The answer to #11 and why that's the correct answer

The solubility of water in diethyl ether has been reported to be 1.468 % by mass.' Assuming that 23.0 mL of diethyl ether were a

melomori [17]

Explanation:

As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.

Hence, amount of diethyl ether present will be calculated as follows.

(100ml - 1.468 ml)

= 98.532 ml

So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.

Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.

Amount of water = $\frac{1.468 ml \times 23 ml}{98.532 ml}$

= 0.3427 ml

Now, when magnesium dissolves in water then the reaction will be as follows.

$Mg + H_{2}O \rightarrow Mg(OH)_{2}$

Molar mass of Mg = 24.305 g

Molar mass of $H_{2}O$ = 18 g

Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.

Amount of Mg = $\frac{24.305 g \times 0.3427 ml}{18 g}$

= 0.462 g

6 0

3 years ago

What would happen to life if earth were in a different location within the solar system

chubhunter [2.5K]

Closer=Burn
Farther=Freeze
We are the perfect distance away from the sun for it to sustain life.

3 0

3 years ago

If expending 3500 kcal is equal to a loss of 1.0 lb, how many days will it take Charles to lose 5.0 lb? Express your answer to t

Trava [24]

Answer:

WAIT

good luck on whatever this is for my dude.

3 0

3 years ago

Thallium-207 decays exponentially with a half life of 4.5 minutes. if the initial amount of the isotope was 28 grams, how many g

Agata [3.3K]

An exponential decay law has the general form: A = Ao * e ^ (-kt) =>

A/Ao = e^(-kt)

Half-life time => A/Ao = 1/2, and t = 4.5 min

=> 1/2 = e^(-k*4.5) => ln(2) = 4.5k => k = ln(2) / 4.5 ≈ 0.154

Now replace the value of k, Ao = 28g and t = 7 min to find how many grams of Thalium-207 will remain:

A = Ao e ^ (-kt) = 28 g * e ^( -0.154 * 7) = 9.5 g

Answer 9.5 g.

7 0

3 years ago

Group the following electron configurations in pairs that would represent similar chemical properties at their atoms;

marishachu [46]

Answer: Pairs are: (a) and (d), (b) and (f), (c) and (e)

Explanation:

In a periodic table, elements are arranged in 18 vertical columns known as groups and 7 horizontal rows known as periods.

Elements arranged in a group show similar chemical properties because of the presence of same number of valence electrons.

Valence electrons are defined as the electrons which are present in the outermost shell of an atom. Outermost shell has the highest value of 'n' that is principal quantum number.

For the given options:

For a:

The given electronic configuration is: $1s^22s^22p^63s^2$

The number of valence electrons in the given configuration are 2

For b:

The given electronic configuration is: $1s^22s^22p^63s^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

For c:

The given electronic configuration is: $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For d:

The given electronic configuration is: $1s^22s^2$

The number of valence electrons in the given configuration are 2

For e:

The given electronic configuration is: $1s^22s^22p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For f:

The given electronic configuration is: $1s^22s^22p^63s^23p^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

Electronic configuration of (a) and (d) will form a pair, (b) and (f) will form a pair, (c) and (e) will form a pair and will have similar chemical properties.

Hence, the pairs are: (a) and (d), (b) and (f), (c) and (e)

4 0

3 years ago