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2 years ago

Which of the following properly lists the metals in order from decreasing to increasing ease of oxidation?

2 answers:

6 0

I know it is either B or D, since I know that K will oxidize in water, and Ag will have a tougher time oxidizing. I am going to lean towards D for my answer.

lora16 [44]2 years ago

6 0

Answer is: D. Ag < Cu < Cr < K.

Reactivity series is an empirical progression of a series of metals, arranged by their reactivity from highest to lowest (alkaline metals have highest reactivity and Noble metals lowest reactivity):

K > Ba > Sr > Na > Ca > Mg > Be > Al > Mn > Zn > Cr> Fe > Cd > Co > Ni > Sn > Pb > H(in acids) > Cu > Hg > Ag.

Metal higher in the reactivity series will displace another and it will be oxidized.

For example, potassium is more reactive than silver.

Chemical reaction: K + AgCl → KCl + Ag.

Potassium is oxidized (from oxidation number 0 to +1) and silver is reduced (from oxidation number +1 to 0).

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3 years ago

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What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?

mars1129 [50]

The concentration of solution in M or mol/L can be calculated using the following formula:

$C=\frac{n}{V}$ .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is $K_{2}SO_{4}$ thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass

Putting the values,

$n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol$

Thus, number of moles of $K_{2}SO_{4}$ will be $0.1061\times 0.5=0.053 mol$ .

The volume of solution is 225 mL, converting this into L,

$1 mL=10^{-3}L$

Thus,

$225 mL=0.225 L$

Putting the values in equation (1),

$C=\frac{(0.053 mol}{0.225 L}=0.236 M$

Therefore, concentration of potassium sulfate solution is 0.236 M.

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3 years ago

Draw a structural formula for the alkene you would use to prepare the alcohol shown by hydroboration/oxidation.

Nastasia [14]

Answer:

See explanation

Explanation:

The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.

This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).

In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.

Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.

Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.

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2 years ago

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finlep [7]

Answer:

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