Answer: The mass of Cu produced is 4.88 g
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ……(1)
Given mass of aluminum = 2.98 g
Molar mass of aluminum = 27 g/mol
Plugging values in equation 1:
\text{Moles of aluminum}=\frac{2.98g}{27g/mol}=0.1104 mol
The given chemical equation follows:
2Al(s)+3CuSO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3Cu(s)
By the stoichiometry of the reaction:
If 2 moles of aluminum produces 3 moles of Cu
So, 0.1104 moles aluminium will produce = \frac{3}{2}\times 0.1104=0.1656mol of Cu
Molar mass of Cu = 63.5 g/mol
Plugging values in equation 1:
\text{Mass of Cu}=(0.1656mol\times 63.5g/mol)=10.516g
The percent yield of a reaction is calculated by using an equation:
\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100 ……(2)
Given values:
% yield of product = 46.4 %
Theoretical value of the product = 10.516 g
Plugging values in equation 2, we get:
46.4=\frac{\text{Actual value of Cu}}{10.516g}\times 100\\\\\text{Actual value of Cu}=\frac{46.4\times 10.516}{100}\\\\\text{Actual value of Cu}=4.88g
Hence, the mass of Cu produced is 4.88 g
Answer:
Specific heat is the amount of energy that must be added, in the form of heat, to one unit of mass of the substance in order to cause an increase of one unit in temperature.
Explanation:
Answer:
Water vapor
Explanation:
The magma consist of dissolved gases when these gases produce the force the volcanic eruption take place. The volcanic gases comes out and their volume is increased tremendously. The gases present in volcano are listed below:
The volcanic gases consist of water vapors, carbon dioxide and sulfur.
These three re the primary gases but the water is present in higher amount.
The percentage of water is 60%.
The carbon dioxide present in 10-40%.
Other gases present in valcano are nitrogen, argon, helium, neon methane and hydrogen.
Answer: The molarity of an HCl solution is 0.218 M if 43.6 mL of a 0.125 M NaOH solution are needed to titrate a 25.0 mL sample of the acid.
Explanation:
Given: 
= 43.6 mL, 
= 0.125 M
= 25.0 mL, 
= ?
Formula used to calculate the concentration of acid is as follows.
Substitute the values into above formula.
Thus, we can conclude that the molarity of an HCl solution is 0.218 M if 43.6 mL of a 0.125 M NaOH solution are needed to titrate a 25.0 mL sample of the acid.
It would be acidic based indicator.
0-6 is acidic
7 is neutral
8-14 is alkaline
