Cu + 2H2SO4 ⟶ CuSO4 + SO2 + 2H20
In left hand side of the equation.
Cu = 1 atom
H = 4 atoms
S = 2 atoms
O = 8 atoms
In right hand side of the equation.
Cu = 1 atom
S = 2 atoms
0 = 8 atoms
H = 4 atoms
• All the atoms are balanced in the left and right side of the equation and it satisfies the law of conservation of mass.
• Equation is balanced and correct.
*See the attachment .
Answer:
There are typically three ways that it is accomplished: use of erythropoietin (EPO) or synthetic oxygen carriers and blood transfusions. While transfusions of large volumes of blood or use of EPO can be detected, microdosing EPO or transfusing smaller volumes of packed red blood cells is much harder to detect.
Answer: acceleration
a = 5.36 m/s²
Explanation: solution attached:
Convert first 60 mi/h to m/s
Use the formula for acceleration
a = vf - vi /t
You have not mention here about those elements but the general concept for this is the:
uneven distribution ------> polar
even distribution -------> non-polar
Answer : The density of air in and is, and respectively.
Explanation :
where,
P = pressure of air = 1 atm
V = volume of air
T = temperature of air = 297 K
The conversion used for the temperature from Fahrenheit to degree Celsius is:
The conversion used for the temperature from degree Celsius to Kelvin is:
n = number of moles
m = mass of air
M = average molar mass of air = 28.97 g/mole
= density of air = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the above formula, we get:
Now we have to calculate density in .
Conversion used :
So,
The density of air in is,
Now we have to calculate density in .
Conversion used :
So,
The density of air in is,