All categories

3 years ago

How does hydrogen bonding affect water molecules

1 answer:

4 0

Since the oxygen likes to hog all the electrons, it gives the hydrogen molecules a slight positive charge and the itself a slightly negative charge.
Positive attracts negative, so there is some sort of attraction between water molecules, though a weak one.

You might be interested in

Write two half‑reactions to describe each of these metal displacement reactions. Include phase symbols.

spin [16.1K]

Mg (s) ---> Mg²⁺ (aq) + 2e

Co²⁺ (aq) + 2e ---> Co(s)

Cu (s) ---> Cu²⁺ (aq) + 2e

Au²⁺ (aq) + 2e ---> Au(s)

7 0

2 years ago

Explain ray diagram and derive first law of reflection from that.

Ivahew [28]

(I posted the picture for a ray diagram and the first law of reflection.) The rays show the first law of reflection since everything gets reflected. Look at the picture for more:

6 0

3 years ago

Read 2 more answers

Which of the following numerical expressions gives the number of particles in 2.0 g of Ne?

mojhsa [17]

Answer:

Number of particles = 2.0 g*(6.0 x 10^23 particles/mol) / 20.18 g/mol

Option C is correct

Explanation:

Step 1: Data give

Mass of Ne = 2.0 grams

Molar mass of neon = 20.18 g/mol

Number of Avogadro = 6.0 *10^23 /mol

Step 2: Calculate number of moles of neon

Moles Ne = Mass of ne / Molar mass of ne

Moles Ne = 2.0 / 20.18 g/mol

Moles Ne = 0.099 moles

Step 3: Calculate nulber of particles

Number of particles = 6.022*10^23 / mol * 0.099 moles = 5.96 *10^22

Number of particles = 6.022*10^23 * (2.0g/ 20.18g/mol)

Number of particles = 2.0 g*(6.0 x 10^23 particles/mol) / 20.18 g/mol

Option C is correct

7 0

3 years ago

if a scientist creates a new rock like substance in a laboratory, why wouldnt this type of material be classified as a true mine

ad-work [718]

The reason why it is not considered this is because the material was made in a lab, not through nature, which is what is required to be considered as a true mineral.

5 0

3 years ago

Read 2 more answers

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

8 0

3 years ago