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2 years ago

What sound is produced when lithium react with water?

Chemistry

1 answer:

Arisa [49]2 years ago

5 0

Answer:

lithium reacts with water, forming lithium hydroxide and highly flammable hydrogen

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The decomposition of dinitrogen pentaoxide has an activation energy of 102 kJ/mol and ΔH°rxn = + 55 kJ/mol. What is the activati

ahrayia [7]

Answer: The activation energy of the reverse reaction is 47 kJ/mol

Explanation:

The chemical equation for the decomposition of dinitrogen pentaoxide follows:

$N_2O_5\rightleftharpoons 2NO_2+\frac{1}{2}O_2$

We are given:

Activation energy of the above reaction (forward reaction) = 102 kJ/mol

Enthalpy of the reaction = +55 kJ/mol

As, the enthalpy of the reaction is positive, the reaction is said to be endothermic in nature.

To calculate the activation energy for the reverse reaction, we use the equation:

$E_a_{\text{(forward)}}=E_a_{\text{(backward)}}+|\Delta H|$

where,

$E_a_{\text{(forward)}}$ = Activation energy of the forward reaction = 102 kJ/mol

$E_a_{\text{(backward)}}$ = Activation energy of the backward reaction = ?

$\Delta H$ = Enthalpy of the reaction = +55 kJ/mol

Putting values in above equation, we get:

$102=E_a_{\text{(backward)}}+55\\\\E_a_{\text{(backward)}}=(102-55)=+47kJ/mol$

Hence, the activation energy of the reverse reaction is 47 kJ/mol

3 0

2 years ago

Determine the [OH-], pH, and pOH of a solution with a [H+] of 0.090 M at 25 °C. [OH-] = pH = pOH = 0 POH =

ValentinkaMS [17]

Answer : The concentration of $OH^-$ ion, pH and pOH of solution is, $1.12\times 10^{-13}M$ , 1.05 and 12.95 respectively.

Explanation : Given,

Concentration of $H^+$ ion = 0.090 M

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

The expression used for pH is:

$pH=-\log [H^+]$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.090)$

$pH=1.05$

The pH of the solution is, 1.05

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-1.05=12.95$

The pOH of the solution is, 12.95

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12.95=-\log [OH^-]$

$[OH^-]=1.12\times 10^{-13}M$

The $OH^-$ concentration is, $1.12\times 10^{-13}M$

5 0

2 years ago

Read 2 more answers

Please explain how to find each answer solution.

svp [43]

Explanation:

1) Based on the octet rule, iodine form an I⁻ ion.

Therefore,

Option E is correct ✔

2) The electronic configuration of the sulfide ion (S²⁻) is :

₁₆S = 1s² 2s² 2p⁶ 3s² 3p⁴ or [Ne] 3s² 3p⁴

₁₈S²⁻ = 1s² 2s² 2p⁶ 3s² 3p⁶ or [Ne] 3s² 3p⁶

Therefore,

Option E is correct ✔

3) valence shell electron of

Halogens = 7

Alkali metal = 1

Alkaline earth metal = 2

Therefore,

Option D is correct ✔

4) Group 2 element lose two electron in order to achieve Noble gas configuration.

And here Group 2 element is Sr

Therefore,

Option B is correct ✔

5) Group 13 element lose three electron in order to achieve Noble gas configuration.

And here Group 13 element is Al

Therefore,

Option B is correct ✔

6) For a given arrangements of ions, the lattice energy increases as ionic radius decreases and as ionic charge increases.

Therefore,

Option A is correct ✔

7 0

2 years ago

Why do acids and bases neutralize eachother?

timofeeve [1]

Answer:

Explanation:

When an acid and a base are placed together, they react to neutralize the acid and base properties, producing a salt. The H(+) cation of the acid combines with the OH(-) anion of the base to form water. The compound formed by the cation of the base and the anion of the acid is called a salt.

5 0

2 years ago

Jai besoi daide

JulijaS [17]

Answer:

i dont know what you are saying

Explanation:

????????????

4 0

2 years ago