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sasho [114]
3 years ago
12

When naming compounds part of the second element's name is dropped and what is added in its place?

Chemistry
1 answer:
FromTheMoon [43]3 years ago
3 0

Answer:

ide is the correct answer

Explanation:

example compound of sodium and hydrogen in sodium hydroxide

and compound of hydrogen and chlorine is hydrogen chloride

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The element chlorine is represented by the symbol ____. (Capitalization is important.)
Semmy [17]
Chlorine is represented by the symbol Cl
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3 years ago
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Which would be the nobel gas preceding Tin?
Alexxandr [17]

Answer: The noble gas preceding tin is Krypton [Kr].

Explanation: Atomic Number of tin is 50 , Atomic Number of neon is 10, Atomic Number of argon is 18, Atomic Number of krypton is 36 and Atomic Number of xenon is 54.

Atomic number tells us the number of electrons present in the atom.

Thus the electronic configuration of tin according to increasing energy of orbitals is written as [Kr]4d^105s^25p^2.



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3 years ago
How do you find the molar mass of calcium carbonate??
exis [7]
Ca: 40.078 g/mol
C: 12.011 g/mol
O: 16.00g/mol

CaCO3:
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7 0
3 years ago
Calculate the ph of the solution resulting from the addition of 85.0 ml of 0.35 m hcl to 30.0 ml of 0.40 m aniline (c6h5nh2). kb
stiv31 [10]

Answer:

pH = 0.81

Explanation:

HCl reacts with aniline, thus:

C₆H₅NH₂ + HCl → C₆H₅NH₃⁺ + Cl⁻

Moles of HCl are:

0.085L × (0.35mol HCl / L) = <em>0.02975mol HCl</em>

Moles of aniline are:

0.030L × (0.40mol HCl / L) = <em>0.012mol aniline</em>

Thus, after reaction, will remain:

0.02975mol - 0.012mol = <em>0.01775mol HCl</em>

Moles of HCl in solution are equal to moles of H⁺, thus, moles of H⁺ are: <em>0.01775mol H⁺</em>

As total volume is 85.0mL + 30.0mL = 115.0mL ≡ <em>0.115L</em>

0.01775mol / 0.115L = 0.1543M

pH of solution = -log[H⁺]

pH = -log 0.1543M

<em>pH = 0.81</em>

<em></em>

3 0
3 years ago
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A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical formula? The first step is to calcula
Georgia [21]
For a 100-g compound, we would have 2 g H, 32.7 g S, and 65.3 g O. We then divide each of these masses by their molar masses:
2 g H / 1.01 g/mol = 1.98 mol H
32.7 g S / 32.07 g/mol = 1.0196 mol S
65.3 g O / 16.00 g/mol = 4.08125 mol O
We then divide each amount of moles by the smallest of them, which is the 1.0196 from S:
1.98 mol H / 1.0196 mol S = 1.94 H ~ 2 H
1.0196 mol S / 1.0196 mol S = 1 S
4.08125 mol O / 1.0196 mol S = 4 O
So the empirical formula of the compound is H2SO4.
7 0
3 years ago
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