Answer:
The required mass to prepare 2.5 L of 1.0 M NaOH solution is 100 g
Explanation:
We do this by preparing the equation:
Mass = concentration (mol/L) x volume (L) x Molar mass
Mass = 1.0 M x 2.5 L x 40 g/mol
Mass = 100 g
Answer:
density.
Explanation:
A cubic centimeter of water weighs 1 gram. Since a cubic centimeter of vegetable oil weighs less than 1 gram, oil will float on water. Corn syrup is more dense than water so 1 cubic centimeter of corn syrup weighs more than 1 gram. Therefore, corn syrup sinks in water.
Answer:
The sp³ orbital hybridization has 25% s-character and 75% p-character.
Explanation:
Hybridization refers to the process of mixing of a given number atomic orbitals to form equal number of new hybrid orbitals. The three types of hybrid orbitals are sp³, sp² and sp.
The<u> four sp³ hybridized orbitals</u> are formed by the mixing of <u>one s orbital and three p orbitals.</u>
Thus s-character in a sp³ hybridized orbital = (1÷4)×100 = 25%
and p-character in a sp³ hybridized orbital = (3÷4)×100 = 75%
<u>Therefore, the sp³ orbital hybridization has 25% s-character and 75% p-character.</u>
The percent yield of the reaction between ammonia gas with oxygen gas is 90.52%.
A chemical reaction between ammonia gas (NH3) with oxygen gas (O2)
NH₃ + O₂ → NO₂ + H₂O
The balanced reaction 4NH₃ + 7O₂ → 4NO₂ + 6H₂O
Calculate the number of moles from the reactant
- Ammonia gas
Molar mass N = 14 gr/mol
Molar mass H = 1 gr/mol
Molar mass NH₃ = 14 + (3 × 1) = 14 + 3 = 17 gr/mol
mass = 28.5 grams
n = m ÷ molar mass = 28.5 ÷ 17 = 1.68 mol - Oxygen gas
Molar mass O = 16 gr/mol
Molar mass O₂ = 16 × 2 = 32 gr/mol
mass = 83.4 grams
n = m ÷ molar mass = 83.4 ÷ 32 = 2.61 mol - n O₂ ÷ coefficient O₂ = 2.61 ÷ 7 = 0.37
n NH₃ ÷ coefficient NH₃ = 1.68 ÷ 4 = 0.42
0.42 > 0.37 it means that the ammonia gas is in excess and the O₂ is limiting.
According to stoichiometry, the number of moles NO₂ with the number of moles O₂ has the ratio with the coefficient in reaction.
- Theoretically the number moles of NO₂
n O₂ : n NO₂ = 7 : 4
2.61 : n NO₂ = 7 : 4
n NO₂ = 4 x 2.61 : 7 = 1.49 mol - The actual number of moles NO₂
Molar mas NO₂ = 14 + (16 × 2) = 14 + 32 = 46 gr/mol
n NO₂ = m ÷ molar mass = 61.9 ÷ 46 = 1.35 mol
The percent yield NO₂ is the ratio of the actual number of moles NO₂ with the theoretical number of moles NO₂ times 100%.
P = (1.35 ÷ 1.49) × 100%
P = 0.9052 × 100%
P = 90.52%
Learn more about stoichiometry here: brainly.com/question/13691565
#SPJ4
Just find the percent mass of oxygen in sucrose again. and then multiply that by 50.00.