Answer:
18.018 seconds.
Explanation:
Given that the half life of Manganese, Mn = 3 seconds. The initial sample mass = 90.0 gram, the final sample mass = 1.40 gram.
The general idea to the question is to look for the time it will take to decay from the initial mass that is 90 gram to 1.40 gram.
Therefore, we will be making use of the formula below;
J(t) = J(o) × (1/2)^t/t(hL).
Where t(hL) is the half life, t is the time taken, J(t)= mass after time,t and J(o) is the initial mass. So, let us slot in the values into the equation above.
1.4 = 90 × (1/2)^ t/3.
1.4/90 = (1/2)^t/3.
t/3 = log(0.5) (1.4/90).
+Please note that the 0.5 of the log is at the subscript).
That is the base 0.5 logarithm of (1.4/90) 0.01556 is 6.0060141295.
t = 3 × 6.0060141295.
t = 18.018 seconds.
Answer:
A its a step by step equation the answer is A
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
Oak tree, because it’s the primary producer
