a. There are 1.85 moles in 2.00 × 10² g of silver (Ag).
b. There are 0.618 moles in 37.1 g of silicon dioxide (SiO₂)
<h3>What is the molar mass?</h3>
The molar mass is the mass in grams of 1 mole of particles, that is, the mass in grams of 6.02 × 10²³ particles. The units are g/mol.
We want to calculate the number of moles represented by different masses of different substances. In each case, the conversion factor between mass and moles is the molar mass.
- a. 2.00 × 10² g of silver (Ag)
The molar mass of silver is 107.87 g/mol.
2.00 × 10² g × (1 mol/107.87 g) = 1.85 mol
- b. 37.1 g of silicon dioxide (SiO₂)
The molar mass of silicon dioxide is 60.08 g/mol.
37.1 g × (1 mol/60.08 g) = 0.618 mol
a. There are 1.85 moles in 2.00 × 10² g of silver (Ag).
b. There are 0.618 moles in 37.1 g of silicon dioxide (SiO₂)
Learn more about molar mass here: brainly.com/question/21334167
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Answer:
12 hydrogen + 12 hydrogen = 24 hydrogen
Hydrogen was transferred from sugar and water on the left to water on the right releasing
Explanation:
Answer:
They are protection from the cell!
Explanation:
They move liquid around the cells and have the same internal structure as each other. They are also found in most microorganisms. Hope this helps :)
A compound differs from an element in that it can be decomposed by a chemical reaction.
A compound, by definition, is made up of two or more elements that have been combined together through a chemical bond.
Thus, the chemical bond or bonds in a compound can be decomposed by a chemical reaction giving rise to the individual elements that make up the compound.
This is unlike an element that can not be decomposed by any chemical reaction.
More on elements and compounds can be found here: brainly.com/question/5997683
Answer:
4
10
Explanation:
The reaction equation is given as;
Ca(OH)₂ → Ca²⁺ + 2OH⁻
Concentration of Ca(OH)₂ = 5 x 10⁻⁵M
Unknown:
pOH of the solution = ?
pH of the solution = ?
Solution:
Solve for the pOH of this solution using the expression below obtained from the ionic product of water;
pOH = ⁻log₁₀[OH⁻]
Ca(OH)₂ → Ca²⁺ + 2OH⁻
1moldm⁻³ 1moldm⁻³ 2 x 1moldm⁻³
5 x 10⁻⁵moldm⁻³ 5 x 10⁻⁵moldm⁻³ 2( 5 x 10⁻⁵moldm⁻³ )
1 x 10⁻⁴moldm⁻³
Therefore;
pOH = -log₁₀ 1 x 10⁻⁴ = 4
Since
pOH + pH = 14
pH = 14 - 4 = 10
