0.1498 mol + 3.1052 mol = 3.255 mol

Next we calculate the mole percent;

mole percent of vinyl acetate X_{V} = moles of vinyl acetate / total  moles

X_{V} = (0.1498 mol / 3.255 mol) × 100

X_{V}  = 4.6%

mole percent of Ethylene X_{E} = moles of Ethylene / total  moles

X_{E}  = (3.1052 mol / 3.255 mol) × 100

X_{E}  = 95.397% ≈ 95.4%

we know that, mean repeat unit molar mass for a sample = ∑X_{i}M_{i}

where X_{i} is the fraction ratio and M_{i} is the molecular  weight

so or the PEVA

mean repeat unit molar mass M = ( X_{V}M_{V}) + ( X_{E}M_{E})

so we substitute

M = ( 4.6% × 86.09) + ( 95.4% × 28.05 )

M = 3.96014 + 26.7597

M = 30.72 g/mol

Therefore, The mean repeat unit molar mass for PEVA is 30.72 g/mol

b)

Degree of polymerization

DP_{n} = \frac{M_{n} }{M}

where M_{n} is the number average molecular weight ( 39,870 g/mol )

so we substitute

DP_{n} = 39,870 g/mol / 30.72 g/mol

DP_{n}  = 1297.85 ≈ 1300   { 3 significance figure }

Therefore, degree of polymerization of the copolymer is 1300

5 0
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