N x 10^exp or N x 10e(exp)
examples: 5.6 x 10^-3 or 5.6 x 10e-3
the most energy to remove an
electron?
Why do these elements have a higher ionization energy?
Answer:
Noble gases
Explanation:
The noble gases are non-metals that requires the highest amount of energy to remove an electron from their shells.
The reason for this difficult is that their electronic configuration confers a stable configuration them.
A.
6CO2 + C6H12O6 --> 6H2O + 6O2
B.
6CO2 + 6H2O --> C6H12O6 + 6O2
C.
C6H12O6 + 6O2 --> 6CO2 + 6H2O
D.
6CO2 + 6O2 --> C6H12O6 + 6H2O
The answer is B. 6CO2+H2O yields C6H12O6+ 6H20.
o helium (He)
o hydrogen (H)
sodium (Na)
O argon (Ar)
Answer: helium(He)
Explanation: Helium has only 2 electrons in the outermost energy level, but all of the other elements in its column have eight.
What occurs when the concentration of H2(g) is increased?
(1) The rate of the forward reaction increases and the concentration of N2(g) decreases.
(2) The rate of the forward reaction decreases and the concentration of N2(g) increases.
(3) The rate of the forward reaction and the concentration of N2(g) both increase.
(4) The rate of the forward reaction and the concentration of N2(g) both decrease.
If the concentration of is increased, the rate of the forward reaction increases and the concentration of decreases.
The principle states that reactions in equilibrium being disturbed by factors such as temperature, the concentration of species in the reaction, etc, will experience a shift in the equilibrium so as to annul the effects of the disturbance.
Thus, adding more of the reactants ( to the reaction will see more products being synthesized. This means more will be consumed and its concentration will decrease accordingly.
More on Le chartelier principle can be found here: brainly.com/question/12099569
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To determine how much of a 144g sample of carbon-14 will remain after 1.719 x 10^4 years, you can use the formula for exponential decay:
\[N(t) = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{T}}\]
Where:
- \(N(t)\) is the remaining amount after time \(t\).
- \(N_0\) is the initial amount.
- \(t\) is the time that has passed.
- \(T\) is the half-life.
In this case, \(N_0\) is 144g, \(t\) is 1.719 x 10^4 years, and \(T\) is the half-life of carbon-14, which is 5,730 years.
Plug these values into the formula:
\[N(t) = 144g \cdot \left(\frac{1}{2}\right)^{\frac{1.719 \times 10^4\text{ years}}{5,730\text{ years}}}\]
Now, calculate:
\[N(t) = 144g \cdot \left(\frac{1}{2}\right)^{\frac{3}{2}}\]
\[N(t) = 144g \cdot \left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right)\]
\[N(t) = 144g \cdot \frac{1}{8}\]
Now, multiply 144g by 1/8 to find the remaining amount:
\[N(t) = \frac{144g}{8} = 18g\]
So, after 1.719 x 10^4 years, only 18g of the 144g sample of carbon-14 will remain.