Five darts strike near the center of the target. Who ever threw the darks is?

Chemistry

2 answers:

Burka [1]1 year ago

7 0

Better than i am and very precice

Salsk061 [2.6K]1 year ago

5 0

I'm not sure how to answer this but I will try, they have a high probability of good eyesight and they either landed on 9, 7, or 5...

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At 700 K, Kp for the following equilibrium is (5.6 x 10-3) 2HgO(s)--> 2Hg(l) + O2(g) Suppose 51.2 g of mercury(II) oxide is p

Vesnalui [34]

Answer: Option (B) is the correct answer.

Explanation:

According to the given reaction equation, formula to calculate $\Delta n$ is as follows.

$\Delta n$ = coefficients of gaseous products - gaseous reactants

= 1 - 0

= 1

Also we know that,

$K_{p} = K_{c} \times (RT)^{\Delta n}$

$5.6 \times 10^{-3} = K_{c} \times (0.0821 \times 700)^{1}$

$K_{c} = 0.097 \times 10^{-3}$

For the equation, $2HgO(s) \rightarrow 2Hg(l) + O_{2}(g)$

Activity of solid and liquid = 1

As, $K_{p} = \frac{P^{2}_{Hg} \times P_{O_{2}}}{P^{2}_{HgO}}$

$5.6 \times 10^{-3} = P_{O_{2}}$

Hence, $P_{O_{2}}$ = 0.0056 atm

Thus, we can conclude that partial pressure of oxygen gas at equilibrium is 0.0056 atm.

5 0

1 year ago

What is [h3o ] in a solution of 0.25 m ch3co2h and 0.030 m nach3co2?

Brilliant_brown [7]

Hello!

To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:

$pH=pKa+log( \frac{[A^{-}] }{[HA]} )$

$pH=4,76+log( \frac{0,030M}{0,25M} ) \\ \\ pH=3,84$

Now, we use the definition of pH and clear [H₃O⁺] from there:

$pH=-log[H_3O^{+}]$

$[H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M$

So, the [H₃O⁺] concentration is 0,00014 M

Have a nice day!

4 0

1 year ago

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Bumek [7]

Answer:

✌3 I think - She should not have multiplied the hydrogen atoms by the coefficient. sorry im not sure but I tried tell, tell me if im wrong Plz!✌