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Firdavs [7]
3 years ago
7

Colligative properties include __________ the freezing point, _____________ the vapor pressure, and ___________ the boiling poin

t. Which of the following sequences of words correctly fills in the blanks?
a) raising, lowering, loweringb) raising, raising, loweringc) lowering, raising, loweringd) lowering, lowering, raising
Chemistry
2 answers:
MrMuchimi3 years ago
7 0
Colligative properties include lowering the freezing point, lowering the vapor pressure, and raising the boiling point.

pashok25 [27]3 years ago
3 0

The answer is: d) lowering, lowering, raising.

The vapor pressure depression (lowering), freezing point depression (lowering) and the boiling point elevation (raising) are the colligative properties od solution.

For example pure water has lower boiling point than solution of water and some salt.

The higher is the molarity of the solution, the lower is freezing point.

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The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh
Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

150.0 mL \times \frac{1.02g}{mL}  = 153 g

Given the specific heat capacity of the solution (c) is 4.184 J/g・°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J  }{\frac{4.184J}{g.\° C }  \times 153g} = 5.87 \° C

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

7 0
2 years ago
What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
Margarita [4]

Answer:

108.6 g

Explanation:

  • 2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:

  • P = 1.00 atm
  • V = 56.0 L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 0 °C ⇒ 0 + 273.2 = 273.2 K

<u>Inputting the data</u>:

  • 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
  • n = 2.5 mol

Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

  • 2.5 mol N₂ * \frac{2molNaN_3}{3molN_2} = 1.67 mol NaN₃

Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:

  • 1.67 mol * 65 g/mol = 108.6 g
6 0
3 years ago
After 20 min, a reactant has decomposed to 85 % of its original concentration. Which order would the reaction need to be to allo
rjkz [21]

Answer : The value of rate constant is, 0.0949\text{ min}^{-1}

Explanation :

First we have to calculate the rate constant, we use the formula :

Expression for rate law for first order kinetics is given by:

k=\frac{2.303}{t}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = time passed by the sample  = 20 min

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process = 100 - 85 = 15 g

Now put all the given values in above equation, we get

k=\frac{2.303}{20}\log\frac{100}{15}

k=0.0949\text{ min}^{-1}

Therefore, the value of rate constant is, 0.0949\text{ min}^{-1}

7 0
3 years ago
PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST
Ivahew [28]

Answer:

up to down(1 to 6)

Explanation:

speed:

1,3,4

Velocity:

everything except 1,3,4

5 0
3 years ago
Read 2 more answers
What is the H+ concentration in a solution with a pH of 1.25?
umka2103 [35]
The H+ concentration in a solution =10^(-1.25)=0.05623413M
3 0
3 years ago
Read 2 more answers
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