Answer is: A. Chemical energy to electromagnetic energy and thermal energy.
Balanced chemical reaction: 2Mg(s) + O₂(g) → 2MgO(s) + energy.
This is chemical change (chemical reaction), because new substance (magnesium oxide MgO) is formed, the atoms are rearranged and the reaction is followed by an energy change (exothermic reaction because energy is released).
Chemical changes (chemical synthesis) is when a substance combines with another (in this example magnesium and oxygen) to form a new substance.
Answer:
39.2 %
Explanation:
The following data were obtained from the question:
Mass of sample = 24 g
Mass of Cl = 14.6 g
Mass of B = 9.4 g
Percentage composition of boron =?
The percentage composition (by mass) of boron in the sample can be obtained as illustrated below:
Percentage composition of boron = mass of B /mass of sample × 100
Percentage composition of boron = 9.4/24 × 100
Percentage composition of boron = 39.2 %
Therefore, the percentage composition (by mass) of boron in the sample is 39.2 %
<span>There are a number of ways
to express concentration of a solution. This includes molality. Molality is
expressed as the number of moles of solute per mass of the solvent. We calculate as follows:
0.200 mol I2 / kg CCl4 ( .750 kg CCl4 ) ( 253.809 g I2 / mol I2) = 38.07 g I2 needed
Hope this helps.
</span>
Answer: d. More than 6.5 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.
Explanation: 
As can be seen from the chemical equation, 2 moles of aluminium react with 3 moles of copper chloride.
According to mole concept, 1 mole of every substance weighs equal to its molar mass.
Aluminium is the limiting reagent as it limits the formation of product and copper chloride is the excess reagent as (14-7.5)=6.5 g is left as such.
Thus 54 g of of aluminium react with 270 g of copper chloride.
1.50 g of aluminium react with=
of copper chloride.
3 moles of copper chloride gives 3 moles of copper.
7.5 g of copper chloride gives 7.5 g of copper.
The answer is 200 g.
If the molar mass of CaCl2 is 110.98 g/mol, this means there are 110.98 g in 1 L of 1 M solution.
Let's find how many g of CaCl2 are present in 0.720 M.
110.98 g : 1 M = x : 0.720 M
x = 110.98 g * 0.720 M : 1 M
x = 79.90 g
So there are 79.90 g in 0.720 M. In other words, in 1 L of 0.720 M solution there will be 79.90 g.
Now, we need to prepare ten beakers with 250 mL of solutions:
10 * 250 mL = 2500 mL = 2.5 L
79.90 g : 1 L = x : 2.5 L
x = 79.90 g * 2.5 L : 1 L
x = 199.75 g ≈ 200 g
