Answer:
part 1
3) 5.47 molecules
4) 11.15 moles
5) 2.95 × 10^23 molecules
6) 7670.9g
part2
1) 216000g
2) 5.71 molecules
3) 0.18 moles
4) 737.40g
5) 2.89×10^23 molecules
Explanation:
part 1
3) number of moles = (6.02 × 10^23)/11.0
=5.47molecules.
4) moles= (6.02 × 10^23)/(5.40×10^23)
= 11.15 moles
5) molar mass of NH3= 14+(3×1)= 14+3=17
number of molecules= (6.02 × 10^23) × 17/35
= (6.02 × 10^23)× 0.49 =2.95 × 10^23 molecules
6) molar mass of N2I6= (2×14) + (6×127)
= 28+762= 790g/mol
mass= 790 × (6.02 × 10^23)/(6.20×10^22)
=790× 9.71= 7670.9g
part 2
1) molar mass of CuBr= 64+80= 144g/mol
mass= 1500 × 144= 216000g
2) molecules = (6.02 × 10^23)/1.055
= 5.71 molecules
3) moles = (6.02 × 10^23)/(3.35×10^24)
= 0.18 moles
4) molar mass of SiO2= 28+ (2×16)= 28+32
=60gmol
mass= 60 × (6.02 × 10^23)/(4.90×10^22)
= 60× 12.29= 737.40g
5) molar mass of CH4= 12+(4×1)= 12+4=16g/mol
number of molecules= (6.02 × 10^23) × 16/33.6
= (6.02 × 10^23)× 0.48
= 2.89×10^23 molecules
Answer:
The compound may have properties that are very different from those of the elements.
Explanation:
For example - common salt, sodium chloride ( NaCl) has very different properties from sodium and chlorine
Answer:
I think it is A.
Explanation:
Hope my answer has helped you!
160 g of SO3 are needed to make 400 g of 49% H2SO4.
<h3>How many grams of SO3 are required to prepare 400 g of 49% H2SO4?</h3>
The equation of the reaction for the formation of H2SO4 from SO3 is given below as follows:
1 mole of SO3 produces 1 mole of H2SO4
Molar mass of SO3 = 80 g/mol
Molar mass of H2SO4 = 98 g/mol
80 g of SO3 are required to produce 98 og 100%H2SO4
mass of SO3 required to produce 400 g of 100 %H2SO4 = 80/98 × 400 = 326.5 g of SO3
Mass of SO3 required to produce 49% of 400 g H2SO4 = 326.5 × 49% = 160 g
Therefore, 160 g of SO3 are needed to make 400 g of 49% H2SO4.
Learn more about mass and moles at: brainly.com/question/15374113
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Carbon is special and unique because it is able to form different compounds with a lot of elements, including itself. When it bonds with itself, this is possible because of the concept of hybridization. It is the mixing of atomic orbitals into a new hybrid orbital. In this case, methane is formed through the sp³ hybridization.
