Water boils at a temperature of _______ degrees on the celsius scale. a. 212 b. 194 c. 100 d. 90

Well, we might as well try a similar question. Magnesium metal reacts with hydrochloric acid to produce hydrogen gas and an aque

Andrei [34K]

Answer: Magnesium Mg

Explanation:

Oxidization is the process by which a substance either gains oxygen or losses electrons.

The chemical reaction of the above is denoted by,

Mg(s) + 2HCl(aq) -----> MgCl2(aq) + H2(g)

Mg went from a 0 to a +2 state which would mean that it lost electrons.

It was therefore oxidized.

Please do react or comment if you need clarification or if the answer helped you. This can help other users as well. Thank you.

5 0

2 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

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

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

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

T = temperature = 500 K

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

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

2 years ago

This reaction appears to be destroying the logs

Alex_Xolod [135]

What exactly is the question you are asking?

5 0

2 years ago

Find the mass of 1 atom of Chlorine in gram.

vazorg [7]

B. 35.45 ! Hope this helps

6 0

2 years ago

Methane and water react to form hydrogen and carbon monoxide, like this:

Usimov [2.4K]

Based on Le Chatelier's principle, if a system at equilibrium is disturbed by changes in the temperature, pressure or concentration, then the equilibrium will shift in a direction to undo the effect of the induced change.

The given reaction is endothermic i.e, heat is supplied:

CH4(g) + H2O (g) + heat ↔ 3H2(g) + CO(g)

a) When the temperature is lowered, heat is being removed from the system. The reaction will move in a direction to produce more heat i.e. to the left.

Hence, the pressure of CH4 will increase and equilibrium will shift to the left

b) When the temperature is raised, heat is being added to the system. The reaction will move in a direction to consume the added heat i.e. to the right.

Hence, the pressure of CO will increase and equilibrium will shift to the right

7 0

2 years ago