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mel-nik [20]
3 years ago
14

g Sucrose (C12H22O11), a nonionic solute, dissolves in water (normal freezing/melting point 0.0°C) to form a solution. If some u

nknown mass of sucrose is dissolved in 150.g of water and this solution has a freezing point of -0.56°C, calculate the mass of sucrose dissolved.
Chemistry
1 answer:
DIA [1.3K]3 years ago
4 0

Answer:

15.4 g of sucrose

Explanation:

Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m

0.56°C / 1.86 m/°C = m → 0.301 mol/kg

m → molality (moles of solute in 1kg of solvent)

Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg

0.301 mol/kg .  0.150kg = 0.045 moles.

We determine the mass of sucrose, by the molar mass:

0.045 mol . 342 g/1mol = 15.4 g

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If the [H+] = 0.01 M, what is the pH of the solution, and is the solution a strong acid, weak acid, strong base, or weak base?
lana66690 [7]

Answer:

2, strong acid

Explanation:

Data obtained from the question. This includes:

[H+] = 0.01 M

pH =?

pH of a solution can be obtained by using the following formula:

pH = –Log [H+]

pH = –Log 0.01

pH = 2

The pH of a solution ranging between 0 and 6 is declared to be an acid solution. The smaller the pH value, the stronger the acid.

Since the pH of the above solution is 2, it means the solution is a strong acid.

5 0
3 years ago
The reason that elements in the same group all have similar chemical properties is because they all have the same
sveta [45]
D-number of electrons in the outer energy level.
4 0
3 years ago
A lead–tin alloy of composition 30 wt% Sn–70 wt% Pb is slowly heated from a temperature of 150°C (300°F).(a) At what temperature
Novay_Z [31]

Answer:

a) 231.9 °C

b) 100% Sn

c) 327.5 °C

d) 100% Pb

Explanation:

This is a mixture of two solids with different fusion point:

Tf_{Pb}=327.5 C

Tf_{Sn}=231.9 C

<u>Given that Sn has a lower fusion temperature it will start to melt first at that temperature. </u>

So the first liquid phase forms at 231.9 °C and because Pb starts melting at a higher temperature, that phase's composition will be 100% Sn.

The mixture will be completely melted when you are a the higher melting temperature of all components (in this case Pb), so it will all in liquid phase at 327.5 °C.

At that temperature all Sn was already in liquid state and, therefore, the last solid's composition will be 100% Pb.

3 0
3 years ago
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
3 years ago
A solution of NaOH is titrated with H2SO4. It is found that 20.05 mL of 0.3564 M H2SO4 solution is equivalent to 43.42 mL of NaO
Darya [45]

Answer : The concentration of NaOH is, 0.336 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL

Putting values in above equation, we get:

2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL

M_2=0.336M

Thus, the concentration of NaOH is, 0.336 M

3 0
3 years ago
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