Percent (%) Composition of CuO
Cu = 1 x 50g - Multiply by one as there is one Cu
O = 1 x 12.5g - Multiply by one as there is one O
CuO = 62.5g
% for Cu = 50g over 62.5 multiplied by 100 = 80%
% for O = 12.5g over 62.5 multiplied by 100 = 20%
Final Answer :
<em>Percent (%) Composition of CuO = </em>80% (Cu) & 20% (O)
The class of compounds that ammonia belongs to is it is categorized as a basic compound. As it posses a pH value or rating above 7, which is characteristic of basic substances, solutions.
Answer: When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is 
.
Explanation:
The word equation is given as maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas.
Now, in terms of chemical formulae this reaction equation will be as follows.
Here, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
To balance this equation, multiply HCl by 4 on reactant side and multiply 
by 2 on product side. Therefore, the equation can be rewritten as follows.
Hence, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.
Thus, we can conclude that when maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is .
Answer:
D
Explanation:
The amount of energy released or absorbed is equal the product of the mass, the specific heat capacity and the temperature change. The temperature change being the difference between the final and initial temperature.
Q = mc∆T
Q = heat energy (Joules, J) m = mass of a substance (kg) c = specific heat (units J/g∙K)
∆ is a symbol meaning "the change in" ∆T = change in temperature (Kelvins, K)
From the data provided in the question, we can deduce that:
Q = 16.7KJ = 16,700J
m = 225g
c = 1.74J/g.k
For the temperature, let the final temperature be f. This means our ∆T = f - 20
16,700 = 225 * 1.74 * (f - 20)
16700 = 391.5 (f - 20)
f - 20 = 16700/391.5
f - 20 = 42.7
f = 20 + 42.7 = 62.7
Answer:
Q = 233.42 J
Explanation:
Given data:
Mass of lead = 175 g
Initial temperature = 125.0°C
Final temperature = 22.0°C
Specific heat capacity of lead = 0.01295 J/g.°C
Heat absorbed by water = ?
Solution:
Heat absorbed by water is actually the heat lost by the metal.
Thus, we will calculate the heat lost by metal.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 22.0°C - 125.0°C
ΔT = -103°C
Q = 175 g × 0.01295 J/g.°C×-103°C
Q = -233.42 J
Heat absorbed by the water is 233.42 J.
