Answer:
The answer should be 1000 kg / m3
Answer:
285g of fluorine
Explanation:
To solve this problem we need to find the mass of Freon in grams. Then, with its molar mass we can find moles of freon and, as 1 mole of Freon, CCl₂F₂, contains 2 moles of fluorine, we can find moles of fluorine and its mass:
<em>Mass Freon:</em>
<em>2.00lbs * (454g / 1lb) = </em>908g of Freon
<em>Moles freon -Molar mass: 120.91g/mol- and moles of fluorine:</em>
908g of Freon * (1mol / 120.91g) =
7.5 moles of freon * (2moles Fluorine / mole Freon): 15 moles of fluorine
<em>Mass fluorine -Atomic mass: 19g/mol-:</em>
15 moles F * (19g / mol) =
<h3>285g of fluorine</h3>
question 1
moles = mass/molar mass of Al(OH)3
convert Kg to g
that is 1.09 x 1000=1090g
moles is therefore=1090g/78(molar mass of Al(OH)3)= 13.974 moles
question 2
moles=2.55g/327.2(molar mass of Pb(CO3)2= 7.79 x 10^-3 moles
from avogadro constant
1moles=6.02 x10^23 formula units
what about 7.79 x 10 ^-3
={(7.79 x 10^-3)moles x ( 6.02 x10^23)} /1 mole=4.69 x10^21 formula units
The weight of aluminum are required to produce 8.70 moles of aluminum chloride is 234.9 g
<h3>
What is the use of aluminium chloride ?</h3>
Aluminum chloride is useful for the treatment of palmar, plantar, and axillary hyperhidrosis.
Aluminum chloride has also been reported to be useful in facial and scalp hyperhidrosis
The balanced chemical equation represents the mole ratio in which the chemicals combine.
In this case, illustrates that 2 mol Al produces 2 mol Al Cl₃, hence these 2 chemicals are in a 1:1 ratio.
Thus, to produce 8.70 mol aluminium chloride, it will require 8.70 mol aluminium.
But this quantity of Al has a mass in grams of
m = n × Mr
= 8.70 mol × 27g/mol
= 234.9 g
Hence, The weight of aluminum are required to produce 8.70 moles of aluminum chloride is 234.9 g
Learn more about mole concept here ;
https://brainly.in/question/12599804
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<span>(NH4)2CO3 -> 96.09 g/mol
(6.995g ammonium carbonate)(1mol ammonium carbonate/ 96.09 g ammonium carbonate) = 0.072796 mol ammonium carbonate
In this conversion, the 'grams' unit is crossed out because it is in both the numerator and the denominator, which leaves the 'mol' unit left.
Looking at the formula (NH4)2CO3, you can look at it as if it were:
2 mol (NH4) + 1 mol (CO3) = 1 mol (NH4)2CO3
For every 1 mol of ammonium carbonate, you have 1 mol of carbonate ions and 2 moles of ammonium ions.
(0.072796 mol ammonium carbonate) = (0.072796 mol carbonate ion)+(0.363981 mol ammonium ion) </span>
