Give me lil reasoning so I know your not lying for points

What us the % composition of oxygen (O) in iron (III) hydroxide, Fe(OH)<br> 3?

S_A_V [24]

Answer:

44.91% of Oxygen in Iron (III) hydroxide

Explanation:

To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:

<em>Molar mass Oxygen / molar mass Fe(OH)3 * 100</em>

<em />

<em>Molar mass Fe(OH)3 and oxygen:</em>

1Fe = 55.845g/mol*1 = 55.845

3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen

3H = 1.008g/mol*3 = 3.024

55.845 + 48.00 + 3.024 =

106.869g/mol is molar mass of Fe(OH)3

% Composition of oxygen is:

48.00g/mol / 106.869g/mol * 100 =

<h3>44.91% of Oxygen in Iron (III) hydroxide</h3>

7 0

3 years ago

Butane (C4H10) is used as a fuel where natural gas is not available. How many grams of butane will fill a 3.50-liter container a

erik [133]

<span>We use the formula PV = nRT. P = 758 torr = 0.997 atm. V = 3.50 L. T = 35.6 C = 308.15 K. R = 0.0821. Rearranging the equation gives up n = PV/Rt and we get .0138 moles of butane. Mass of 0.0138 moles of butane = .0138 x 58.12 = 8.02g.</span>

6 0

3 years ago

The chemical equation is Mg(s) + O2(g) → MgO(s)

vredina [299]

Answer:

The magnesium will burn until consumed entirely. There is much more oxygen available in the atmosphere than needed to consume the magnesium. Thus the magnesium is the limiting reactant because it determines the amount of product formed.

Explanation:

Mg produces less amount of MgO than O2; therefore Mg is the limiting reagent. O2 produces more amount of MgO than Mg; therefore O2 is the excess reagent.

4 0

3 years ago

What is the volume of 1g of ice in cm3?

Amanda [17]

Answer:

What is the volume of 1g of ice in cm3?

Explanation:

3 0

3 years ago

Washing soda, a compound used to prepare hard water for washing laundry, is a hydrate, which means that a certain number of wate

nadya68 [22]

Answer:

moles H₂O = 10

Explanation:

The mass of Na₂CO₃⋅xH₂O is 3.837 g and the mass of Na₂CO₃ is 1.42g

Therefore the mass of xH₂O is 3.837 - 1.42 = 2.417 g

The molar mass of Na₂CO₃ is 106 g/mol and for H₂O is 18 g/mol

The moles of Na₂CO₃ and H₂O in the sample are:

Na₂CO₃ = 1.42 / 106 = 0.01340 moles

H₂O = 2.417 / 18 = 0.1343

Now using rule of three :

1 mole of Na₂CO₃ has x moles of H₂O

0.01340 moles of Na₂CO₃ has 0.1343 moles of H₂O

x = 1 * 0.1343 / 0.01340 = 10

4 0

3 years ago