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galben [10]
3 years ago
12

How many moles are in 153’pp grams of KClO3

Chemistry
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

n = 1.24 moles

Explanation:

Given that,

Mass = 153 grams

Molar mass of KClO₃ = 122.55 g/mol

We need to find the number of moles.

We know that,

No. of moles = given mass/molar mass

So,

n=\dfrac{153}{122.55 }\\\\n=1.24

So, there are 1.24 moles in 153 g of KClO₃.

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Short note on Gabriel synthesis​
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Explanation:

The Gabriel synthesis is a chemical reaction that transforms primary alkyl halides into primary amines. Traditionally, the reaction uses potassium phthalimide. ... The alkylation of ammonia is often an unselective and inefficient route to amines. In the Gabriel method, phthalimide anion is employed as a surrogate of H2N−.

7 0
3 years ago
Given the noble gas configuration of an element: [Ar] 4s2, 3d5, what is the element?
monitta

Question

<em>Given the noble gas configuration of an element: [Ar] 4s2, 3d5, what is the element? </em>

Answer:

<em>B.) Argon</em>

Hope this helps!

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What is the number of electrons that can be held in the first orbit closest to the nucleus
Juliette [100K]
The first shell can hold up to 2 electrons, the second shell can hold up to 8 (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on.
5 0
3 years ago
first used the term cell when he used a (5)_____ to observe box-like structures when he was examining cork under a microscope. (
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Answer:

sorry if its wrong.

Explanation:

(5) a cell was used to refer to these tiny units of life

(6) Robert Hooke

(7)  cell walls

(8) modern cell theory. im so sorry if im wrong.

3 0
3 years ago
The vapor pressure of benzene is 73.03 mm Hg at 25°C. How many grams of estrogen (estradiol), C18H24O2, a nonvolatile, nonelectr
alexira [117]

Answer:

14.9802 grams of estrogen must be added to 216.7 grams of benzene.

Explanation:

The relative lowering of vapor pressure of solution containing non volatile solute is equal to mole fraction of solute.

\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}

Where:

p_o = Vapor pressure of pure solvent

p_s = Vapor pressure of the solution

n_1 = Number of moles of solvent

n_2 = Number of moles of solute

p_o = 73.03 mmHg

p_s= 71.61 mmHg

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\frac{73.03 mmHg-71.61 mmHg}{73.03 mmHg}=\frac{n_2}{2.7739 mol+n_2}

n_2=0.05499 mol

Mass of 0.05499 moles of estrogen :

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14.9802 grams of estrogen must be added to 216.7 grams of benzene.

6 0
3 years ago
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