Answer:
In the presence of UV light, ethane will react with bromine in a substitution reaction. UV light is the condition under which the reaction will occur so it is written above the arrow in the chemical equation. As the reaction proceeds, the intensity of the re-brown colour of the bromine water decreases.
Answer:
density=mass/volume
Explanation:
density is defined as mass per unit volume.
Answer: The chemical equation for the formation one mole of 
from its elements in their standard states is 
Explanation:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
The chemical equation for the formation one mole of from its elements in their standard states is:
where (s) represents solid state and (g) represents gaseous state.
Answer:
The correct answer is: 6.6 g MgO
Explanation:
First we have to write and balance the chemical reaction as follows:
2Mg(s) + O₂(g) → 2MgO(s)
That means that 2 moles of Mg(s) react with 1 mol of O₂(g) to give 2 moles of MgO(s). If Mg is totally consumed and a mass of O₂ remains unaltered after reaction, t<em>he limiting reactant is Mg</em>. We use the limiting reactant to calculate the mass of product.
According to the balanced chemical equation, 2 moles of Mg(s) produce 2 moles of MgO(s).
2 moles Mg = 2 mol x molar mas Mg= 2 mol x 24.3 g/mol = 48.6 g Mg
2 moles MgO= 2 mol x (molar mass Mg + molar mass O) = 2 mol x (24.3 g/mol + 16 g/mol) = 80.6 g MgO
The stoichiometric ratio is 80.6 g MgO/48.6 g Mg. So, we multiply this ratio by the mass of consumed Mg (4.0 g) in order to obtain the produced mass of MgO:
4.0 g Mg x 80.6 g MgO/48.6 g Mg = 6.63 g MgO
6.6 grams of magnesium oxide are formed.
Answer:
A is Mg^2+
B is K^+
C is Se^2-
D is Cl^-
Ca^2+ lies between Mg^2+ and K^+
S^2- lies between K^+ and Se^2-
Explanation:
I have attached an image of the four spheres.
Remember that, cations are smaller in size than the corresponding atom atom because the formation of a cation often involves the removal of an entire shell. Anions are larger in size than the corresponding atom because an electron is added thereby expanding the electron cloud and size of the anion.
From left to right let us designate the spheres as ABCD.
A is Mg^2+
B is K^+
C is Se^2-
D is Cl^-
Ca^2+ lies between Mg^2+ and K^+
S^2- lies between K^+ and Se^2-
