Oxygen = 16
Iron = 55.8
16 x 27.6% = 4.4 /4 = 1.1
55.8 x 72.4% = 40.4 /4 = 10.1
1 oxygen and 10 iron, so Fe10 O
The engine displacement in cubic nanometer (nm³) is 8.785×10²⁴ nm³
<h3>Data obtained from the question </h3>
- Displacement in in³ = 536 in³
- Displacement in nm³ =?
<h3>Conversion scale </h3>
1 in³ = 1.639×10²² nm³
With the above convesion scale we can obtain the displacement in nm³
<h3>How to determine the displacement in nm³</h3>
1 in³ = 1.639×10²² nm³
Therefore,
536 in³ = 536 × 1.639×10²²
536 in³ = 8.785×10²⁴ nm³
Thus, the displacement in nm³ is 8.785×10²⁴ nm³
Learn more about conversion:
brainly.com/question/2139943
amount of electrons on their outer shell: 1 electron.
Answer:
c) B, C
Explanation:
NaOH(aq) + HBr(aq) -----> NaBr(aq) +H2O(l)
1) concentration of acid CA= 0.05 M
Concentration of base CB= 0.1 M
Volume of acid VA= 25.00ml
Volume of base VB= unknown
Number of moles of acid NA= 1
Number of moles of base NB= 1
CAVA/CBVB = NA/NB
CAVANB =CBVBNA
VB= CAVANB/CB NB
VB= 0.05 × 25 × 1/ 0.1 ×1
VB= 12.5 ML
2.
Answer:
The net force on a vehicle that is accelerating at a rate of 1.5
The net force on a vehicle that is accelerating at a rate of 1.5 m/s2 is 1,800 newtons.
Explanation: