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3 years ago

If a solution has a pOH of 8.71, what is the [H+]?

Chemistry

1 answer:

TEA [102]3 years ago

6 0

Answer: $5.1\times 10^{-6}M$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH+pOH=14$

Putting in the values:

$pH=14-8.71=5.29$

$5.29=-log[H^+]$

$[H^+]=5.1\times 10^{-6}$

Thus $[H^+]$ is $5.1\times 10^{-6}M$

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What does the large number in front of the molecule represent?

madam [21]

A large number in front of a compound designates how many units there are of that compound. Parentheses can be used to designate a special structure, where other molecules are attached to the larger, complex molecule.

4 0

2 years ago

Read 2 more answers

Copper is plated on zinc by immersing a piece of zinc into a solution containing copper(II) ions. In the plating reaction, coppe

Aloiza [94]

Hey there!:

Copper plating on Zinc will occur via this simple reaction:

Cu²⁺ + 2 e⁻ ⇌ Cu

When a species gains electron, it is reduced. Its oxidation state is decreased. For example, in the above reaction Cu2+ gained 2 electron to get reduced to Cu. Its oxidation state changed from +2 to 0. Hence it is a reduction reaction.

So, the correct answer is :

The copper II ions gain two electrons and are reduced.

Answer B

Hope this helps!

8 0

3 years ago

Determine how many liters of hydrogen adjusted to STP there are in a 50.0 liter steel cylinder if the pressure inside is 100.0 a

Dvinal [7]

Answer : The volume of hydrogen gas at STP is 4550 L.

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 100.0 atm

$P_2$ = final pressure of gas at STP = 1 atm

$V_1$ = initial volume of gas = 50.0 L

$V_2$ = final volume of gas at STP = ?

$T_1$ = initial temperature of gas = $27.0^oC=273+27.0=300K$

$T_2$ = final temperature of gas at STP = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get:

$\frac{100.0atm\times 50.0L}{300K}=\frac{1atm\times V_2}{273K}$

$V_2=4550L$

Therefore, the volume of hydrogen gas at STP is 4550 L.

3 0

3 years ago

Study the graph in Figure 6.32 and answer to these questions. (see attached image)

Dimas [21]

Hey there :)

We can see that the solubility of salt increases with increasing temperature. This happens with most substances.

To find out the maximum mass of copper sulfate that can be dissolved in water at these temperatures, just interpret the graph.

Considering Y-axis as g copper sulfate/100 g water and the X-axis as the temperature in °C:-

1)

a: 0 °C - 14 g of copper sulfate/100 g of water

b: 50 °C - 34 g of copper sulfate/100 g of water

c: 90 °C - 66 g of copper sulfate/100 g of water

2) From the graph, we can infer that temperature affects the solubility of the salt.

Answered by Benjemin360 :)

3 0

2 years ago

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans

Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of $H_2SO_4$ (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

$H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of $H_2SO_4(aq)$ ? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of $H_2SO_4$ (aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_1,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL$

Putting values in above equation, we get:

$2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M$

0.044 M is the molarity of $H_2SO_4$ (aq).

4 0

2 years ago