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Ann [662]
3 years ago
14

What may be expected when K < 1.0? Choose the THREE correct statements. The concentration of one or more of the reactants is

small. The concentration of one or more of the products is small. The reaction will not proceed very far to the right. The reaction will generally form more reactants than products.
Chemistry
1 answer:
mestny [16]3 years ago
4 0

Answer:

The concentration of one or more of the products is small.

The reaction will not proceed very far to the right.

The reaction will generally form more reactants than products  

Explanation:

We often write

K =[Products]/[Reactants]

Thus, if K is small

  • We have fewer products than reactants
  • We have more reactants than products
  • The position of equilibrium lies to the left

A. is wrong. Usually, if K < 1, the concentration of reactants is greater than that of the products.

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When a solution of NaI reacts with a solution of AgNO3, what is the net ionic equation?
Leni [432]

Answer:

Ag⁺(aq) + I⁻(aq) → AgI(s)

Explanation:

Net ionic equation is a way to write a chemical equation in which you are listing only the species that are participating in the reaction.

In the reaction:

AgNO₃(aq) + NaI(aq) → AgI(s) + NaNO₃(aq).

The ionic equation is:

Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + I⁻(aq) → AgI(s) + Na⁺(aq) + NO₃⁻(aq).

Now, listing only the species that are participating in the reaction:

<h3>Ag⁺(aq) + I⁻(aq) → AgI(s)</h3>

3 0
3 years ago
An unknown acid solution has PH 3.4. 66% of the acid is ionized. Whats the pka?
juin [17]

Answer:

pKa=3.58

Explanation:

Hello,

In this case, since the pH defines the concentration of hydrogen:

pH=-log([H^+])

[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}

And the percent ionization is:

\% \ ionization=\frac{[H^+]}{[HA]}*100\%

We compute the concentration of the acid, HA:

[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%}  *100\%\\\\

[HA]=6.03x10^{-4}

Thus, the Ka is:

Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\  \\Ka=2.63x10^{-4}

So the pKa is:

pKa=-log(Ka)=-log(2.63x10^{-4})\\\\pKa=3.58

Regards.

5 0
3 years ago
If a bug is traveling 5 meters across the floor in 5 seconds. How fast did it<br> travel?
Sergio039 [100]

Answer:

5 seconds

Explanation:

Speed x Time. So t=ds. t=51=5.

4 0
2 years ago
PLEASE ANSWER FAST AND YOU WILL GET BRAINLIESTTTTTTT!!!!!!! And POINTSSSSSS!! PLEASE ANSWER FAST AND YOU WILL GET points and BRA
stiks02 [169]

Answer:

B on both questions

Explanation:

6 0
3 years ago
Read 2 more answers
If the initial temperature of a movable cylinder was 50 degrees Celsius
slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0
3 years ago
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