Avegadro's number = 6.02 x10^23 atoms
so
3.5g x 1mol/63.55g Cu x 6.02 x 10^23/ 1mol=3.32 x 10^22 atoms
An occluded front forms when a warm air mass is caught between two cooler air masses.
4.22g
1. Work out the equation and balance
2AgNo3 +BaCl2 ---> 2AgCl + Ba(No3)2
2. Work out formula mass of silver nitrate (should = 169.9)
3. Calculate the number of moles by doing moles = mass / formula mass so 5 divided by 169.9 = 0.0294 moles.
4. Check ratio - here the ratio is 2:2 i.e. 2 moles of silver nitrate to 2 moles of silver chloride so the moles will be the same so the moles of silver chloride is also 0.0294
5. Work out the formula mass of AgCl (always ignore big numbers at the start when working out formula mass) = 143.4
6. Work out mass by doing equation moles = mass/formula mass so
mass = moles x formula mass
mass = 0.0294 x 143.4
mass = 4.22g (3sf or 2dp)
Answer:
A. N2S3
yep u r right
Explanation:
Dinitrogen Trisulfide N2S3 Molecular Weight -- EndMemo.
