The volume of the balloon at 85.0°C would be 2.3L
Answer:
Filtration
Explanation:
Filtration is used to separate solid substances from liquids or large molecules from small molecules. Since, Acid is a liquid and metal oxide is solid so Filtration can be used to separate these two.
Answer:
2192.64 PSI.
Explanation:
- From the general law of ideal gases:
<em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the container in L (V = 1650 L).
n is the no. of moles of the gas in mol (n = 9750 mol).
R is the general gas constant (R = 0.082 L.atm/mol.K).
T is the temperature of the gas in (T = 35°C + 273 = 308 K).
∴ P = nRT/V = (9750 mol)(0.082 L.atm/mol.K)(308 K)/(1650 L) = 149.2 atm.
- <u><em>To convert from atm to PSI:</em></u>
1 atm = 14.696 PSI.
<em>∴ P = 149.2 atm x (14.696 PSI/1.0 atm) = 2192.64 PSI.</em>
<span>atomic weights: Al = 26.98, Cl = 35.45
In this reaction; 2Al = 53.96 and 3Cl2 = 212.7
Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed.
Step 2:
(a) Ratio of Al:Cl = 2.70/4.05 = 0.6667
since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537.
so Cl is limiting
(b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced.
From Step 1:
212.7g of Cl will produce 266.66g AlCl3
212.7g = 266.66g
4.05g = x
x = 5.08g of AlCl3 can be produced
(c)
Al:Cl = 0.2537
Al:Cl = Al:4.05 = 0.2537
mass of Al used in reaction = 4.05 x 0.2537 = 1.027g
Excess reactant = 2.70 - 1.027 = 1.67g
King Leo · 9 years ago</span>
Answer:
See explanation
Explanation:
The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.
This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).
In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.
Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.
Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.