Answer:
oxygen is limiting reactant
Explanation:
Given data:
Mass of phosphorus = 25.0 g
Mass of oxygen = 50.0 g
What is limiting reactant ?
Solution:
Chemical equation:
P₄ + 5O₂ → P₄O₁₀
Number of moles of P₄:
Number of moles = mass/molar mass
Number of moles = 25.0 g/ 123.89 g/mol
Number of moles = 0.20 mol
Number of moles of O₂:
Number of moles = mass/molar mass
Number of moles = 50.0 g/ 32 g/mol
Number of moles = 1.56 mol
now we will compare the moles of reactants with product:
P₄ : P₄O₁₀
1 : 1
0.20 : 0.20
O₂ : P₄O₁₀
5 : 1
1.56 : 1/5×1.56 = 0.312 mol
Less number of moles of product are formed by the oxygen thus it will act as limiting reactant.
Answer is in picture below.
Use 100 grams of the compound:
ω(Cl) = 85.5% ÷ 100%.
ω(Cl) = 0.855; mass percentage of the chlorine in the compound.
m(Cl) = 0.855 · 100 g.
m(Cl) = 85.5 g; mass of chlorine.
m(C) = 100 g - 85.5 g.
m(C) = 14.5 g; mass of carbon.
n(Cl) = m(Cl) ÷ M(Cl).
n(Cl) = 85.5 g ÷ 35.45 g/mol.
n(Cl) = 2.41 mol; amount of chlorine.
n(C) = 14.5 g ÷ 12 g/mol.
n(C) = 1.21 mol; amount of carbon.
n(Cl) : n(C) = 2.41 mol : 1.21 mol = 2 : 1.
This compound is dichlorocarbene CCl₂.
To be used as a conductor for heat
Answer:
9.62 × 10² J
Explanation:
In this problem, we need to convert 230 calories to joules. For this, we must know the conversion from calories to joules.
1 Calorie = 4.184 Joules
To find Joules in 230 calories, multiply 4.184 by 23. So,
230 Calorie = (230 × 4.184) Joules
= 962.32 Joules
or
230 Calories = 9.62 × 10² J
So, the correct option is (A).
Answer:
150.7815
Explanation:
In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.
A common request on this site is to convert grams to moles. To complete this calculation, you have to know what substance you are trying to convert. The reason is that the molar mass of the substance affects the conversion. This site explains how to find molar mass.