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PtichkaEL [24]
2 years ago
15

If a 28.0 L balloon with a temperature of

Chemistry
1 answer:
Alexeev081 [22]2 years ago
6 0

Answer:

5.6 L

Explanation:

We can apply Charles' Law here since our pressure is constant (will not change inside the refrigerator) and we are relating change in volume with change in temperature:

V₁ / T₁ = V₂ / T₂ where V₁ and T₁ are initial volume and temperature, and V₂ and T₂ are final volume and temperature. Let's plug in what we know and solve for the unknown:

28.0 L / 25 °C = V₂ / 5 °C => V₂ = 5.6 L

5.6 L is our new volume (at 5 °C).

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Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g) Equilibrium concentrations: [H2] = 0.322 M and [CH3
Annette [7]

Answer:

The equilibrium value of [CO] is 1.04 M

Explanation:

Chemical equilibrium is the state to which a spontaneously evolving  chemical system, in which a reversible chemical reaction takes place.  When this situation is reached, it is observed that the  concentrations of substances, both reagents and reaction products,  they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.

Reagent concentrations  and products in equilibrium are related by the equilibrium constant Kc. Being:

aA + bB ⇔ cC + dD

Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }

Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case:

Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }

You know:

  • Kc= 14.5
  • [H₂]= 0.322 M
  • [CH₃OH] =1.56 M

Replacing:

14.5=\frac{1.56}{[CO]*0.322^{2} }

Solving:

[CO]=\frac{1.56}{14.5*0.322^{2} }

[CO]= 1.04 M

The equilibrium value of [CO] is 1.04 M

8 0
3 years ago
Since acids have 1 more proton (H+ - ions) than base, and the acid gives it away, doesn't that mean that they switch roles? Acid
andreev551 [17]

Answer:

In an acid-base equilibrium, acid becomes a conjugate base and base becomes a conjugate acid.

Explanation:

Let's remember the Bronsted-Lowry theory to answer this specific question. According to the theory, acid is a proton donor, while a base is a proton acceptor.

Consider an acid in a form HA (aq) and base in a form of B (aq). Since acid is a proton donor, it will donate its hydrogen ion to the base, B. The resultant products would be A^{-} (aq) and BH^{+} (aq).

Remember that an acid-base reaction is an equilibrium reaction. This means we may also look at this proton transfer reaction from the product side towards the reactants. Summarizing what has been said, we may write the equilibrium as:

HA (aq) + B (aq) ⇄ BH^{+} (aq) + A^{-} (aq)

Now acid, HA, donates a proton to become a conjugate base. The conjugate base, if we look from the reverse equation side, is actually a base, since it can accept a proton to become HA. Similarly, B accepts a proton to become a conjugate acid. Looking from the reverse reaction, it can now donate a proton, so in reality we can consider it a base.

To summarize, your logic is correct.

6 0
3 years ago
A cart accelerating slower when mass was increased.<br> 1st Law - 3rd Law
svet-max [94.6K]

Answer:

Newton's second law

Explanation:

It is mentioning acceleration and mass

Newton's second law's equation = F = m*a

Hope u understood

Please mark brainliest

Thank You

8 0
2 years ago
What type of bond is composed of a metal and anonmetal?
sesenic [268]

Answer:

anonmetal

Explanation:

6 0
3 years ago
Read 2 more answers
The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati
a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.80M^{-1}s^{-1}

t = time taken  = 95 s

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.86 M

Now put all the given values in above equation, we get:

0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0
3 years ago
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