<span>ALPO4 can be broken up into AL/P/O4.
There is one AL mole, one P mole, and four O moles. Together, that is a total of six moles -- 1 + 1 + 4 = 6. You can find the moles for each section by breaking the atoms up by type. There are 4 O because the number 4 follows the O. There are none in front of AL or P, so they are singular.</span>
Answer:
THE HEAT NEEDED TO CHANGE 3KG OF WATER FROM 10 C TO 80 C IS 877.8kJ OR 877,800 J.
Explanation:
Mass = 3.0 kg = 3 * 1000 = 3000 g
Initial temperature = 10 C
Final temperature = 80 C
Change in temperature = 80 - 10 = 70 C
Specific heat of water = 4.18 J/g C
Heat needed = unknown
Heat is the amount of energy in joules needed to change a gram of water by 1 C.
Heat = mass * specific heat * change in temperature
Heat = 3000 g * 4.18 J/g C * 70 C
Heat = 877 800 Joules
Heat = 877.8 kJ.
The heat needed to change 3 kg mass of water from 10 C to 80 C is 877,800 J or 877.8 kJ.
Answer:
Group 1 ( The Leftmost)
More Information :
The alkali metals are six chemical elements in Group 1, the leftmost column in the periodic table.
They are :
- lithium (Li)
- sodium (Na)
- potassium (K)
- rubidium (Rb)
- cesium (Cs)
- francium (Fr).
Answer:
0.147 billion years = 147.35 million years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of Potassium-40 is 1.25 billion years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(1.25 billion years) = 0.8 billion year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
<em></em>
where, k is the rate constant of the reaction (k = 0.8 billion year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of (Potassium-40) ([A₀] = 100%).
[A] is the remaining concentration of (Potassium-40) ([A] = 88.88%).
- At the time needed to be determined:
<em>8 times as many potassium-40 atoms as argon-40 atoms. Assume the argon-40 only comes from radioactive decay.</em>
- If we start with 100% Potassium-40:
∴ The remaining concentration of Potassium-40 ([A] = 88.88%).
and that of argon-40 produced from potassium-40 decayed = 11.11%.
- That the ratio of (remaining Potassium-40) to (argon-40 produced from potassium-40 decayed) is (8: 1).
∴ t = (1/k) ln([A₀]/[A]) = (1/0.8 billion year⁻¹) ln(100%/88.88%) = 0.147 billion years = 147.35 million years.
Answer:
The answer is D
Explanation:
All of these solutions are correct.
