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3 years ago

Please help me #6!!!!!!!!

Chemistry

1 answer:

Vesnalui [34]3 years ago

3 0

I believe that the answer is 12 because there is already 3 O molecules and since its in parentheses with 3 outside it that means that there are 3 of those CO molecules meaning that for every 1 CO there will be 3 O’s so 3, four times Is 12

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Is salt and water suspension

mars1129 [50]

Answer:

nah

Explanation:

l though Most Salt are soluble in water (Properties of Ionic Compound) , Not all are soluble ,which means mixing a salt and water can produce either a Solution (homogeneous mixture) or a Suspension (heterogeneous mixture).

6 0

2 years ago

Read 2 more answers

Two factors that determine kinetic energy are

Anna11 [10]

Your answer is mass and speed! Please give me brainlist :)

4 0

3 years ago

Read 2 more answers

Due to the small and highly electronegative nature of fluorine, the oxyacids of the this element are much less common and less s

Molodets [167]

Answer:

HOFO = (0, 0, +1, -1)

Explanation:

The formal charge (FC) can be calculated using the following equation:

$FC = V - N - \frac{1}{2}B$

Where:

V: are the valence electrons

N: are the nonbonding electrons

B: are the bonding electrons

The arrange of the atoms in the oxyacid is:

H - O₁ - F - O₂

Hence, the formal charge (FC) on each of the atoms is:

H: FC = 1 - 0 - 1/2*(2) = 0

O₁: FC = 6 - 4 - 1/2*(4) = 0

F: FC = 7 - 4 - 1/2*(4) = +1

O₂: FC = 6 - 6 - 1/2*(2) = -1

We can see that the negative charge is in the oxygen instead of the most electronegative element, which is the F. This oxyacid is atypical.

I hope it helps you!

3 0

2 years ago

A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal

SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻× $\frac{1molI_{3}^-}{2molS_{2}O_{3}^-}$ = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻× $\frac{2molCe^{4+}}{1molI_{3}^-}$ = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)× $\frac{140,116g}{1mol}$ = 0,0625 g of Ce(IV)

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = 1,812 wt%

I hope it helps!

5 0

3 years ago

A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro

Shkiper50 [21]

Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

$NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)$

So, $OH^{-}$ is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of $OH^{-}$ increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant.

Therefore excess amount of $OH^{-}$ combines with $NH_{4}^{+}$ to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.

5 0

3 years ago