bromine, a liquid at room temperature, has a boiling point of 58°C and a melting point of –7.2°C. Bromine can be classified as a
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Answer:
it's a pure substance
Explanation:
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Use exponential notation to express the number 385,500 ha. one significant figure. b. two significant figures. c. three significant figures. d. five significant figures.
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What is inert pair effect
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2C8H18+25O2----> 16CO2+18H2O when balanced. 10/114.229 g/mol = .087543... mol, converting to moles by dividing by molar mass. The ratio is 25/2, using the integers in front of the molecules in the balanced equation. so you just multiply .087543... by the ratio. That's 1.09429, which you multiply by the molar mass of O2, 31.9988 g/mol. The answer is 35.02 g O2
35.2 grams of oxygen are required to react with 10.0 grams of octane
Complete combustion of Hydrocarbons with Oxygen will be obtained by CO₂ and H₂O compounds.
If O₂ is insufficient there will be incomplete combustion produced by CO and H and O
Hydrocarbon combustion reactions (specifically alkanes)
For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:
Balancing C atoms, H and last atoms O atoms
Octane combustion reaction
C₈H₁₈ + O₂ ---> CO₂ + H₂O
To equalize the reaction equation we give the reaction coefficients with variables a, b and c while the most complex compounds, namely Octane, we give the number 1
So the reaction becomes
C₈H₁₈ + aO₂ ---> bCO₂ + cH₂O
C atom on the left 8, right b, so b = 8
left H atom = 18, right 2c so 2c = 18 ---> c = 9
Atom O on the left 2a, right 2b + 2c, so 2a = 2b + 2c
2a = 2.8 + 9
2a = 16 + 9
2a = 25
a = 25/2
The equation becomes:
C₈H₁₈ + 25/2O₂ ---> 8CO₂ + 9H₂O or be
2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O
To find the mass O2, we find the mole first from the mole ratio with Octane
- 1. We are looking for the octane mole
Mr. octane = 8.Ar C + 18.Ar H
Mr. octane = 8.12 + 18.1
Mr octane = 96 + 18
Mr octane = 114
known octane mass: 10 grams then the mole:
mol = gram / Mr
mol = 10/ 114
mol = 0.088
- 2. We look for the O₂ mole
because of the ratio of the reaction coefficient between O2 and octane = 25: 2 then the mole of O₂ =
25/2 x 0.088 = 1,1
So that the mass O₂ =
mole. Mr = 1.1 32
mass = 35.2 grams
a combustion reaction
the type of chemical reaction
The combustion of alkanes, alkenes, and alkynes
Complete combustion of a sample of a hydrocarbon
Keywords: Complete combustion of Hydrocarbons, alkanes, equalize the reaction equation
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0-6 is an acid
8-14 is a base
7 is neutral
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30 +273 = 303 K
P₁* V₁ / T₁ = P₂*V₂/ T₂
1.00 * 50.0 = 1 * V₂ / 273
V₂ = 1.00 * 50.0 * 273 / 303 * 1 = V₂
V₂ =13650 / 303
V₂ = 45.0 L
hope this helps!
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The property best used when determining the strength of an ionic bond in a solid is:
LATTICE ENERGY
The lattice energy is the energy released during the deposition of gaseous ions of opposite charges. Deposition is the transformation of gas to solid.
Explanation:
The main types of solids are ionic, molecular, covalent, and hard. Ionic solids consist of absolutely and negatively energized ions held concurrently by electrostatic forces; the power of the bonding is indicated in the lattice energy. Ionic solids manage to have high melting points and are fairly hard. lattie energy is a stratagem of the energy received in the crystal lattice of a compound, similar to the energy that would be delivered if the element ions were produced together from infinity.
Answer:
A. lattice energy
Explanation:
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When you consume proteins, carbs, and fats they are chemically changed into energy.
b. Solution a must have a higher solute concentration than solution
b. Solution a and solution b must have the same concentration of solutes. Solution b must be pure water.
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The answer is solution a must have a lower solute concentration than solution b.
That is when water is moving across a membrane from solution a into solution b, then solution a must have a lower solute concentration than solution b.
When solution a have a lower solute concentration than solution b, then water moves across a membrane from solution a into solution b.