Answer: D) 36.0 km/hr, downstream
Explanation:
For downstream motion of the boat, the actual velocity of the boat is the sum of velocity of the water current and the velocity of the boat due to pushed by wind.
Velocity of water current, v = 15 km/h
Velocity of the boat going downstream, u = 21 km/h
Actual velocity of the boat = v'
v' = v + u
⇒v' = 15 km/h + 21 km/h
⇒u = 21 km/h +15 km/h = 36.0 km/h downstream
Thus, the correct answer is option D.
Answer : 0.026 moles of oxygen are in the lung
Explanation :
We can solve the given question using ideal gas law.
The equation is given below.

We have been given P = 21.1 kPa
Let us convert pressure from kPa to atm unit.
The conversion factor used here is 1 atm = 101.3 kPa.

V = 3.0 L
T = 295 K
R = 0.0821 L-atm/mol K
Let us rearrange the equation to solve for n.



0.026 moles of oxygen are in the lung
Acetaldehyde is an organic compound (a compound containing C atoms) composed of a carbonyl group. On the other hand, a carbonyl group is a functional group containing C = O. The hybrid orbitals of a compound determines the number pi and s orbitals in the electronic configuration. For a single bond, there are two s orbitals. For double bonds, on the other hand, the number of s orbital bond is 1 while the number of pi bonds is 2. For triple bonds, there are three pi bonds present in the cloud.
Thus for a c = O bond, the atomic orbital configuration is sp3 containing 1 s orbital and 2 pi bonds.
Answer:
104.84 moles
Explanation:
Given data:
Moles of Boron produced = ?
Mass of B₂O₃ = 3650 g
Solution:
Chemical equation:
6K + B₂O₃ → 3K₂O + 2B
Number of moles of B₂O₃:
Number of moles = mass/ molar mass
Number of moles = 3650 g/ 69.63 g/mol
Number of moles = 52.42 mol
Now we will compare the moles of B₂O₃ with B from balance chemical equation:
B₂O₃ : B
1 : 2
52.42 : 2×52.42 = 104.84
Thus from 3650 g of B₂O₃ 104.84 moles of boron will produced.
Answer:
The answer to your question is: 1, 2, 1, 2
Explanation:
1 Fe(s) + 2 Na⁺(aq) → 1 Fe²⁺(aq) + 2 Na(s)
Fe⁰ - 2e⁻ ⇒ Fe⁺² Oxidases
Na⁺ + 1 e⁻ ⇒ Na⁰ Reduces
1 x ( 1 Fe⁰ ⇒ 1 Fe⁺²) Interchange number of
2 x ( 2Na⁺ ⇒ 2 Na⁰ ) electrons