*pls hurry, will give brainlyest or whatever!!*

I have a few questions from my homework and I was wondering if I can get some help

sashaice [31]

Well it breaks down into small parts

3 0

3 years ago

Question 10 OT 20

Airida [17]

Answer:

B) is reduced.

Explanation:

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Consider the following reactions.

4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced and it is oxidizing agent.

CO + H₂O → CO₂ + H₂

the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized and it is reducing gent.

Oxidizing agents:

Oxidizing agents oxidize the other elements and itself gets reduced.

Reducing agents:

Reducing agents reduced the other element are it self gets oxidized.

8 0

3 years ago

What is the pH of a KOH solution that has [H ] = 1. 87 × 10–13 M? What is the pOH of a KOH solution that has [OH− ] = 5. 81 × 10

vlada-n [284]

pH is the hydrogen ion concentration and pOH is the hydroxide ion concentration in the solution. pH KOH is 12.73, pOH KOH is 2.24 and pH NaCl is 7.

pH is the negative log of the hydrogen ion concentration and pOH is the negative log of the hydroxide ion concentration.

The relation between the pH and pOH can be given as, $\rm pOH = 14 - pH$

The pH of KOH can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the first case, the concentration of the KOH is $1. 87 \times 10^{-13}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 87 \times 10^{-13}\;\rm M ]\\\\&= 12.73\end{aligned}$

Hence, the pH of KOH is 12.73.

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pOH of KOH can be calculated by the formula,

$\rm pOH = \rm -log [OH^{-}]$

The hydroxide concentration of the KOH solution is $5. 81 \times 10^{-3}\;\rm M$

Substituting value in the equation:

$\begin{aligned} \rm pOH &= \rm -log [OH^{-}]\\\\&= \rm -log [5. 81 \times 10^{-3}\;\rm M ]\\\\&= 2.24 \end{aligned}$

Hence, the pOH of KOH is 2.24

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The pH of NaCl can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the third case, the concentration of the NaCl is $1. 00\times 10^{-7}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 00 \times 10^{-7}\;\rm M ]\\\\&= 7 \end{aligned}$

Hence, the pH of KOH is 7.0.

Therefore, KOH is basic and NaCl is approximately neutral.

Learn more about pH and pOH here:

brainly.com/question/13885794

3 0

3 years ago

Molarity to percent by mass. Convert 1.672 mol/L MgCl2(aq) solution to percent by mass of MgCl2 in the solution. The solution de

nignag [31]

Answer:

$\%m/m=14\%$

Explanation:

Hello!

In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:

$[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}$

Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:

$[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14$

Which is also the by-mass fraction and in percent it turns out:

$\%m/m=0.14*100\%\\\\\%m/m=14\%$

Best regards!

6 0

3 years ago

What did timbuktu contributed to mali's important as a kingdom

Alex73 [517]

Timbuktu was considered a MASSIVE trade center in Mali. That is how all of the foreign goods came into the kingdom. It also had education centers which was very important in Mali.

5 0

3 years ago

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