A) -0.5(9.8)*t^2 = -25(t-2) - 0.5(9.8)(t-2)^2
-4.9t^2 = -25t + 50 - 4.9(t^2-4t+4)
0 = -25t+50+19.6t - 19.6
5.4t = 30.4
t = 5.62962963 s
b) h = -4.9(5.62962963)^2
h = -155.2943759
the building is 155.2943759 m high
c) speed 0of first stone
= at
= 9.8*5.62962963
= 55.17037037 m/s
speed of second stone
= v + at
= 25+9.8*3.62962963
= 60.57037037 m/s
Answer:
Where are the question's???
Explanation:
Answer:
1.- Chemical change
2.- Because the atoms in the image are tranformed into a new molecule
3.- Yes, it does
4.- Due to the amount of atoms are the same both in products and reagents
Explanation:
Using the Rydberg formula, the spectral line of H - atom is suitable for this purpose is Paschen, ∞ → 3.
- Using the Rydberg formula;
1/λ = RH(1/nf^2 - 1/ni^2)
Given that;
λ = wavelength
RH = Rydberg constant
nf = final state
ni = initial state
- When final state = 3 and initial state = ∞
Then;
1/λ = 1 × 10^7 m-1 (1/3^2 - 1/ ∞^2)
1/λ = 1 × 10^7 m-1 (1/3^2 )
λ = 900 nm
Hence, the correct answer is Paschen, ∞ → 3
Learn more about the Rydberg formula; brainly.com/question/17753747
Answer:
Qsp > Ksp, BaCO3 will precipitate
Explanation:
The equation of the reaction is;
Na2CO3 + BaBr2 -------> 2NaBr + BaCO3
Since BaCO3 may form a precipitate we can determine the Qsp of the system.
Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles
concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M
Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles
concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M
Hence;
[Ba^2+] = 6.7 * 10^-5 M
[CO3^2-] = 9.1 * 10^-4 M
Qsp = [6.7 * 10^-5] [9.1 * 10^-4]
Qsp = 6.1 * 10^-8
But, Ksp for BaCO3 is 5.1*10^-9.
Since Qsp > Ksp, BaCO3 will precipitate
