Answer:
no
Explanation:
los materiales sinteticos son aquellos materiales q estan hechos de polimeros sintetilizados o de pequeñas moleculas
Answer:
5SiO2 + 2CaC2 = 5Si + 2CaO + 4CO2
Explanation:
balancing equations is a lot of trial and error. My strategy to approaching this equation was to get the O's balanced. After trying several combonations I found that I needed 10 O's on each side of the equation for the other elements to match up. After I balanced the O's, I balanced my C's to 4 on each side. Then I balanced my Ca's to have 2 on each side. And last but not least I balanced my Si to have 5 on each side.
Given parameters:
Volume of CuSO₄ = 250mL
Concentration of CuSO₄ = 2.01M
Unknown:
Mass of CuSO₄.5H₂O = ?
To solve this problem, we must write the chemical relationship between both species.;
CuSO₄.5H₂O → CuSO₄ + 5H₂O
Now that we know the expression, it is possible to solve for the unknown mass.
First find the number of moles of CuSO₄;
Number of moles = Concentration x Volume
Take 250mL to L so as to ensure uniformity of units;
Volume = 250 x 10⁻³L
Input the parameters and solve for number of moles;
Number of moles = 250 x 10⁻³ x 2.01 = 0.5mol
From the equation;
1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O
So 0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O
Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16) = 249.6g/mole
Mass of CuSO₄.5H₂O = number of moles x molar mass
= 0.5 x 249.6
= 124.8g
The mass of CuSO₄.5H₂O is 124.8g
Combustion reaction for menthol is as follows;
CxHyOz + O₂ ---> xCO₂ + H₂O
Mass of CO₂ formed - 28.16 mg
Therefore number of moles formed - 28.16/ 44 g/mol = 0.64 mmol
Mass of water formed - 11.53 mg
number of water moles formed - 11.53 mg/18 g/mol = 0.64 mmol
From CO₂,
1 mol of CO₂ - 1 mol of C and 2 mol of O
therefore number of C moles - 0.64 mmol
O moles - 1.28 mmol
from H₂O
1 mol of H₂O - 2 mol of H and 1 mol of O
number of H moles - 1.28 mmol
O moles - 0.64 mmol
Mass of menthol initially - 10 mg
in reactions, the masses of products are equal to the masses of reactants. The excess mass to the products formed is due to O₂ in air
Original mass of menthol - 10 mg
mass of water and CO₂ - 11.53 mg + 28.16 mg = 39.69
Difference in mass - 39.69 - 10 = 29.69 mg
This difference comes from O moles in air - 29.69 mg/ 16 g/mol = 1.8556 mmol
then O moles coming from menthol - (1.28 + 0.64) - 1.8556 = 0.064 mmol
In menthol
C moles - 0.64 mmol
H moles - 1.28 mmol
O moles - 0.064 mmol
ratios of C:H:O
C H O
0.64 1.28 0.064
x1000 x1000 x1000 to get whole numbers
640 1280 64
10 20 1
Simplest ratio of C:H:O is 10:20:1
therefore empirical formula of menthol is C₁₀H₂₀O
Answer:
1. c) shiny
2) True. Reactivity is a chemical property.
