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3 years ago

Question 15 (1 point)

Chemistry

1 answer:

zimovet [89]3 years ago

6 0

Answer:

Option B. 0.5 N.

Explanation:

Data obtained from the question include:

Mass (m) = 1 Kg

Acceleration (a) = 0.5 m/s²

Force (F) =?

Force is defined as the product of mass (m) and acceleration (a). Mathematically, it is expressed as:

Force (F) = mass (m) × acceleration (a)

F = ma

Using the above equation, we can obtain the force exerted on the floor as follow:

Mass (m) = 1 Kg

Acceleration (a) = 0.5 m/s²

Force (F) =?

F = ma

F = 1 × 0.5

F = 0.5 N

Thus, the car exerted a force of 0.5 N on the floor.

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The following Lewis diagram represents the valence electron configuration of a main-group element. This element is in group 2A A

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Answer: The symbol of the ion formed is $Ca^{2+}$

Explanation:

An ion is formed when a neutral atom looses or gains electrons.

When an atom looses electrons, it results in the formation of positive ion known as cation.
When an atom gains electrons, it results in the formation of negative ion known as anion.

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

The element present in Group 2-A and in period 4 is Calcium (Ca)

Electronic configuration of Ca atom: $1s^22s^22p^63s^23p^64s^2$

This atom will loose 2 electrons to attain stable electronic configuration similar to Argon element (noble gas)

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The correct answer is B. Electrons will be added to an atom starting from the lowest possible level or sublevel.

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Learn how changes in binding free energy affect binding and the ratio of unbound and bound molecules.

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C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

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F)[D'] / [ED'] = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)

E + D ⇄ ED → K_a = [ED] / [D][E] (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED] (dissociation constant)

[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

=130/25 = 5.20

[D]/[ED] = 5.20

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

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ΔG_1 = -8184.633 * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

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-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED'] E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

[D'] / [ED'] = 0.0579

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