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Jobisdone [24]
3 years ago
13

PLEASEEEEE PLEASEEEEE ANSWER THIS QUESTION ILL GIVE BRAINLIEST AND 15 POINTSSSSS PLEASEEEEEEEEEEE!!!!!!!!!!

Chemistry
2 answers:
torisob [31]3 years ago
7 0
Hey hey calm down what’s the question tho
lozanna [386]3 years ago
6 0
Hi what’s the question.... um
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A solution containing 3.90 g of an unknown nonelectrolyte liquid and 9.60 g water has a freezing point of −3.33 °C. The Kf = 1.8
pantera1 [17]

Answer:

The molar mass is 227 g/mol

Explanation:

Step 1: Data given

Mass of unknown nonelectrolyte = 3.90 grams

Mass of water = 9.60 grams

Freezing point of the solution = -3.33 °C

Kf = 1.86°C/m

Nonelectrolyte has a van't Hoff factor = 1

Step 2:

If you know the number of moles, and you know that is equivalent to 1.00 g, you can get molar mass.

∆T =i*m*K f

⇒ ∆T  = difference in temperature between freezing point of solution and pure water = 3.33 °C

⇒ Van't hoff factor of the nonelectrolyte = 1

⇒ molality = moles nonelectrolyte / mass water

⇒ Kf = freezing point constant = 1.86 °C/m

3.33 = (1)(m)(1.86)

m = 1.79 molal = 1.79 moles / kg H2O

Step 3: Calculate moles nonelectrolyte

molality = moles / mass H2O

moles = molality * mass H2O

Moles = 1.79 molal * 0.0096

Moles = 0.017184‬ moles

Step 4: Calculate molar mass of nonelectrolyte

Molar mass = mass / moles

Molar mass = 3.90 grams / 0.017184 moles

Molar mass = 227 g/mol

The molar mass is 227 g/mol

5 0
3 years ago
A(n) ____________________ is a compound that turns litmus blue and is often found in soaps and detergents.
ehidna [41]
B. base
_______________________________________________________________
8 0
3 years ago
How. does knowing the weather forecast affect people making decisions each day?
Sladkaya [172]
Just an example.. This can affect people because it can cause people to bring umbrellas wear rain coats of it were to rain.
5 0
3 years ago
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Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a
coldgirl [10]

The final temperature, t₂ = 30.9 °C

<h3>Further explanation</h3>

Given

24.0 kJ of heat = 24,000 J

Mass of calorimeter = 1.3 kg = 1300 g

Cs = 3.41 J/g°C

t₁= 25.5 °C

Required

The final temperature, t₂

Solution

Q = m.Cs.Δt

Q out (combustion of compound) = Q in (calorimeter)

24,000 = 1300 x 3.41 x (t₂-25.5)

t₂ = 30.9 °C

3 0
3 years ago
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What is the oxidation number of boron in Na(B(NO3)4)
Vitek1552 [10]

Answer:

the oxidation state od boron in sodium boron hydride is (+3).

7 0
2 years ago
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