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2 years ago

Does sugar can dissolve in ccl4

Chemistry

1 answer:

wel2 years ago

8 0

Answer:

Sugar is a polar compound , in the solid state of it Hydrogen bond strongly binds the molicules of sugar together. Being a non polar solvent and low polarity difference between Cl and C atom C—Cl bond in CCl4 is unable to form hydrogen bond. That's why sugar can't be soluble in CCl4.

Explanation:

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Examine the incomplete statement.

hjlf

Answer: Option (c) is the correct answer.

Explanation:

In liquids, molecules are held by slightly less strong intermolecular forces of attraction as compared to solids.

Hence, molecules of a liquid are able to slide past each other as they have more kinetic energy than the molecules of a solid.

As a result, liquids are able to occupy the shape of container in which they are placed. Also, liquids have fixed volume but no fixed shape.

Thus, we can conclude that liquids have a variable shape and a fixed volume.

5 0

2 years ago

Read 2 more answers

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

Answer: The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

8 0

2 years ago

A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What i

In-s [12.5K]

Answer:

81.26% is the percent yield

Explanation:

Based on the reaction:

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.

To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:

Actual yield (0.366g) / Theoretical yield * 100

Moles CaCl₂ = Moles CaCO₃:

0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃

Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:

0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃

Percent yield = 0.366g / 0.450g * 100

81.26% is the percent yield

3 0

2 years ago

Please help me!!!!

miv72 [106K]

Answer:

0.8g/ml

Explanation:

d=m/v

d=20/25

d=0.8

7 0

3 years ago

Determine the mass of CuSO4 • 5H20 that must be used to prepare 250mL of 2.01 M CuSO4(aq).

mario62 [17]

Given parameters:

Volume of CuSO₄ = 250mL

Concentration of CuSO₄ = 2.01M

Unknown:

Mass of CuSO₄.5H₂O = ?

To solve this problem, we must write the chemical relationship between both species.;

CuSO₄.5H₂O → CuSO₄ + 5H₂O

Now that we know the expression, it is possible to solve for the unknown mass.

First find the number of moles of CuSO₄;

Number of moles = Concentration x Volume

Take 250mL to L so as to ensure uniformity of units;

Volume = 250 x 10⁻³L

Input the parameters and solve for number of moles;

Number of moles = 250 x 10⁻³ x 2.01 = 0.5mol

From the equation;

1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O

So 0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O

Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16) = 249.6g/mole

Mass of CuSO₄.5H₂O = number of moles x molar mass

= 0.5 x 249.6

= 124.8g

The mass of CuSO₄.5H₂O is 124.8g

5 0

3 years ago

Does sugar can dissolve in ccl4​

Does sugar can dissolve in ccl4