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3 years ago

I would appreciate your help! Due today! ( I don't know if this is even chemistry)

Chemistry

2 answers:

Vinil7 [7]3 years ago

6 0

Answer:

6+69

Explanation:

Tresset [83]3 years ago

3 0

This is the photo of the first page

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In the hypothetical reaction below, substance A is consumed at a rate of 2.0 mol/L·s. If this reaction is at dynamic equilibrium

Goshia [24]

<span>The rate of substance B being consumed should be equal to that of the substance A. Dynamic equilibrium is the balance in a process that is continuing. It is achieved in a reaction when the forward rate of reaction and the backward rate of reaction is at the same value or equal.</span>

3 0

4 years ago

A beaker contains 30.0 mL of 0.1 M silver nitrate solution. What is the minimum volume of 0.1 M ammonium carbonate solution that

kenny6666 [7]

Answer:

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Explanation:

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3 0

3 years ago

For the reaction:

Rom4ik [11]

Answer:

Half-life at 310 K is 6.54 × 10³ s.

Explanation:

Let's consider the following reaction:

2 N₂O₅(g) → 4 NO₂(g) + O₂(g)

The rate law is:

(Δ[O₂]/Δt) = k . [N₂O₅]

Since [N₂O₅] is raised to the power of 1, the reaction order is 1.

For a first-order reaction:

$t_{1/2}=\frac{ln2}{k}$

where,

$t_{1/2}$ is the half-life

$k$ is the rate constant

At 300 K,

$(t_{1/2})_{1}=\frac{ln2}{k_{1}}\\k_{1}=\frac{ln2}{(t_{1/2})_{1}} =\frac{ln2}{2.50 \times 10^{4} s} =2.77 \times 10^{-5} s^{-1}$

We can use two-point Arrhenius equation to solve for k₂ at 310 K

$ln\frac{k_{2}}{k_{1}} =\frac{-Ea}{R} (\frac{1}{T_{2}} -\frac{1}{T_{1}} )\\ln\frac{k_{2}}{k_{1}} =\frac{-103.3kJ/mol}{8.314 \times 10^{-3} kJ/mol.K} .(\frac{1}{310K}-\frac{1}{300K} )\\ln\frac{k_{2}}{k_{1}}=1.34\\k_{2}=1.06 \times 10^{-4} s^{-1}$

At 310 K,

$(t_{1/2})_{2}=\frac{ln2}{k_{2}}=\frac{ln2}{1.06 \times 10^{-4} s^{-1} } =6.54 \times 10^{3} s$

4 0

4 years ago

Describe a trip to space and how the human body will react

dlinn [17]

Answer:

The Human body will react in lots of different ways, look down below

Explanation:

Bone mass decreases
The Body grows slightly taller

That's all I know at the moment. Hope I could help!

7 0

3 years ago

Calculate the pH at 25 degrees celsius of a 0.39 M solution of pyridinium chloride (c5h5nhcl) . Note that pyridine (c5h5n) is a

lutik1710 [3]

Answer:

$pH=2.3$

Explanation:

Hello!

In this case, since pyridinium chloride has a pKb of 8.77 which is a Kb of 1.70x10⁻⁹ and therefore a Ka of 5.89x10⁻⁵ which means it tends to be acidic, we write its ionization via:

$C_5H_5NHCl(aq)+H_2O(l)\rightleftharpoons C_5H_5NCl^-(aq)+H_3O^+(aq)$

Because it is a Bronsted base which donates one hydrogen ion to the water to produce hydronium. Thus, we write the equilibrium expression with the aqueous species only:

$Ka=\frac{[C_5H_5NCl^-][H_3O^+]}{[C_5H_5NHCl]}$

In terms of the reaction extent $x$ , we write:

$5.89x10^{-5}=\frac{x*x}{0.39-x}$

Thus, solving for $x$ we obtain:

$x_1=-0.0048M\\\\x_2=0.0048M$

Clearly the solution is 0.0048 M because to negative values are not allowed, therefore, since it equals the concentration of hydronium which defines the pH, we write:

$pH=-log([H_3O^+])=-log(0.0048)\\\\pH=2.3$

Best regards!

5 0

3 years ago