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Artist 52 [7]
3 years ago
8

What is the charge of a monatomic ion? [A] oxidation number [B] electronegativity [C] polarity [D] negative [E] positive

Chemistry
2 answers:
mart [117]3 years ago
8 0
[B] electronegativity.
eduard3 years ago
5 0
I believe <span>[B] electronegativity</span>
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Chemistry problem. Is alcohol soluble in water?
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Alcohol is completely soluble in water in any amounts. That is because every alcohol contains carbon chain and OH group which are non-polar and polar. Any substances or solutes that will be mixed with the alcohol you will just get a simple solution.
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3 years ago
Hands on mole activity
Amanda [17]

Explanation:

one mole= 6.02x10^23

Everything going from a mole will always be multiplied and everything going into mole will be divided.

4 0
3 years ago
Please help me. The question is blow thanks.
lesantik [10]
I'm pretty sure it's C amplitude and wavelength hope this helps
3 0
3 years ago
Please help asap 18 points if help and right answer please ​
barxatty [35]

Answer:

Lithium = 8%

Bromine = 92%

Explanation:

To calculate percent composition, you must:

- Calculate the molar mass.

- Divide the subtotal for each element's mass by the molar mass.

- Convert to a percentage

With that being said, given LiBr (Lithium Bromide), calculate the molar mass:

Lithium has an atomic weight of 7, and there is one Lithium atoms in LiBr. Bromine has an atomic weight of 80, and there is one Bromine atom in LiBr:

1(7)\\1(80)

Add the two products:

80+7

=87

The molar mass of LiBr is about 87 grams.

With that information, divide the subtotal of Lithium by the molar mass, and then multiply by 100 to convert to a percentage:

\frac{7}{87} × 100=

8.0459%

(Round to nearest percentage):

8%

Therefore, the percent composition of Lithium in the compound Lithium Bromide is about 8%.

Now, divide the subtotal of Bromine by the molar mass, and then multiply by 100 to convert to a percentage:

\frac{80}{87} × 100=

91.9540

(Round):

92%

Therefore, the percent composition of Bromine in the compound Lithium Bromide is about 92%.

6 0
2 years ago
According to the following reaction, how many grams of ammonia will be formed upon the complete reaction of 31.2 grams of hydrog
Aneli [31]

Answer:

176.8 g of ammonia, NH3.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6 g

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34 g.

From the balanced equation above,

6 g of H2 reacted to produce 34 g of NH3.

Finally, we shall determine the mass of ammonia, NH3 produced by reacting 31.2 g of H2.

This can be obtained as follow:

From the balanced equation above,

6 g of H2 reacted to produce 34 g of NH3.

Therefore, 31.2 g of H2 will react to produce = (31.2 x 34)/6 = 176.8 g of NH3.

Therefore, 176.8 g of ammonia, NH3 were obtained from the reaction.

8 0
3 years ago
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