I have no idea!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
25.97oC
Explanation:
Heat lost by aluminum = heat gained by water
M(Al) x C(Al) x [ Temp(Al) – Temp(Al+H2O) ] = M(H2O) x C(H2O) x [ Temp(Al+H2O) – Temp(H2O) ]
Where M(Al) = 23.5g, C(Al) = specific heat capacity of aluminum = 0.900J/goC, Temp(Al) = 65.9oC, Temp(Al+H2O)= temperature of water and aluminum at equilibrium = ?, M(H2O) = 55.0g, C(H2O)= specific heat capacity of liquid water = 4.186J/goC
Let Temp(Al+H2O) = X
23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)
21.15(65.9-X) = 230.23(X-22.3)
1393.785 - 21.15X = 230.23X – 5134.129
230.23X + 21.15X = 1393.785 + 5134.129
251.38X = 6527.909
X = 6527.909/251.38
X = 25.97oC
So, the final temperature of the water and aluminum is = 25.97oC
Answer:

Explanation:
Hello there!
In this case, according to the given information, it turns out necessary for us remember that the first-order kinetics is given by:

Whereas the 27.5% complete means A/Ao=0.275, and thus, we solve for the rate constant as follows:

Then, we plug in the variables to obtain:

Regards!
NaHCO3 is the right answer
<h3>Answer:</h3>
18.75 grams
<h3>Explanation:</h3>
- Half-life refers to the time taken by a radioactive material to decay by half of the original mass.
- In this case, the half-life of element X is 10 years, which means it takes 10 years for a given mass of the element to decay by half of its original mass.
- To calculate the amount that remained after decay we use;
Remaining mass = Original mass × (1/2)^n, where n is the number of half-lives
Number of half-lives = Time for the decay ÷ Half-life
= 40 years ÷ 10 years
= 4
Therefore;
Remaining mass = 300 g × (1/2)⁴
= 300 g × 1/16
= 18.75 g
Hence, a mass of 300 g of an element X decays to 18.75 g after 40 years.