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4 years ago

I need help on #'s 4 and 5

Chemistry

1 answer:

kompoz [17]4 years ago

3 0

The answer for four is frequency.

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How many grams of solid barium sulfate form when 25.0 ml of 0.160 m barium chloride reacts with 68.0 ml of 0.055 m sodium sulfat

GalinKa [24]

Hope this will help you.

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4 years ago

What element produces violet light of approximately 412 nm?

Natali [406]

a. Hydrogen

Hydrogen has a line at 410 nm.
Mercury has a line at 405 nm.
Sodium and neon have no lines near 412 nm.

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3 years ago

Read 2 more answers

A small diamond, which is made of pure carbon, contains too many carbon atoms to count individually. Which is closest to the num

8090 [49]

The correct answer is option C, 5.02 x 10²² carbon atoms

Atomic mass of C = 12 g/mol

According to Avogadro, 1 mole of C has 6.023 x 10²³C atoms

Now 1 mole of C is equal to 12 g

Therefore, 12 g of C = 6.023 x 10²³ C atoms

1 g of C = $\frac{6.023 x 10^2^3}{12}$ C atoms = 5.02 x 10²² C atoms

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3 years ago

Is this a scientific model? Use complete sentences to explain why or why not

Fittoniya [83]

Answer:

it is

Explanation:

8 0

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,

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

8 0

3 years ago