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4 years ago

Last question based on my previous question

Chemistry

1 answer:

nexus9112 [7]4 years ago

6 0

Answer:

Explanation:

1) White tile can be placed underneath the conical flask to make easier spotting the end point of color change.

2) Because it will affect the number of molecules of reactant in the flask.

Acid is added slowly into alkali solution mixed with indicator

The color of indicator changed. So, the mixture has neutralised.

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FILL IN THE BLANKWORD BANK (can use more than once): less, increases, decreases, greaterNO2Cl(g)+NO(g)⇌NOCl(g)+NO2(g)1.Disturbin

Jlenok [28]

Explanation:

If we have the following reaction at equilibrium:

aA + bB ⇄ cC + dD

where a, b, c, and d are the stoichiometric coefficients for the reacting species A, B, C, and D. For the reaction at a particular temperature:

Kc=([C]^c *[D]^d)/([A]^a *[B]^b)

where Kc is the equilibrium constant, which holds that for a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentrations has a constant value, Kc (the equilibrium constant). Note that although the concentrations may vary, as long as the reaction in in equilibrium and temperatura don't change, the value of K remains constant.

For reactions that have not reached equilibrium, we obtain the reaction quotient (Qc), instead of the equilibrium constant by substituting the initial concentrations into the equilibrium constant expression.

Qc=([Co]^c *[Do]^d)/([Ao]^a *[Bo]^b)

To determine the direction in wich the net reaction will proceed to reach equilibrium, que compare the values of Qc and Kc.

Qc < Kc: To reach equilibrium, reactants must be converted to products (→)
Qc = Kc: The initial concentrations are equilibrium concentrations. The system in at equilibrium.
Qc > Kc: To reach equilibrium, products must be converted to reactants (←)

Solution:

We have the following reaction:

NO2Cl(g)+NO(g)⇌NOCl(g)+NO2(g)

So:

Kc=([NOCl]^1*[NO2]^1)/([NO2Cl]^1 *[NO]^1)

=([NOCl][NO2])/([NO2Cl][NO])

1. In the equation above, [NO2Cl] it's in the denominator, so if we increase it's numericall value by adding NO2Cl decreases Qc to a value less than Kc.

(From the chemical point of view, if we disturb the equilibrium adding NO2Cl (a reactant), to reach equilibrium again the system proceeds from left to right (→) consuming this reactant.)

2. To reach a new state of equilibrium (where Qc = Kc), Qc therefore increases wich means that the denominator of the expression for Qc decreases (in order to increase the denominator as mention above).

3. To accomplish this, the concentration of reageants decreases (reagents are being consumed), and the concentration of prodcuts increases (products are being formed).

4 0

3 years ago

How do the inner transition metals properly fit into the rest of the table?

SashulF [63]

They fit in by atomic mass and amount of valence electrons categorized in families

8 0

4 years ago

Safrole was once used as a flavoring in root beer, until it was banned in 1960. what is the vapor pressure of a solution prepare

BartSMP [9]

Answer is: 48,25 torr.
Raoult's Law: p = x(solv) · p(solv)
p - vapour pressure of a solution.
x(solv) - mole fraction of the solvent.
p(solv) - vapour pressure of the pure solvent.
n(ethanol) = 950g ÷ 46,07g/mol = 20,62 mol.
x(solv) = moles of solvent ÷ total number of moles
x(solv) = 20,62 ÷ 21,77 = 0,965.
p = 0,965 ·50,0 torr = 48,25 torr.

4 0

3 years ago

Read 2 more answers

Which of the following changes to a system WILL NOT result in an increase in the system’s pressure?

Sophie [7]

Answer:

B. All the rest raise P.

Explanation:

6 0

3 years ago

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

Sauron [17]

Answer:

The concentration of COF₂ at equilibrium is 0.296 M.

Explanation:

To solve this equilibrium problem we use an ICE Table. In this table, we recognize 3 stages: Initial(I), Change(C) and Equilibrium(E). In each row we record the concentrations or changes in concentration in that stage. For this reaction:

2 COF₂(g) ⇌ CO₂(g) + CF₄(g)

I 2.00 0 0

C -2x +x +x

E 2.00 - 2x x x

Then, we replace these equilibrium concentrations in the Kc expression, and solve for "x".

$Kc=8.30=\frac{[CO_{2}] \times [CF_{4}] }{[COF_{2}]^{2} } =\frac{x^{2} }{(2.00-2x)^{2} } \\8.30=(\frac{x}{2.00-2x} )^{2} \\\sqrt{8.30} =\frac{x}{2.00-2x}\\5.76-5.76x=x\\x=0.852$

The concentration of COF₂ at equilibrium is 2.00 -2x = 2.00 - 2 × 0.852 = 0.296 M

6 0

3 years ago