Answer:
Aluminium atoms = 4.13 *10^22 aluminium atoms
The correct answer is E
Explanation:
Step 1: Data given
Mass of Al2O3 = 3.50 grams
Molar mass of Al2O3 = 101.96 g/mol
Number of Avogadro = 6.022 * 10^23 /mol
Step 2: Calculate moles Al2O3
Moles Al2O3 = mass Al2O3 / molar mass Al2O3
Moles Al2O3 = 3.50 grams / 101.96 g/mol
Moles Al2O3 = 0.0343 moles
Step 3: Calculate moles Aluminium
In 1 mol Al2O3 we have 2 moles Al
in 0.0343 moles Al2O3 we have 2*0.0343 = 0.0686 moles Al
Step 4: Calculate aluminium atoms
Aluminium atoms = moles aluminium * Number of Avogadro
Aluminium atoms = 0.0686 * 6.022 * 10^23
Aluminium atoms = 4.13 *10^22 aluminium atoms
The correct answer is E
C2H6O + O2 ---> C2H4O2 + H2O
using the molar masses:-
24+ 6 + 16 g of C2H6O produces 24 + 4 + 32 g C2H4O2 (theoretical)
46 g produces 60g
60 g C2H4O2 is produced from 46g C2H6O
1g . .................................46/60 g
700g ................................. (46/60) * 700 Theoretically
But as the yield is only 7.5%
the required amount is ((46/60) * 700 ) / 0.075 = 7155.56 g
= 7.156 kg to nearest gram. Answer
<span>The instructor should be questioned to see if the filtrate is able to be recycled. This precipitate can contaminate the filtrate, rendering it useless for repeated experiments. If it is able to be recycled, a second pass through the filter might be required to remove the precipitate.</span>
Answer: The questions looks unclear
Explanation: Periodic table is a table that contains elements arranged according to their increasing atomic number.
1. D belongs to group 4
E. Belongs to group 7
B belongs to group 1
A belongs to group 8. A noble gas.
R belongs to group 3. K belongs to group 6 C belongs to group 1. H belongs to group 8
Answer:6.719Litres of Cl2 gas.
Explanation:According to eqn of rxn
2Na +Cl2=2NaCl
P=689torr=689/760=0.91atm
T=39°C+273=312K
according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl
But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW
MW of NaCl=23+35.5=58.5g/mol
n=28g(mass given of NaCl)/58.5
n=0.479moles of NaCl
Going back to the reaction,
if 1moles of Cl2 produces 2moles of NaCl
x moles of Cl2 will give 0.479moles of NaCl.
x=0.479*1/2
x=0.239moles of Cl2.
To find the volume, we use ideal ggas eqn,PV=nRT
V=nRT/P
V=0.239*0.082*312/0.91
V=6.719Litres
