Answer:
After 18 hours, the amount of pure technetium that will be remaining is 12.5 grams
Explanation:
To solve the question, we note that the equation for half life is as follows;

Where:
N(t) = Quantity of the remaining substance = Required quantity
N₀ = Initial radioactive substance quantity = 100 g
t = Time duration = 18 hours
= Half life of the radioactive substance = 6 hours
Therefore, plugging in the values, we have;

Therefore, after 18 hours, the amount of pure technetium that will be remaining = 12.5 grams.
Answer:
A for me
Explanation:
I dont know what you mean by this as this is an opinon question
Answer:
3.91 L
Explanation:
Using the ideal gas law equation as follows:
PV = nRT
Where:
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
Based on the information given in this question,
P = 5.23 atm
V= ?
n = 0.831 mol
T = 27°C = 27 + 273 = 300K
Using PV = nRT
V = nRT/P
V = (0.831 × 0.0821 × 300) ÷ 5.23
V = 20.47 ÷ 5.23
V = 3.91 L
Answer:
The products are carbon dioxide and water
Explanation:
Step 1: Data given
Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant.
Step 2: The complete combustion of C3H7OH:
For the combustion of 1-propanol, we need O2.
The products of this combustion are CO2 and H2O.
C3H7OH + O2→ CO2 + H2O
On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3
C3H7OH + O2→ 3CO2 + H2O
On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.
C3H7OH + O2→ 3CO2 + 4H2O
On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).
To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.
2C3H7OH + 9O2→ 6CO2 + 8H2O
For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O
The products are carbon dioxide and water