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3 years ago

A sample of grape juice has a hydroxide ion concentration of 1.4 × 10-10 M

2 answers:

5 0

The relation between hydroxide ion (OH-) and hydronium ion (H3O+) is given by:

pH + pOH = 14, which is equivalent to [H3O+] [OH-] = 10^ -14

=> [H3O+] = (10^ -14) / [OH-] = (10^ -14) / (1.4*10^-10) = 0.71 * 10^ - 4 = 0.071 * 10^-5

Answer: 0.071 * 10 ^ -5

finlep [7]3 years ago

4 0

Answer:

7.1 * 10x6^-5M

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Explanation:

A redox reaction is a reaction when oxidation states (or numbers) change during reaction.

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Other planets besides the earth have a lithosphere, but none of them has ________ in their lithosphere, as earth does.

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3 years ago

Read 2 more answers

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

Answer: The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

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3 years ago

Which activity might help to increase the validity of this experiment?

MArishka [77]

A should be the answer because the more you test an experiment the more data you have to rely on changing the experiment would cause you to have different outcomes making the results different and unreliable so B, C, and D is not going to be the answer Hope this helps

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3 years ago

3. Discuss the difference between molarity and molality, state the units of each, state the symbol for each, and give an example

dangina [55]

Molality(m) or molal concentration is a measure of concentration and it refers to amount of substance in a specified amount of mass of the solvent. Used unit for molality is mol/kg which is also sometimes denoted as 1 molal. It is equal to the moles of solute (the substance being dissolved) divided by the kilograms of solvent (the substance used to dissolve).

Molarity(M) or molar concentration is also a measure of concentration and represents the amount of substance per unit volume of solution(number of moles per litre of solution. Used unit for molarity is mol/L or M. A solution with a concentration of 1 mol/L is equivalent to 1 molar (1 M).

Molality is preferred when the temperature of the solution varies, because it does not depend on temperature, (neither number of moles of solute nor mass of solvent will be affected by changes of temperature), while molarity changes as temperature changes(volume of solution changes as temperature changes).

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3 years ago