Answer:
D, because he discovered electrons
Explanation:
Answer:
Option e.
Explanation:
Molarity is the concentration that indicates moles of solute in 1 L of solution.
We have another concentration, percent by mass.
Percent by mass indicates mass of solute in 100 g of solution.
Our solute is HNO₃, our solvent is water.
17.5 g of nitric acid is the mass of solute. We can convert them to moles:
17.5 g . 1mol / 63g = 0.278 moles
We do not have volume of solution. We assume the mass is 100 g because the percent by mass but we need density to state the volume.
Density = Mass / Volume
Mass / Density = Volume
Once we have the volume, we need to be sure the units is in L, to determine molarity
M = mol /L
The intended sense is that of a reaction that depends on absorbing heat if it is to proceed. The opposite of an endothermic process is an exothermic process, one that releases "gives out" energy in the form of heat
Answer:
uuuh 4.25191 × 10-21 grams????
Explanation:
That will make a gold-202 nucleus.
<h3>Explanation</h3>
Refer to a periodic table. The atomic number of mercury Hg is 80.
Step One: Bombard the 
with a neutron 
. The neutron will add 1 to the mass number 202 of 
. However, the atomic number will stay the same.
- New mass number: 202 + 1 = 203.
- Atomic number is still 80.
.
Double check the equation:
- Sum of mass number on the left-hand side = 202 + 1 = 203 = Sum of mass number on the right-hand side.
- Sum of atomic number on the left-hand side = 80 = Sum of atomic number on the right-hand side.
Step Two: The 
nucleus loses a proton 
. Both the mass number 203 and the atomic number will decrease by 1.
- New mass number: 203 - 1 = 202.
- New atomic number: 80 - 1 = 79.
Refer to a periodic table. What's the element with atomic number 79? Gold Au.
.
Double check the equation:
- Sum of mass number on the left-hand side = 203 = 202 + 1 = Sum of mass number on the right-hand side.
- Sum of atomic number on the left-hand side = 80 = 79 + 1 = Sum of atomic number on the right-hand side.
A gold-202 nucleus is formed.
