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SpyIntel [72]
3 years ago
7

When 35.47 g of sodium hydroxide react with boric acid (H3BO3), how many moles of sodium borate will be produced?

Chemistry
1 answer:
belka [17]3 years ago
4 0

Answer:

5.83 g

Explanation:

First, you must start with a balanced equation so you can see the mole ratios.

NaOH + H₃BO₃ --> NaBO₂ + 2H₂O

You can see that it takes 1 mole of sodium hydroxide to form 1 mole of sodium borate. 1:1 ratio

Now you must calculate how many moles of NaOH 35.47 g equals.

Na = 22.99 amu

O = 15.99 amu

H = 1.008 amu

NaOH = 39.997 amu

35.47 g ÷ 39.997 amu = 0.08868 moles of NaOH

Since it's a 1:1 ratio, the same number of moles of NaBO₂ is created. Now you must convert moles to grams.

Na = 22.9 amu

B = 10.81 amu

2 O = 31.998 amu

NaBO₂ = 65.798 amu

0.08868 moles x 65.798 = 5.83 g

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Arrange the following substances in order of decreasing magnitude of lattice energy. Rank the compounds in order of decreasing m
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The lattice energy of the compounds is distributed in the following decreasing order of magnitude: MgO > CaO > NaF > KCl.

<h3>KCl or NaF, which has a higher lattice energy?</h3>

The lattice energy increases with increasing charge and decreasing ion size.(Refer to Coulomb's Law.)MgF2 > MgO.Following that, we can examine NaF and KCl (both of which have 1+ and 1-charges), as well as atomic radii.NaF will have a larger LE than KCl since Na is smaller then K and F was smaller than Cl.

<h3>MgO or CaO, which has a larger lattice energy?</h3>

MGO is more difficult than CaO, hence.This is because "Mg" (two-plus) ions are smaller than "Ca" (two-plus) ions in size.MgO has higher lattice energy as a result.

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1 year ago
Salt is often added to water to raise the boiling point to heat food more quickly. if you add 30.0g of salt to 3.75kg of water,
sammy [17]

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

<h3>What is the boiling-point elevation?</h3>

Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.

  • Step 1: Calculate the molality of the solution.

We will use the definition of molality.

b = mass solute / molar mass solute × kg solvent

b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m

  • Step 2: Calculate the boiling-point elevation.

We will use the following expression.

ΔT = Kb × m × i

ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C

where

  • ΔT is the boiling-point elevation
  • Kb is the ebullioscopic constant.
  • b is the molality.
  • i is the Van't Hoff factor (i = 2 for NaCl).

The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:

100 °C + 0.140 °C = 100.14 °C

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

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