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Ket [755]
2 years ago
13

Water is a polar molecule. What does that mean?

Chemistry
2 answers:
Dimas [21]2 years ago
8 0
Answer: it is a molecule with no charge
aalyn [17]2 years ago
7 0
The answer. Is c I hope this helps
You might be interested in
The Lewis dot notation for two atoms is shown.
Iteru [2.4K]

Answer:

Mg donates two electrons to O

Explanation:

Lewis dot notation uses dots and crosses to represent valence electrons on atoms.

Magnesium is a metal and would donate or lose electrons during bonding.

Oxygen is a non metal and would gain electrons during bonding.

The correct option is;

Mg donates two electrons to O

5 0
2 years ago
Find the empirical formula of the following compounds:
Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus \frac{mol phosphorus}{ 30.97 g phosphorus}= 0.0293 mol

Moles bromine 6.99 g bromine\frac{mol bromine}{79.90 g bromine}=0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

#SPJ4

8 0
1 year ago
How many grams of O are in 675 g of Na2O
PIT_PIT [208]
I think it is .00005815
6 0
3 years ago
The solubility product of AgCl is 1.4 x 10-4 at 100°C. Calculate the solubility of AgCl in
marshall27 [118]

Answer:

AgCl=Ag+ + Cl-

1.4×10^-4 =x × x

x²=1.4×10^-4

x=√(1.4×10^-4)

x=0.012

5 0
2 years ago
POINTS AND BRAINIEST ANSWER!!!!
denis23 [38]

Answer:

hydrgen = i think it is 4

oxygen = i think it is 3

Explanation:

3 0
3 years ago
Read 2 more answers
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