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3 years ago

Bob rides his horse 52 km in 3 hours 15 minutes. What is his average speed in kilometers per hour?

Chemistry

1 answer:

laila [671]3 years ago

7 0

I think the answer is 16.51

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3 years ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.

Back to this question. $[\text{H}^{+}]$ is given. 25 °C implies that $K_w = 10^{-14}$ . As a result,

$\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}$ .

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3 years ago

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Beryllium, Magnesium, Calcium....etc have two valence electrons

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2 years ago

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Answer:

A. 0.0440 moles/day

Explanation:

First, let's figure out how many moles 33.23 grams of silver is. We do this by dividing the number of grams by the molar mass of silver, which is 107.87 g/mol:

33.23 g Ag ÷ 107.87 g/mol = 0.3081 mol Ag

Now, let's divide this by 7 to get the rate per day:

0.3081 mol Ag ÷ 7 days = 0.0440 mol/day

Thus, the answer is A.

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3 years ago

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3 years ago

Bob rides his horse 52 km in 3 hours 15 minutes. What is his average speed in kilometers per hour?​

Bob rides his horse 52 km in 3 hours 15 minutes. What is his average speed in kilometers per hour?