C is the answer.
It says on the third picture that Bohr refined Rutherford's model by giving distinct orbits for the electrons with distinct radii.
I believe they are called Energy Levels or Energy Orbitals.
Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L
Answer:
The sample of lead has a volume of 11.1 cm³
Explanation:
<u>Step 1:</u> Data given
x cm³ lead has a density of 11.3 g/cm³
it has the same mass as 330cm³ of a piece of redwood with density 0.38g/cm³
<u>Step 2</u>: Calculate mass of the piece of redwood
Density = mass/volume
mass = density * volume
Mass of the piece of redwood = 0.38 g/cm³ * 330cm³ = 125.4 grams
Since the sample of lead has the same mass, it also has a mass of 125.4 grams
<u>Step 3</u>: Calculate volume of the lead
Density = mass/ volume
Volume = mass/ density
Volume of lead = 125.4g / 11.3g/cm³ = 11.097 cm³≈11.1 cm³
The sample of lead has a volume of 11.1 cm³