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3 years ago

PLEASE HELP ME ASAP PLEASE!!!

Chemistry

1 answer:

Sliva [168]3 years ago

4 0

Answer:

$\large \boxed{\text{1763 psi}}$

Explanation:

We can use Dalton's Law of Partial Pressures:

Each gas in a mixture of gases equals its pressure independently of the other gases

$\begin{array}{rcl}p_{\text{NO2}} + p_{\text{CO2}} & = & p_{\text{tot}} \\p_{\text{NO2}} + \text{795 psi} & = &\text{2558 psi} \\p_{\text{NO2}} & = &\text{2558 psi - 795 psi} \\& = & \textbf{1763 psi}\\\end{array}\\\text{The partial pressure of nitrogen dioxide is $\large \boxed{\textbf{1763 psi}}$}$

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A race car is driven by a professional driver at 99 . What is this speed in and ? 1 mile = 1.61 kilometers 1 hour = 60 minutes E

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Answer: The speed is equivalent to 159.39 kilometers per hour or 2.65 kilometers per minute.

Explanation:

Given, The speed of a race car = 99 miles/ hour

To convert the speed into kilometers per hour and kilometers per minute

Since 1 mile = 1.61 kilometers

So, Speed of car = (99 ) x (1.61 )

= 159.39 kilometers per hour.

Also, 1 hour = 60 minutes

Then, Speed of car = (159.39) ÷60

= 2.6565≈2.65 kilometer per minute.

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The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago