<span>6.03 moles.
1 molecule of butane contains 4 carbon atoms and ten hydrogen atoms.
The molar mass is 4 times the atomic mass of carbon, 12 g/mol, plus 10 times the atomic weight of hydrogen, 1 g/mol.
Molar mass = 4 * 12 g/mol + 10 * 1 g/mol = 58 g/mol.
This means that 1 mole of butane has a mass of 58 g.
To figure out how many moles are in a sample of butane, divide the mass of sample in grams by 58 grams
Number of moles in sample = 350 g / 58 g/mol = 6.03 moles.</span>
Ethylene- C2H4 = 85.7% Carbon and 14.3% Hydrogen
Find the atomic masses for each element and multiply it by the number of atoms in the compound, then add.
C- 12.0 * 2= 24.0
H- 1.00 * 4= 4.00
-----------------------
28.0
Take the masses for each element and divide it by the total mass. Then change the answer to get the percent.
C 24.0 / 28.0= .857 = 85.7%
H 4.00 / 28.0= .143 = 14.3%
<h3>
Ethylene is 85.7% Carbon and 14.3% Hydrogen </h3>
<span>Here are some
pH < 7
Sour taste (though you should never use this characteristic to identify an acid in the lab)
Reacts with a metal to form hydrogen gas Increases the H+ concentration in water
Donates H+ ions<span>
Turns blue litmus indicator red</span></span>
Answer:
0, l is n-1 always, ml is l to -l
The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below
write the reacting equation
Co(NO3)2 + Li2S = 2LiNO3 + COS
find the moles of CO(NO3)2 = molarity x volume
= 130 ml x 0.160=20.8 moles
since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles
volume of Lis is therefore = moles of Lis/ molarity of LiS
= 20.8/0.160 = 130 Ml
