Answer:
0.7g of HCl
Explanation:
First, let us write a balanced equation for the reaction between HCl and Al(OH)3.
This is illustrated below:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:
Molar Mass of Al(OH)3 = 27 + 3(16+1)
= 27 + 3(17) = 27 + 51 = 78g/mol.
Molar Mass of HCl = 1 + 35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g
Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
From the equation above,
78g of Al(OH)3 reacted with 109.5g of HCl.
Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl
Answer:
9.28 g/L
Explanation:
We will be using the ideal gas law to solve this problem:
PV = nRT where P is the pressure (atm)
V is the volume (L)
R is the gas constant 0.08205 Latm/Kmol
T is the temperature (K)
n is the number of moles
The number of moles is the mass divided by the molecular weight, and from here we can solve for the density. (Note here we use the atomic weight of radon since its is a monoatomic noble gas)
PV = m/AW RT ⇒ P = (m/V ) RT/AW ⇒ P AW /RT =D
0.950 atm x 222.0 g/mol / [( 0.08205 Latm/Kmol ) x 277 K ] = D
9.28 g/L = D
I learned this in class! the answer is c
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