Ask question

Ask question

All categories

3 years ago

15

Given the reaction below, which is the oxidized substance?

2 answers:

IRINA_888 [86]3 years ago

8 0

I would say the answe is C

tresset_1 [31]3 years ago

3 0

Answer:

Mg ²⁺

Explanation:

Τhe metal loses electrons and in forming Mg²⁺ ,it loses 2 electrons and hence oxidized.

Mg(s) ⇒ Mg²⁺ + 2e⁻

You might be interested in

The energy exchange with the environment during a chemical reaction is called the heat of reaction.

JulijaS [17]

This is a hard question

6 0

3 years ago

What is the representative particle for barium chloride?

KIM [24]

Barium chloride is salt with ionic bonds. Barium chloride contains one barium cations (charge of ion is 2+) and two chlorine anions (charge for ion is 1-).
The representative particle<span> of an ionic compound is its formula unit (</span>lowest whole number ratio of ions in an ionic compound), for barium chloride that is BaCl₂.

8 0

2 years ago

Read 2 more answers

What are non examples of tissue? Help please

antiseptic1488 [7]

Epithelial

I know just one

7 0

2 years ago

Which sub-layer thins out into space, where there is no air

KengaRu [80]

<span>The "exosphere" is the most distant and tenuous "layer" of our atmosphere.</span>

3 0

2 years ago

Read 2 more answers

Be sure to answer all parts.The equilibrium constant, Kc, for the formation of nitrosyl chloride from nitric oxide and chlorine,

djverab [1.8K]

<u>Answer:</u> The reaction proceeds in the forward direction

<u>Explanation:</u>

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

Relation of $K_p\text{ with }K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta n_g}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $6.5\times 10^4$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $35^oC=[35+273]K=308K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=6.5\times 10^4\times (0.0821\times 500)^{-1}\\\\K_p=1583.43$

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

The expression of $Q_p$ for above equation follows:

$Q_p=\frac{(p_{NOCl})^2}{p_{Cl_2}\times (p_{NO})^2}$

We are given:

$p_{NOCl}=1.76atm$

$p_{NO}=1.01atm$

$p_{Cl_2}=0.42atm$

Putting values in above equation, we get:

$Q_p=\frac{(1.76)^2}{0.42\times (1.01)^2}=7.23$

We are given:

$K_p=1583.43$

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium

As, $K_p>Q_p$ , the reaction will be favoring product side.

Hence, the reaction proceeds in the forward direction

4 0

2 years ago