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2 years ago

What happens when the concentration of water inside a cell is lower than the concentration of water outside the cell?

Chemistry

1 answer:

adell [148]2 years ago

4 0

Answer:

Water outside the cell will flow inwards by osmosis to attain equilibrium

Explanation:

In the hypotonic environment, the concentration of water is greater outside the cell and the concentration of solute is higher inside. A solution outside of a cell has a lower concentration of solutes relative to the cytosol.

If concentrations of dissolved solutes are greater inside the cell, the concentration of water inside the cell is correspondingly lower. As a result, water outside the cell will flow inwards by osmosis to attain equilibrium.

Osmosis is a process by which molecules of a solvent tend to pass from a less concentrated solution into a more concentrated one through a semipermeable membrane.

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If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as th

qaws [65]

This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

$COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)$

$K_c=4.74\times 10^4$

Concentration at equilibrium:

$[COBr_2]=1.58\times 10^{-6}M$

$[Co]=2.78\times 10^{-3}M$

$[Br_2]=2.51\times 10^{-5}M$

If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

$COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)$

The expression for reaction quotient will be :

$Q=\frac{[CO][Br_2]}{[COBr_2]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}$

The given equilibrium constant value is, $K_c=4.74\times 10^4$

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When $Q>K_c$ that means product > reactant. So, the reaction is reactant favored.

When $Q$ that means reactant > product. So, the reaction is product favored.

When $Q=K_c$ that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the $Q$ that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

3 0

3 years ago

Which symbol is associated with structure C? A. _ B.+ C.X D. none

IgorLugansk [536]

B I’m guessing not sure

6 0

2 years ago

Someone please help! this is the last question I only need help with B.

ludmilkaskok [199]

1. mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. Na₂CO₃ as a limiting reactant

<h3>Further explanation</h3>

Given

Reaction

2 Al(NO₃)₃ + 3 Na₂CO₃ → Al₂(CO₃)₃ + 6 NaNO₃

Required

mol ratio

Limiting reactant

Solution

The reaction coefficient in the chemical equation shows the mole ratio of the components of the compound involved in the reaction (reactants and products)

1. From the equation mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. mol : coefficient of Al(NO₃)₃ : Na₂CO₃ = 2 mole/2 : 2 mole/3 = 1 : 0.67

Na₂CO₃ as a limiting reactant (smaller)

6 0

2 years ago

If a 45 mM phosphate solution(solution A) had an absorbance of 1.012. What would be the absorbance if 11 mL of solution A was us

timofeeve [1]

Answer:

0.550

Explanation:

The absorbance (A) of a substance depends on its concentration (c) according to Beer-Lambert law.

A = ε . l . c

where,

ε: absorptivity of the species

l: optical path length

A 45 mM phosphate solution (solution A) had an absorbance of 1.012.

A = ε . l . c

1.012 = ε . l . 45 mM

ε . l = 0.022 mM⁻¹

We can find the concentration of the second solution using the dilution rule.

C₁ . V₁ = C₂ . V₂

45mM . 11mL = C₂ . 20.0 mL

C₂ = 25 mM

The absorbance of the second solution is:

A = (ε . l ). c

A = (0.022 mM⁻¹) . 25 mM = 0.55 (rounding off to 3 significant figures = 0.550)

8 0

3 years ago

What is the importance of knowing capacity, appreciation and experimental error in measuring instruments?

Tanzania [10]

Answer:

Understanding” refers to the ability of the individual to comprehend the ... The “Appreciation” component of decision-making capacity involves the ability ... Note that in addition to explaining the importance of making and ... we excluded instruments that did not measure all four components detailed earlier).

Explanation:hope it helps

8 0

2 years ago