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3 years ago

How do cations form?

1 answer:

5 0

For cations, the loss of an electron leaves them with a net positive charge, whereas for anions, the addition of an electron leaves them with a net negative charge. Understanding the processes behind this, including the ionization energy and electron affinity of different atoms, helps you see why certain atoms become ions more easily than others and what causes it to happen.

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What is an isotope?

gizmo_the_mogwai [7]

An isotope of an element is another element that has the same number of protons but a different number of neutrons (for instance $^4_2He$ and $^3_2He$ )

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3 years ago

Read 2 more answers

I have a chemistry question that I am needing help on please?

aliina [53]

What country and grade are you?
Please tell me! I’m going to help you

8 0

3 years ago

Consider 2 NOCl(g) \Longleftrightarrow⟺ 2 NO(g) + Cl2 (g) At 25oC under conditions other than equilibrium, there are 1.20 moles

nekit [7.7K]

Answer:

a) Q = 6.1875x10⁻³

b) The direction of the reaction is to form the products.

c) [Cl₂]e = 0.094 M

[NO]e = 0.190 M

[NOCl]e = 0.140 M

Explanation:

a) Q is the reaction quotient, and for a generic reaction aA + bB ⇄ cC + dD it is

$Q = \frac{[C]^cx[D]^d}{[A]^ax[B]^b}$

Which is the same equation for Kc, but in Kc expressions, the concentrations are in the equilibrium. Q is calculated at any time. So, for the reaction given

2NOCl(g) ⇄ 2NO(g) + Cl2(g)

$Q = \frac{[Cl2]x[[NO]^2}{[NOCl]^2}$

[Cl₂] = 0.220/5.00 = 0.044 M

[NO] = 0.450/5.00 = 0.090 M

[NOCl] = 1.20/5.00 = 0.240 M

Q = (0.044)x(0.090)²/(0.240)²

Q = 6.1875x10⁻³

b) Q < Kc, which means that there are fewer products to what are needed to the equilibrium. So the direction of the reaction is to form the products.

2NOCl(g) ⇄ 2NO(g) + Cl2(g)

0.240 0.090 0.044 Initial

-2x +2x +x Reacts (stoichiometry is 2:2:1)

0.240-2x 0.090+2x 0.044+x Equilibrium

$Kc = \frac{(0.044+x)x(0.090+2x)^2}{(0.240 - 2x)^2}$

$1.86x10^{-1} = \frac{(0.044+x)*(8.1x10^{-3} +0.36x + 4x^2)}{(0.0576 - 0.96x +4x^2)}$

3.564x10⁻⁴+0.01584x+0.176x²+8.1x10⁻³x+0.36x²+4x³ = 0.010714-0.17856x+0.744x²

4x³ - 0.208x² + 0.2025x - 0.01036 = 0

Solving this third grade equation in a computer program:

x = 0.05 M

So:

[Cl₂]e = 0.044 + 0.05 = 0.094 M

[NO]e = 0.090 + 2x0.05 = 0.190 M

[NOCl]e = 0.240 - 2x0.05 = 0.140 M

7 0

3 years ago

Identify the bronsted-lowry acid and the bronsted-lowry base in this reaction on the left side of each of the following equation

Dahasolnce [82]

Answer: $NH_4^+$ is an acid, $CN^-$ is a base, $NH_3$ is conjugate base and $HCN$ is conjugate acid

Explanation:

According to Bronsted and Lowry's theory:

An acid is defined as a proton donor while a base is defined as a proton acceptor.

In a chemical reaction, an acid loses a proton to form a conjugate base while a base accepts a proton to form conjugate acid.

For the given chemical reaction:

$NH_4^+(aq)+CN^-(aq)\rightleftharpoons HCN(aq)+NH_3(aq)$

$NH_4^+$ is losing a proton thus it is an acid to form $NH_3$ which is its conjugate base

$CN^-$ is gaining a proton thus it is a base to form $HCN$ which is its conjugate acid

Hence, $NH_4^+$ is an acid, $CN^-$ is a base, $NH_3$ is conjugate base and $HCN$ is conjugate acid

7 0

3 years ago

For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow

lora16 [44]

Answer:

Rate constant of the reaction is $3.3\times 10^{-3} M^{-2} s^{-1}$ .

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

$R=k[A]^a[B]^b[C]^c$

Rate(R) of the reaction in trail 1 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.30 M$

$R=9.0\times 10^{-5} M/s$

$9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c$ ...[1]

Rate(R) of the reaction in trail 2 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.90 M$

$R=2.7\times 10^{-4} M/s$

$2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c$ ...[2]

Rate(R) of the reaction in trail 3 ,when :

$[A]=0.60 M,[B]=0.30 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c$ ...[3]

Rate(R) of the reaction in trail 4 ,when :

$[A]=0.60 M,[B]=0.60 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c$ ...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

$R=k[A]^2[B]^0[C]^1$

Rate constant of the reaction = k

$9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1$

$k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}$

$k=3.3\times 10^{-3} M^{-2} s^{-1}$

7 0

3 years ago