I think the answer would be d but not 100% sure
The metal is lead, Pb.
One unit of the oxide contains one atom of O (16.00 u).
∴ Mass of oxide = 16.00 u O × (100 u MO/7.17 u O) = 223.15 u MO
Mass of M = mass of MO – mass of O = 223.15 u -16.00 u = 207.2 u
The only element with an atomic mass of 207.2 u is lead (Pb) and the formula of the oxide is PbO.
Answer:
C₁₁H₁₂NO₄
Explanation:
In order to determine the empirical formula of doxycycline, we need to follow a series of steps.
Step 1: Determine the centesimal composition
C: 59.5 mg/100 mg × 100% = 59.5%
H: 5.40 mg/100 mg × 100% = 5.40%
N: 6.30 mg/100 mg × 100% = 6.30%
O: 28.8 mg/100 mg × 100% = 28.8%
Step 2: Divide each percentage by the atomic mass of the element
C: 59.5 /12.0 = 4.96
H: 5.40/1.00 = 5.40
N: 6.30/14.0 = 0.450
O: 28.8/16.0 = 1.80
Step 3: Divide all the numbers by the smallest one
C: 4.96/0.450 = 11
H: 5.40/0.450 = 12
N: 0.450/0.450 = 1
O: 1.80/0.450 = 4
The empirical formula of doxycycline is C₁₁H₁₂NO₄
Answer:
Lead(III) Fluoride PbF3 Molecular Weight -- EndMemo.
Explanation:
paki brainly po tysm
<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
