Answer: option C) gather information and identify stakeholders
Explanation:
The sales, distribution and advertisement of alcoholic beverages requires information on consumer protection, health risks, and environmental factors. And Garrett would simply get such from the relevant stakeholders like regulatory agencies.
Thus, Garrett should first gather information and identify stakeholders
<span>0.48 grams.
Not a well worded question since it's assuming I know the reactions. But I'll assume that since there's just 1 atom of copper per molecule of Cu(NO3)2, that the reaction will result in 1 atom of copper per molecule of Cu(NO3)2 used. With that in mind, we will have 0.010 l * 0.75 mol/l = 0.0075 moles of copper produced.
To convert the amount in moles, multiply by the atomic weight of copper, which is 63.546 g/mol. So
0.0075 mol * 63.546 g/mol = 0.476595 g.
Round the results to 2 significant figures, giving 0.48 grams.</span>
Answer:
C3H7OH → C3H6 + H20
Explanation:
If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.
Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.
The Correct Answer Is <span>Both Processes Produce Energy From Mass. </span>
Answer:
4.66 x 10^8 yr
Explanation:
The age of the rock can be calculated using the equation:
ln (N/N₀) = - kt where N is the quantiy of radioisotope decayed and N₀ is the initially quantity present of the radioisotope; k is the decay constant, and t is the time.
Now from the data , we have 78 argon-40 atoms for every 22 potassium-40 atoms, we can deduce that originally we had 22 + 78 = 100 atoms of potassium-40 so this is our N₀.
When we look at the equation, we see that k is unknown, but we can calculate it from the half-life which is given by the equation:
k = 0.693/ t half-life = 0.693/ 1.3 x 10⁹ yr = 5.33 x 10⁻¹⁰ yr⁻¹
Now we are in position to answer the question.
ln ( 78/100 ) = - (5.33 x 10⁻¹⁰ yr⁻¹ ) t
- 0.249 = - 5.33 x 10⁻¹⁰ yr⁻¹ t
0.249/ 5.33 x 10⁻¹⁰ yr⁻¹ = t
4.66 x 10^8 yr
