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serg [7]
3 years ago
15

A cucumber is placed in a concentrated salt solution. What will most likely happen? Select one: a. Water will flow from the cucu

mber to the solution. b. Water will flow from the solution to the cucumber. c. Salt will flow into the cucumber. d. Salt will precipitate out. e. No change will occur.
Chemistry
2 answers:
lesantik [10]3 years ago
6 0

Answer: Option (a) is the correct answer.

Explanation:

Osmosis is defined as a process in which solvent molecules pass through a semipermeable membrane from a region of less concentration to a region of more concentration. As a result, it equals the concentration on each side of the membrane.

Therefore, when a cucumber is placed in a concentrated salt solution then molecules of cucumber will travel towards the concentrated salt solution in order to balance the concentration.

Thus, we can conclude that when a cucumber is placed in a concentrated salt solution then water will flow from the cucumber to the solution.

antiseptic1488 [7]3 years ago
3 0

Answer:

a.- Water will flow from the cucumber to the solution

Explanation:

This is an example of osmosis in difference of concentration of solutions. In the salty solution there is more salt molecules than inside the cucumber. Trying to make up for the same salt concentration inside that outside the cucumber, it will tend to flow some of the water that is inside the cucumber tissues to the environment that surrounds it.

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Identify the three types of metamorphism.
Elena-2011 [213]

Answer:

Three types of metamorphism exist: contact, dynamic, and regional. Metamorphism produced with increasing pressure and temperature conditions is known as prograde metamorphism.

4 0
3 years ago
The vapor pressure of a substance is measured over a range of temperatures. A plot of the natural log of the vapor pressure vers
Gala2k [10]

Answer:

28.7664 kJ /mol

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c

Where,

P is the vapor pressure

ΔHvap  is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

The graph of ln P and 1/T gives a slope of - ΔHvap/ R and intercept of c.

Given :

Slope = -3.46×10³ K

So,

- ΔHvap/ R = -3.46×10³ K

<u>ΔHvap = 3.46×10³ K × 8.314×10⁻³ kJ /mol K = 28.7664 kJ /mol</u>

<u></u>

6 0
3 years ago
What is the empirical formula for a compound that contains 79.86 % iodine and 20.14 % oxygen by mass?
serg [7]

Answer:

IO₂

Explanation:

We have been given the mass percentages of the elements that makes up the compound:

Mass percentage given are:

Iodine = 79.86%

Oxygen = 20.14%

To calculate the empirical formula which is the simplest formula of the compound, we follow these steps:

> Express the mass percentages as the mass of the elements of the compound.

> Find the number of moles by dividing through by the atomic masses

> Divide by the smallest and either approximate to nearest whole number or multiply through by a factor.

> The ratio is the empirical formula of the compound.

Solution:

I O

% of elements 79.86 20.14

Mass (in g) 79.86 20.14

Moles(divide by

Atomic mass) 79.86/127 20.14/16

Moles 0.634 1.259

Dividing by

Smallest 0.634/0.634 1.259/0.634

1 2

The empirical formula is IO₂

7 0
3 years ago
which is a false statement about vinger. A. its a pure surprise B. it's a physical combined c. it is made up of more than one pu
victus00 [196]
I would think the answer is D
6 0
3 years ago
You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of
Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}

Amount of BaSO₄ from K₂SO₄ is;

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4} = X\times\frac{233.38 }{142.04} = 1.643·X

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4} = Y\times\frac{233.38 }{174.259 } = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0
3 years ago
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