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:
3 half-lives
Explanation:
The half-life is the time that it takes to a radioactive element to decay to half of its initial amount.
Let's suppose we start with 64 g of the radioactive element.
- After 1 half-life, the mass of the element will be 32 g.
- After 2 half-lives, the mass of the element will be 16 g.
- After 3 half-lives, the mass of the element will be 8 g.
This would be 8.010 * 10^-2
Answer:
C: The C horizon is a subsurface horizon. It is the least weathered horizon. Also known as the saprolite, it is unconsolidated, loose parent material. The master horizons may be followed by a subscript to make further distinctions between differences within one master horizon.
Explanation:
