One mole of hydrogen peroxide contains 6.02 x 10^23 molecules of hydrogen peroxide. And each molecule contains 4 atoms, so the answer is 4 x 6.02 x 10^23.
Hey there!:
Molar mass:
CHCl3 = ( 12.01 * 1 )+ (1.008 * 1 ) + ( 35.45 * 3 ) => 119.37 g/mol
C% = ( atomic mass C / molar mass CHCl3 ) * 100
For C :
C % = (12.01 / 119.37 ) * 100
C% = ( 0.1006 * 100 )
C% = 10.06 %
For H :
H% = ( atomic mass H / molar mass CHCl3 ) * 100
H% = ( 1.008 / 119.37 ) * 100
H% = 0.008444 * 100
H% = 0.8444 %
For Cl :
Cl % ( molar mass Cl3 / molar mass CHCl3 ):
Cl% = ( 3 * 35.45 / 119.37 ) * 100
Cl% = ( 106.35 / 119.37 ) * 100
Cl% = 0.8909 * 100
Cl% = 89.9%
Hope that helps!
Answer: -40
Explanation: Percent error is calculated by subtracting the value you actually recieved from the literature value (175 in your case) SO your answer should be 135-175=-40.
Answer : Option (A) Accelerator 2 model has the lowest percentage of energy lost as waste.
Solution : Given,
For Accelerator 1 model,
Input energy = 2078.3 J
Wasted energy = 663.1 J
Output energy = 1415.2 J
For Accelerator 2 model,
Input energy = 7690.0 J
Wasted energy = 2337.5 J
Output energy = 5353.5 J
For Accelerator 3 model,
Input energy = 4061.9 J
Wasted energy = 2259.6 J
Output energy = 1802.3 J
Formula used for lowest percentage of energy lost as waste is:
% energy lost as waste = (Total energy wasted / Total input energy ) × 100
For Accelerator 1 model,
% energy lost as waste =
= 31.90%
For Accelerator 2 model,
% energy lost as waste =
= 30.39%
For Accelerator 3 model,
% energy lost as waste =
= 55.62%
So, we conclude that the Accelerator 2 model has the lowest percentage of energy lost as waste.
Answer:
1.63 × 10²⁴ atoms.
Explanation:
To calculate the number of atoms (N) contained in 2.7moles of carbon, we multiply the number of moles (n) by Avogadro's number (6.02 × 10²³).
That is, N = n × nA
Where;
N = number of atoms
n = number of moles (mol)
nA = Avogadro's numbe
N = 2.7 × 6.02 × 10²³
N = 16.254 × 10²³
N = 1.63 × 10²⁴ atoms.
Hence, there are 1.63 × 10²⁴ atoms in 2.7moles of Carbon.