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labwork [276]
2 years ago
14

PLEASE HELP!!!!! Calcium hydroxide is _____. an acid a base a neutral

Chemistry
2 answers:
bogdanovich [222]2 years ago
7 0
Some strong bases like calcium hydroxide aren't very soluble in water. That doesn't matter - what does dissolve is still 100% ionised into calcium ions and hydroxide ions. Calcium hydroxide still counts as a strong base because of that 100% ionisation.
Bingel [31]2 years ago
4 0
I believe it is a base!
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a compound contains two elements, sulfur and oxygen. in a sample of the compound, the mass of sulfur is 6.41 grams. the mass of
Scrat [10]
Percent composition is mass of element over mass of compound times 100. 6.41 + 9.60 = 16.01 g of compound, so % composition of sulfur is 6.41/16.01 x100 or 40.0%. % oxygen is 9.60/16.01 x 100 or 60.0%.
4 0
3 years ago
Which phrase best describes the path of a light wave?
VikaD [51]

Answer: D

Explanation:

Light waves travel in straight paths called rays. Unlike sound, where waves have to travel through matter to be heard, light waves do not have to travel through matter to be seen. Instead, rays travel in a straight path until they hit an object. A ray's straight path is the path of light

7 0
3 years ago
Read 2 more answers
Convert 2.39 x 1022 molecules of water into moles of water
torisob [31]

Answer:

0.0397\ \text{mol}

Explanation:

Number of molecules of water = 2.39\times 10^{22}

N_A = Avogadro's number = 6.022\times 10^{23}\ \text{mol}^{-1}

Number of moles is given by

\dfrac{2.39\times 10^{22}}{6.022\times 10^{23}}

=0.0397\ \text{mol}

The number of moles of water is 0.0397\ \text{mol}.

6 0
3 years ago
How many grams of sodium sulfide can be produced when 45.3 g Na react with 105 g S?
erastova [34]
Balance the equation: 2Na + S --> Na2S
Using the given amount of the reactants in the reaction, calculate the amount of the product:
45.3g Na x (1 mol/22.99 g)= 1.97 mol of Na
105f S x (1 mol/ 32.06g) = 3.28 mol of S
The limiting reactant would be Na:
<span>1.97 mol Na x (1 mol Na2S/ 2 mol Na) x (78.04g/mol) = 76.87g of Na2S produced</span>
8 0
3 years ago
Given 4.80g of ammonium carbonate, find:
V125BC [204]

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

<em>1) Number of moles of the compound:</em>

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

<em>2) Number of moles of ammonium ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

<em>∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ </em>= (2.0)(0.05 mol) =  <em>0.1 mol.</em>

<em>3) Number of moles of carbonate ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

<em>4) Number of moles of hydrogen atoms:</em>

  • Every 1.0 mol of (NH₄)₂CO₃  contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

<em>∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO</em>₃ = (8.0)(0.05 mol) = <em>0.4 mol.</em>

<em>5) Number of hydrogen atoms:</em>

  • It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

<u><em>Using cross multiplication:</em></u>

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

<em>∴ The no. of atoms in  0.4 mol of H atoms</em> = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = <em>2.4 x 10²³ molecules.</em>

8 0
3 years ago
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