Questions:
The questions or computes to do are:
<span>a- a massa, em
kg, de cada placa de alumínio;
b- a quantidade mínima de viagens
necessárias para que apenas um veículo de transporte entregue o material
solicitado ao cliente.
Dado: densidade do alumínio = 2,7 g/cm3
Answer:
a) mass in kg of every aluminum plate
Dimensions of every aluminum plate: </span>
<span>2 M X 50 Cm X 2cm
Volume: 200 cm * 50 cm * 2 cm = 20,000 cm^3
Mass:
density = mass / volume => mass = density * volume = 2.7 g/cm^3 * 20,000 cm^3 = 54,000 g = 54 kg.
Answer: the mass of everyplate of aluminum is 54 kg.
b) number of travels required for one truck deliver all the material:
number of travels = amount requested / amount that a truck can deliver in one travel.
amount requested: 100 plates
mass of 100 plates = 100 plates * 54 kg / plate = 5,400 kg
limit of transport per travel: 3 tons = 3,000 kg
number of travels = 5,400 kg / 3,000 kg/travel = 1.8 travels => 2 travels.
Answer: at least 2 travels.
</span>
Answer:
Ununoctium
Explanation:
It's at the very end of the periodic table
The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
Answer is: mass oxygen is 129,28 grams.
Chemical reaction: 2C₁₈H₃₈ + 55O₂ → 36CO₂ + 38H₂O.
m(C₁₈H₃₈) = 37,5 g.
n(C₁₈H₃₈) = m(C₁₈H₃₈) ÷ M(C₁₈H₃₈).
n(C₁₈H₃₈) = 37,5 g ÷ 254 g/mol.
n(C₁₈H₃₈) = 0,147 mol.
From chemical reaction: n(C₁₈H₃₈) : n(O₂) = 2 : 55.
n(O₂) = 4,04 mol.
m(O₂) = 4,04 mol · 32 g/mol.
m(O₂) = 129,28 g.
