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3 years ago

What’s the PH and OH- of 6.0 x 10^-10?

Chemistry

1 answer:

Zarrin [17]3 years ago

8 0

Answer:

The answer to your question is pH = 9.2; [OH] = 1.58 x 10⁻⁵

Explanation:

Data

[H⁺] = 6.0 x 10⁻¹⁰ M

pH = ?

pH definition. pH measures the hydrogen concentration. pH measures the acidity and the alkalinity of a solution. A solution is acid if the pH goes from 0 to 6.9, a neutral solution has a pH of 7 and an alkali solution has a pH from 7.1 to 14.

pOH measures the [OH⁻] concentration

Formula

pH = -log[H⁺]

pH + pOH = 14

-Substitution

pH = -log[6 x 10⁻¹⁰]

-Simplification

pH = - (-9.2)

pH = 9.2

-Calculate pOH

pOH = 14 - pH

pOH = 14 - 9.2

pOH = 4.8

-Calculate the concentration of OH⁻

pOH = -log[OH⁻]

[OH⁻] = -antilog4.8

[OH] = 1.58 x 10⁻⁵

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What is the difference between an anion and a cation

Svetlanka [38]

Cations and anions are both ions. The difference between a cation and an anion is the net electrical charge of the ion. Ions are atoms or molecules which have gained or lost one or more valencee electron giving the ion a net positive or negative charge. Cations are ions with a net positive charge.

please mark as brainliest <3

7 0

2 years ago

Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.5kJ)=-787kJ$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.8kJ)=-571.6kJ$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)$

$\Delta H=52.4kJ$

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

7 0

3 years ago

The fish died after living in the aquarium for many years

miv72 [106K]

In an aquarium the water quantity is limited and fish excrete ammonia through their gills and body, this dissolves in water and creates toxins.

<u>Explanation</u>:

In an aquarium the water quantity is limited and fish excrete ammonia through their gills and body, this dissolves in water and creates toxins. Over some time some bacteria develop in water which converts this ammonia into nitrites and then into nitrates. Till this process is complete, the aquarium remains a death trap.

There are some reasons for a fish to die early,

This one is common for beginners. They don’t have any idea about the nitrogen cycle. And they simply buy a fish tank and fish on the same day, go home and set it up.

Ammonia spikes - Even in a cycled tank, you can have occasional ammonia spikes due to overfeeding. Incompatible tank mates - If you put a too docile fish with a very aggressive cichlid, it will get harassed and eventually die due to stress.

4 0

3 years ago

An archer pulls her bowstring back 0.370 m by exerting a force that increases uniformly from zero to 264 N. (a) What is the equi

Arte-miy333 [17]

Answer:

713.51 N/m

Explanation:

Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.

From hook's law,

F = ke ...........................Equation 1

Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.

Make k the subject of the equation,

k = F/e ............................ Equation 2

Given: F = 264 N, e = 0.37 m.

Substitute into equation 2

k = 264/0.37

k = 713.51 N/m

Hence the spring constant of the bow = 713.51 N/m

3 0

2 years ago

Ca(OH)₂(s) precipitates when a 1.0 g sample of CaC₂(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g sa

Alex

Answer:

D) Ca(OH)₂ will not precipitate because Q < Ksp

Explanation:

This is a question in which we will employ the reaction quotient Q to determine whether a precipitate will form.

Here we have first a chemical reaction in which Ca(OH)₂ is produced:

CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂ is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product constant for Ca(OH)₂ is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

Thus the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M = 0.002 M ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q= [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is (D)

6 0

2 years ago