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nataly862011 [7]
2 years ago
6

For which of the following properties does sodium have a larger value than rubidium? Select all that apply.

Chemistry
1 answer:
Doss [256]2 years ago
4 0

Answer:

Ionization energy

Electronegativity

Explanation:

-due to its smaller ionic radius....the electron in the outter most shell tends to expierence a stronger nuclear attraction...which makes it harder to remove the electron from the sodium atom

-Rubidium has lesser ionization energy because its (i) affected by its larger ionic radius which tends to lessen the nuclear attraction ...hence making it easier to remove the electron...(ii)and also by the screening effect done by the inner shells, which also tends to lessen the nuclear attraction.

Sodium has a higher electronegativity than rubidium;

Electronegativity is the charge density of electrons in an atom...in which its high when the atomic radius is smaller...

So hence due to the sodium atomic radius being smaller...it tends to have a higher charge density than rubidium....which then gives it a higher electronegativity value

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HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.
Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, C_A = 6.0mol/dm³

volume of HA, V_A  = 25.0cm³, = 0.025dm³

Concentration of HB, C_B = 3.0mol/dm³

volume of HB, V_B = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

C_iVi = C_fV_f

where;

C_i is initial concentration

V_i is initial volume

C_f is final concentration of the solution

V_f is final volume of the solution

C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0
2 years ago
How could one rapidly separate red #40 from zinc oxide? indicate every step.
tensa zangetsu [6.8K]
Red #40 is soluble in water while zinc oxide is not.
So the easiest way to separate them is as follows:
1- add water to the mixture until red #40 is dissolved in the water
2- filter to separate the zinc oxide 
4- heat the solution of red #40 and water until water evaporates and red#40 remains.
6 0
1 year ago
Read 2 more answers
When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t
lisov135 [29]

Answer:

1.59mol/L

Explanation:

Data obtained from the question include:

Mass of MgCl2 = 151g

Volume of water(solvent) = 1L

Now, let us calculate the number of mole of MgCl2. This is illustrated below:

Molarity Mass of MgCl2 = 24 + (2x35.5) = 24 + 71 = 95g/mol

Mass of MgCl2 = 151g

Number of mole of MgCl2 =?

Number of mole = Mass /Molar Mass

Number of mole of MgCl2 = 151/95

Number of mole of MgCl2 = 1.59mole

Now we can calculate the molarity of MgCl2 as follow:

Mole = 1.59mole

Volume = 1L

Molarity =?

Molarity = mole /Volume

Molarity = 1.59/1

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8 0
1 year ago
What is the law of conservation of energy as it applies to exothermic dissolution processes?Energy given off by the system durin
Pani-rosa [81]

Answer:

The energy given off by the system during dissolution is less than the energy absorbed by the surroundings.

Explanation:

Energy is used to break bonds in reactants, and energy is released when new bonds form in products. Endothermic reactions absorb energy, and exothermic reactions release energy. The law of conservation of energy states that matter cannot be created or destroyed.

4 0
1 year ago
Suppose that a 1.0-L reaction vessel initially contains 1.0 mol of O3 and 1.0 mol of O2. What fraction of the O3 will have react
NemiM [27]

Answer:

Explanation:

given reaction

                        2O₃        ⇄      3O₂

 initial                1                      1

    final           1-x                   1+ 1.5x

rate of reaction = k  [ O₃ ]² / [ O₂ ]

initial rate = k . 1² / 1 = k

final rate = k (1-x )² / (1+ 1.5x) = k/2

(1-x )² / (1+ 1.5x) = 1/2

2x² - 5.5 x + 1 = 0

x = 0.196

So fraction of ozone reacted = .196

8 0
1 year ago
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