Answer:
126000J
Explanation:
From the question given, we obtained the following information:
M = 1500g
C = 4.2J/g°C
ΔT = 20°C
Q =?
Q= MCΔT
Q = 1500 x 4.2 x 20 =
Q = 126000J
<u>Given:</u>
Calculated density values-
Aluminum = 2.7 g/cm3
Copper = 9.0 g/cm3
Iron = 7.9 g/cm3
Titanium = 4.8 g/cm3
Unknown sample mass = 9.5 g
Sample volume = 2.1 cm3
<u>To determine:</u>
The identity of the unknown sample
<u>Explanation:</u>
'Density' is a physical parameter which can be used to identify the nature of the unknown substance.
Density = Mass/Volume
For the unknown sample
Density = 9.5 g/2.1 cm3 = 4.52 g/cm3
This matches closely with the calculated density of titanium
Ans: The unknown substance is made of titanium
good sir it seems your answer appears to be B
Answer:
Aluminum metal
Explanation:
In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.
First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:


Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.
Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.
Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):
Notice that the overall cell potential upon summing is:

Meaning we obey the law of galvanic cells.
Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.
Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
Explanation : Given,
Mass of oxygen in sulfur dioxide = 3.49 g
Mass of sulfur in sulfur dioxide = 3.50 g
Mass of oxygen in sulfur trioxide = 9.00 g
Mass of sulfur in sulfur trioxide = 6.00 g
Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.
Mass of oxygen per gram of sulfur for sulfur dioxide = 
Mass of oxygen per gram of sulfur for sulfur dioxide = 
and,
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.