Answer:
127.3° C, (This is not a choice)
Explanation:
This is about the colligative property of boiling point.
ΔT = Kb . m . i
Where:
ΔT = T° boling of solution - T° boiling of pure solvent
Kb = Boiling constant
m = molal (mol/kg)
i = Van't Hoff factor (number of particles dissolved in solution)
Water is not a ionic compound, but we assume that i = 2
H₂O → H⁺ + OH⁻
T° boling of solution - 118.1°C = 0.52°C . m . 2
Mass of solvent = Solvent volume / Solvent density
Mass of solvent = 500 mL / 1.049g/mL → 476.6 g
Mol of water are mass / molar mass
76 g / 18g/m = 4.22 moles
These moles are in 476.6 g
Mol / kg = molal → 4.22 m / 0.4766 kg = 8.85 m
T° boling of solution = 0.52°C . 8.85 m . 2 + 118.1°C = 127.3°C
Answer: 700 has one significant figure which is 7.
Explanation: These are some rules for significant figures
•All non-zero digits are significant: 1,2,3,4,5,6,7,8,9
•Zero between non-zero digits are significant: They are three significant figures in 203.
•Leading zeros are not significant: There are two significant figures in 0.56.
•Trailing zero to the right of decimal are significant. There are four significant figures in 62.00
•Trailing zeros in a whole number with the decimal shown are significant: This makes "700." three significant figures.
•Trailing zeros in a whole number with no decimal shown are not significant: This makes 700 one significant figure.
Answer:
A is the molecular formula for xylose because shows the actual number of atoms in the compound: Formula B is the empirical formula for xylose because it shows the smallest whole-number ratio for the different atoms in the compound: Formula A is the molecular formula for xylose because shows the arrangement of atoms in the compound: Formula B is the structurab formula for xylose because it shows the smallest whole-number ratio for the different atoms in the compound: Formula A is the empirical formula for xylose because it shows the actual number of atoms in the compound: Formula B is the molecular formula for xylose because it shows the smallest whole-number ratio for the different atoms in the compound: Formula A is the structural formula for xylose because it shows the arrangement of atoms in the compound: Formula B is the empirical formula for xylose because it shows the smallest whole-number ratio for the different atoms in the compound.
I would always start by balancing your carbons, and then balancing the rest from there.
1. C2H5OH + O2 —> CO2 + H2O - You have two carbons on the left and one on the right. Multiply CO2 by 2.
C2H5OH + O2 —> 2CO2 + H2O
Now balance hydrogen. You have 6 on the left and 2 on the right. Multiply H2O by 3.
C2H5OH + O2 —> 2CO2 + 3H2O
Now balance oxygen. You have 3 on the left and 7 on the right. You need 4 more on the left. Don’t multiply the C2H5OH by anything because that will change the numbers of everything else too. Multiply O2 by 3 instead.
C2H5OH + 3O2 —> 2CO2 + 3H2O
Check that all atoms are now balanced, and you’re good.
2. Same process as before.
First carbons - C3H8 + O2 —> 3CO2 + H2O
Then hydrogens - C3H8 + O2 —> 3CO2 + 4H2O
Then oxygens - C3H8 + 5O2 —> 3CO2 + 4H2O
3. Same again.
Carbons) C6H12O6 + O2 —> 6CO2 + H2O
Hydrogens) C6H12O6 + O2 —> 6CO2 + 6H2O
Oxygens) C6H12O6 + 6O2 —> 6CO2 + 6H2O
4. The general reaction for a combustion reaction is a hydrocarbon reacting with oxygen to produce carbon dioxide and water.
We need the reading for this I think