Answer:
If it contains only 2 elements it will begin with “hydro” as in
hydrochloric acid (HCl). If it contains more than 2 elements it is named based on the polyatomic (oxy) anion, as in nitric acid (HNO3), nitrous acid (HNO2), sulfuric acid (H2SO4). If the element has more than one possible charge, the value of the charge comes after the element name and before the word ion. Thus, Fe 2+ is the iron two ion, while Fe 3+ is the iron three ion.
Explanation:
In this item, we are instructed that the copper ion is to be reacted with the ion of the chlorine. The ion for copper is positively charged (cation) because it has to lose some of its electron in order to achieve stability. As the charge suggests, the number of electrons is 2.
The counterion (anion) is the chloride which can be written as Cl⁻¹. Since the charge of Cl is only -1, two of this anion should be allowed to react with Cu²⁺ so that the formed molecule is neutral. The formula is therefore CuCl₂.
<em>ANSWER: CuCl₂</em>
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Answer:
The enthalpy change during the reaction is -7020.09 kJ/mole.
Explanation:
First we have to calculate the heat gained by the calorimeter.
where,
q = heat gained = ?
c = specific heat =
= final temperature =
= initial temperature =
Now put all the given values in the above formula, we get:
The heat gained by water present in calorimeter. = q'
where,
q' = heat gained = ?
m = mass of water =
c' = specific heat of water =
= final temperature =
= initial temperature =
q ' = 12,060.38 J
Now we have to calculate the enthalpy change during the reaction.
where,
= enthalpy change = ?
Q = heat gained = -(q+q') = -(2,485.25 J + 12,060.38 J)= -14,545.63 J
Q = -14.54563 kJ
n = number of moles fructose =
Therefore, the enthalpy change during the reaction is -7020.09 kJ/mole.
Answer:
The answer to your question is: 83.9 %
Explanation:
Data
Cu = 31.8 g
S = 50 g
CuS = 40 g
yield = ?
Equation
Cu + S ⇒ CuS
MW Cu = 64 g
MW S = 32 g
MW CuS = 96 g
Ratio (theoretical/experimental)
Experimental 50/31.8 = 1.57
Theoretical 32/64 = 0.5 limiting reactant Cu
64 g of Cu ------------------ 96 g of CuS
31.8 g ------------------- x
x = (31.8 x 96) / 64
x = 47.7 g of CuS
% yield = (40/47.7) x 100
= 83.9 %