Answer:
433 m
Explanation:
Since the fall represents motion under gravity, we use the equation
s = ut - 1/2gt² where s = height of cliff or distance bowling ball falls through, u = initial velocity of bowling ball = 0 m/s(since it starts from rest), t = time = 9.4 s and g = acceleration due to gravity = -9.8 m/s².
So, substituting the values of the variables into the equation, we have
s = 0 m/s × 9.4 s - 1/2 × 9.8 m/s² × (9.4 s)²
s = 0 m - 1/2 × -9.8 m/s² × 88.36 s²
s = 1/2(865.928 m)
s = 432.964
s ≅ 433 m
Answer:
K = 10
Explanation:
Using Hess's law, it is possible to obtain the equilibrium constant, K, of a reaction using K of similar reactions. For example:
<em> If A ⇄ B K = X</em>
B ⇄ A K = 1/X
2A ⇄ 2B K = X².
Thus, if A(g) ⇄ 2B(g) K = 0.010
2B(g) ⇄ A(g) K = 1 / 0.010; K = 100
B(g) ⇄ A(g) K = √100 = 10
<h3>K = 10</h3>
<span>The atomic number of a neutral atom is equal to the number of protons and the number of electrons of the atom. The atomic weight meanwhile is equal to the sum of the number of protons and number of neutrons.Hence we can say that the number of protons is 8 and the number of neutrons is 10. </span>
Answer:
760 uM
Explanation:
<em>A biochemist carefully measures the molarity of magnesium ion in 47, mL of cell growth medium to be 97 uM. Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 6.0 mL. Calculate the new molarity of magnesium ion in the cell growth medium Be sure your answer has the correct number of significant digits.</em>
The problem here is that the amount of magnesium ion remains the same irrespective of the volume.
Amount of magnesium in the growth medium = <em>molarity x volume</em>
= 97 x
x 47 x
= 4.559 x 
Then, the volume reduced to 6.0 mL, the new molarity becomes;
<em>molarity = mole/volume </em>
= 4.559 x
/6 x
= 7.598333 x
M = 759.83333 uM
To the correct number of significant digits = 760 uM
If there is any unused Chemicals left in the container it should be properly disposed and chemicals taken out of the original container should not be poured back into the container because it can be <span>contaminated</span>