<h3>
Answer:</h3>
1 x 10^13 stadiums
<h3>
Explanation:</h3>
We are given that;
1 stadium holds = 1 × 10^5 people
Number of iron atoms is 1 × 10^18 atoms
Assuming the stadium would carry an equivalent number of atoms as people.
Then, 1 stadium will carry 1 × 10^5 atoms
Therefore,
To calculate the number of stadiums that can hold 1 × 10^18 atoms we divide the total number of atoms by the number of atoms per stadium.
Number of stadiums = Total number of atoms ÷ Number of atoms per stadium
= 1 × 10^18 atoms ÷ 1 × 10^5 atoms/stadium
= 1 × 10^13 Stadiums
Thus, 1 × 10^18 atoms would occupy 1 × 10^13 stadiums
B. 0.9 <span>
</span>D.Light intensity has no effect on whether electrons are emitted or not.
and
A. X=1.9eV,Y=0.2eV
I already took the gizmo so I know these are right. The first one I got wrong b/c there was no graph and the last one I didn't understand. Basically for the last one you calculate the work function for the metals and find their difference.
Hope this helps.
Answer:
Maintaining a high starting-material concentration can render this reaction favorable.
Explanation:
A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:
ΔG = ΔG° + R.T.lnQ
where,
ΔG° is the standard Gibbs free energy
R is the ideal gas constant
T is the absolute temperature
Q is the reaction quotient
To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.
0.355M x 0.0282L= 0.01 moles of H2SO4. Remember sulphuric acid is diprotic so it will release 2 from each molecule.
<span>So moles of protons = 0.01 x 2 = 0.02 moles of H+ </span>
<span>For neutralization: moles H+ = moles OH- </span>
<span>Therefore moles of NaOH = 0.02 </span>
<span>conc = moles / volume </span>
<span>Conc NaOH = 0.02 / 0.025L = 0.8M </span>
