A mass of mercury occupies 0. 950 L. What volume would an equal mass of ethanol

Question

A mass of mercury occupies 0. 950 L. What volume would an equal mass of ethanol occupy? The density of mercury is 13.546 g/mL and the density of ethanol is 0.789 g/mL. 0.0 553 L 0.0613 L 16.3 L 18.1 L Combustion of hydrogen releases 142 kJ per gram of hydrogen reacted. How many kilocalories of energy arc released by the combustion of 16.0 ounces of hydrogen? 1.31 kcal 19.2 kcal 1.54 times 10^4 kcal 2.69 times 10^5 kcal What is the ground-state electron configuration of Se^2-? [Ar] 3d^10 4s^2 4p^2 [Ar] 3d^10^10 4s^2 4p^4 [Ar] 3d^12 4s^2 4p^4 [Ar] 3d^10 4s^2 4p^6 Of the following, which atom has the smallest atomic radius? Na S K Se Which pair of reactants will produce a precipitate when mixed together? HCl(aq) and Ca(OH(aq) HCl(aq) and K_2CO_3(aq) HCl(aq) and Na_2S(aq) HCI(aq) and AgNO_3(aq) Which of the following represents the change in electronic configuration that is associated with the first ionization energy of magnesium? [Ne]3s^1 3p^1implies [Ne]3s^1 + e^- [Ne]3s^2 implies [Ne]3s^1 3p^1 [Ne] 3s^2 implies [Ne] 3s^1 + e^- [NE)3S^2 + e^- implies [Ne] 3s^2 3p^1 What is the general trend in ionization energy and electron affinity values? Both decrease as one traverses a period from left to right and both decrease as one descends a group. Both decrease as one traverses a period from left to right and both increase as one descends a group Both increase as one traverses a period from left to right and both decrease as one descends a group Both increase as one traverses a period from left to right and both increase as one descends a group.

Explanation / Answer

Solution:- (6)

volume of mercury = 0.950 L

density of mercury = 13.546 g/ml

and density of ethanol = 0.789 g/ml

From the given density and volume we will calculate the mass of mercury and then there would exactly same mass of ethanol be there as is mentioned.

we know that, mass = volume x density

So, mass of mercury = 0.950 L x (1000ml/1L) x (13.546g/ml) = 12868.7g

Now we have mass and density of ethanol so it's volume could easily be calculated as volume = mass/density

volume of ethanol = 12868.7g x (1ml/0.789ml) x (1L/1000ml) = 16.31 L

So, the correct choice is (C) 16.3 L

(7)

combustion energy for hydrogen is given as 142 kJ/g

mass of hydrogen is given as 16.0 ounces. So, we would convert these ounces into pounds and then kg and finally into grams.

(1 lb = 16 oz, 1 kg = 2.20 lb and 1 kg = 1000g)

So, 16.0 oz x (1lb/16oz) x (1kg/2.20 lb) x (1000g/1kg) = 454.5 g

now we multiply these grams by the energy given as kJ/gram and convert kJ into kcal.

(1 kcal = 4.184 kJ)

so, 454.5 g x (142kJ/g) x (1kcal/4.184kJ) = 15425 kcal or 1.54 x 104 kcal

Hence, the correct choice is (C) 1.54 x 104 kcal.

(8)

Atomic number of Se is 34. Since there is -2 charge on Se so the total electrons would be 34 +2 = 36

First we will write the electronic configuration and then look for the noble gas configuration.

To write the electronic configuration we use the Aufbau principle. It tells us about the order of shells and subshells in increasing order of energy.

The increasing order is like 1s < 2s < 2p < 3s < 3p <4s < 3d < 4p and so on....

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

Now we look for the noble gas that comes exactly before Se and it is Ar with atomic number 18. If we look at the electronic configuration then 18 electrons are completed upto 3p. So, in place of 1s2 2s2 2p6 3s2 3p6 we would write [Ar] next to it would write the remaining part of the electronic configuration. So, the right one would be..

[Ar] 4s2 3d10 4p6. It could also be written as [Ar] 3d10 4s2 4p6

Hence, the correct choice is (D) [Ar] 3d10 4s2 4p6.

(9)

Aomic radius decreases on moving left to right in a period and increases on moving top to bottom in a group.

Na and K are the elements of first group and K is below Na so the size is smaller for Na than K.

S and Se are elements of 16th group and Se is below S so the size of S is smaller than Se.

Now if we compare Na and S then they are present in same period(third period) and S is on right side. So, the smallest size is of S in all the give ones.

Hence, the correct choice is (B) S.

(10)

Let's first write the reactions for all of them and then we would check out the solubility rules.

(A) 2HCl(aq) + Ca(OH)2(aq) --------> CaCl2(aq) + 2H2O(l)

(B) 2HCl(aq) + K2CO3(aq) -------> 2KCl(aq) + H2O(l) + CO2(g)

(C) 2HCl(aq) + Na2S(aq) ---------> 2NaCl(aq) + H2S(aq)

(D) HCl(aq) + AgNO3(aq) -------> HNO3(aq) + AgCl(s)

From solubility rules, It is only AgCl which forms a precipitate. So, the correct choice is (D).

(11)

Atomic number of Mg is 12 and it's electronic configuration is 1s2 2s2 2p6 3s2. The noble gas that comes before Mg is Ne and its electronic configuration is 1s2 2s2 2p6. So, In short hand notation electronic configuration could be written as [Ne]3s2.

First ionization energy means the energy requitred to remove one electron from a gaseous atom.

So, [Ne]3s2 --------> [Ne]3s1 + e-

So, the correct choice is (C) [Ne]3s2 --------> [Ne]3s1 + e-

(12)

we know that atomic radius decreases as we move left to right and increases top to bottom. As the size decreases it is hard to remove electron from valence shell dure to large nuclear charge. Same time It's also hard to add electron to the smaller molecules since there becomes repulsion between outer electrons and the coming electron.

So, the correct choice is (C) Both increases as one traverses a period from left to right and both decreases as one descends a group.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.