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MAXImum [283]
3 years ago
6

Eaterfication experiment ​

Chemistry
1 answer:
Schach [20]3 years ago
8 0

Answer:

Do you mean Esterification experiment?

Explanation:

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1 mole is equal to 1 moles Ammonium Sulfate, or 132.13952 grams
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[QUESTION TO EVERY SCARY MOVIE FANATIC] What is the most scary movie you ever seen? If you did see the most scary movie ever, wh
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How many moles are 5.55 x 102 atoms of Mg?<br> M2) an&gt; NU X 02.0
cupoosta [38]

Explanation:

number of atoms = moles × avegadro number

so 5.55 × 10^2 = moles × 6.023 × 10^23

moles = 5.55 × 10^2 ÷ 6.023 × 10^23 = 9.214×10−22 moles

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3 years ago
Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estima
Rzqust [24]

Answer:

See explanation below

Explanation:

To do this, we will use the following expression to calculate the [H⁺]:

[H⁺] = Kw / [OH⁻]

[H⁺] is the same as [H₃O⁺]. So we have the [OH⁻] so, let's replace every value into the above expression to calculate the hydronium concentration and say if it's acidic, basic or neutral. This can be known because if the [H⁺] > 1x10⁻⁷ M the solution is acidic. If it's [H⁺] < 1x10⁻⁷ M the solution is basic, and if it's [H⁺] = 1x10⁻⁷ M the solution is neutral.

a) [H⁺] = 1x10⁻¹⁴ / 6x10⁻¹² = 1.67x10⁻³ M. Acidic.

b) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁹ = 1.11x10⁻⁶ M. Acidic.

c) [H⁺] = 1x10⁻¹⁴ / 8x10⁻¹⁰ = 1.25x10⁻⁵ M. Acidic.

d) [H⁺] = 1x10⁻¹⁴ / 7x10⁻¹³ = 0.0143 M. Acidic.

e) [H⁺] = 1x10⁻¹⁴ / 2x10⁻² = 5x10⁻¹³ M. Basic.

f) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁴ = 1.11x10⁻¹¹ M. Basic.

g) [H⁺] = 1x10⁻¹⁴ / 5x10⁻⁵ = 2x10⁻¹⁰ M. Basic.

h) [H⁺] = 1x10⁻¹⁴ / 1x10⁻⁷ = 1x10⁻⁷ M. Neutral.

Part B.

In this part, we'll use the following expression and replace the given values:

[OH⁻] = Kw / [H⁺]

Replacing the values:

[OH⁻] = 1x10⁻¹⁴ / 5.2x10⁻⁵

[OH⁻] = 1.92x10⁻¹⁰ M

PArt C:

In this case, we will use expression of part A, and replace the given values:

[H⁺] = 1x10⁻¹⁴ / 2.7x10⁻²

[H⁺] = 3.7x10⁻¹³ M

8 0
3 years ago
A 1.525g sample of a compound between nitrogen and hydrogen contains 1.333 g of nitrogen. Calculate its empirical formula. The e
vredina [299]

Answer:

NH2.

Explanation:

The mass of hydrogen in the sample = 1.525 - 1.333 = 0.192g.

Dividing the 2 masses by the  relative atomic mass of hydrogen and nitrogen:

H: 0.192 / 1.008 = 0.1905

N: 1.333 / 14.007 = 0.09517

The ratio of N to H =  0.09517 : 0.1905

=  1 : 2.

So the empirical formula is  NH2.

5 0
3 years ago
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