B.Elements
Explanation: they cannot be separated
Answer:
21.8 grams.
Explanation:
Molar mass data from a modern periodic table:
How many moles of MgO will be produced if Mg is the limiting reactant?
Number of moles of Mg:
.
The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.
How many moles of MgO will be produced if O₂ is the limiting reactant?
Number of moles of O₂:
.
The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. 
of MgO will be produced if O₂ is in excess.
How many moles of MgO will be produced?
0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.
What's the mass of 0.541284 moles of MgO?
Formula mass of MgO:
.
Mass of 0.541284 moles of MgO:
.
Light bulb -it produces heat and light
toaster - it produces heat and the metal strips glow producing light
Answer:
Glucose will move from the solution B to the solution A
Explanation:
Given that:
Solution A contains 1% glucose, and,
Solution B contains 5% glucose
Diffusion is the net movement of the substance from the region of the higher concentration to the region of the lower concentration.
Thus, solution B contains more concentration of glucose as compared to solution A. <u>By the process of diffusion, the particle moves from higher concentration to lower concentration and thus, glucose will move from solution B to solution A.</u>
Answer:
13.44dm^3
Explanation:
To calculate this we first need to know the number of mole produced. We will first need to balance the equation to know the theoretical mole ratio.
C2H6 (g) + 3.5O2 (g) → 2CO2 (g) + 3H2O (g)
From the balanced equation, we can deduce that one mole of ethane yielded 2 moles of carbon iv oxide. We use this information to calculate the actual number of moles yielded.
24g were reacted. Now to know the number of moles reacted, we simply divide the mass by the molar mass. The molar mass of ethane is 2(12) +6(1)= 40g/mol
The number of moles is thus 24/40 = 0.6 moles
Like we said earlier, one mole yielded 2 moles of carbon iv oxide, hence, 0.6 moles will yield 0.6 * 2 = 0.12 moles of carbon iv oxide.
Now, at stp, one mole of a gas occupies a volume of 22.4dm^3 thus, 0.6 mole will occupy 0.6 * 22.4 = 13.44dm^3
