Answer:
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
Explanation:
The half-life time = the time required for a quantity to reduce to half of its initial value. Half of it's value = 50%.
To calculate the half-life time we use the following equation:
[At]=[Ai]*e^(-kt)
with [At] = Concentration at time t
with [Ai] = initial concentration
with k = rate constant
with t = time
We want to know the half-life time = the time needed to have 50% of it's initial value
50 = 100 *e^(-8.7 *10^-3 s^- * t)
50/100 = e^(-8.7 *10^-3 s^-1 * t)
ln (0.5) = 8.7 *10^-3 s^-1 *t
t= ln (0.5) / -8.7 *10^-3 = 79.67 seconds
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
Answer:
D.
Explanation:
The reaction is losing potential energy, which means that the reaction is losing that energy as heat. Exothermic is the loss of energy. Therefore it will be D.
The ER takes up a lot of space in some cells<span>. The endoplasmic reticulum may be “rough” or “smooth.” ER that has no attached ribosomes is called smooth endoplasmic reticulum. </span>
I think it’s A I’m not 100% sure but I mean it’s worth a try
Answer:
<em>3.27·10²³ atoms of O</em>
Explanation:
To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.
The chemical formula for sodium sulfate is <em>Na₂SO₄, </em>and its molar mass is approximately 142.05
.
We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.
19.3g <em>Na₂SO₄</em> · 
· 
·
After doing the math for this dimensional analysis, you should get a quantity of approximately <em>3.27·10²³ atoms of O</em>.
