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3 years ago

Why does H2O have a lower melting point than MgO

Chemistry

1 answer:

Vesnalui [34]3 years ago

4 0

Answer:

Magnesium Oxide (MgO) has a higher melting point than Sodium Chloride (NaCl) which is around 2,800 degrees Celcius. This is because its Mg2+ and O2- ions have a greater number of charges, and so form stronger ionic bonds, than the Na+ and Cl– ions in Sodium Chloride.

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The amount of water vapor in a given volume of air is ————

xenn [34]

Answer:

Humidity is the amount of moisture (water vapor) in the air. It can be expressed as absolute humidity or relative humidity. Absolute humidity is the mass of water vapor divided by a unit volume of air (grams of water/cm3 of air).

6 0

3 years ago

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The Ksp of silver bromide at 25°C (298 K) is 5.0 x 10-13. Write the solubility expression hydrogen oxygen for this compound and

Rufina [12.5K]

AgBr(s) <—> Ag+(aq) + Br-(aq)

Ksp = [Ag+][Br-]

5 0

3 years ago

If a ballon at 3.4 atm at 301k, what is the pressure when cooled to 212k

stepladder [879]

Answer:

2.39 atm

Explanation:

- Use Gay-Lussac's law

- P2 = P1T2/T1

- Fill in with our values

- Hope this helped! Please let me know if you need further explanation.

8 0

3 years ago

What transition energy corresponds to an absorption line at 527 nm?

Ede4ka [16]

Answer:

E = 3.77×10⁻¹⁹ J

Explanation:

Given data:

Wavelength of absorption line = 527 nm (527×10⁻⁹m)

Energy of absorption line = ?

Solution:

Formula:

E = hc/λ

h = planck's constant = 6.63×10⁻³⁴ Js

c = speed of wave = 3×10⁸ m/s

by putting values,

E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 527×10⁻⁹m

E = 19.89×10⁻²⁶ Jm /527×10⁻⁹m

E = 0.0377×10⁻¹⁷ J

E = 3.77×10⁻¹⁹ J

4 0

3 years ago

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- Equilibrium shifts for slightly soluble compounds The reaction for the formation of a saturated solution of silver bromide (Ag

RideAnS [48]

Answer:

1) Favor formation of reactant

2) Favor formation of product

3) Favor formation of product

4) Favor formation of product

5) Favor formation of product

Explanation:

The reaction is $AgBr(s)--->Ag^{+}(aq)+Br^{-}(aq)$

The reaction is endothermic

1. Adding hydrobromic acid (HBr)

It will increase the amount of bromide ion. Bromide ion is one of the product so it will favor formation of reactant.

2. Lowering the concentration of bromide (Br-) ions

As we are decreasing the the concentration of product it will favor formation of more products

3. Heating the concentrations

If we are heating the mixture, it will increase the temperature and it will favor endothermic reaction. Thus will favor formation of product.

4. Adding solid silver bromide (AgBr)

If we are adding silver bromide it means we are increasing the concentration of reactant, it will favor formation of products

5. Removing silver (Ag+) ions

Removing silver ions means we are decreasing the concentration of product thus it will favor formation of products.

3 0

3 years ago