Q = mcΔθ
67.5 = m x 0.45 x (28.5 - 21.5)
M = 67.5 / 3.15
= 21.4 g
Answer:
0.0344 moles and 1.93g.
Explanation:
Molarity is defined as the ratio between moles of a solute (In this case, KOH), and the volume. With molarity and volume we can solve the moles of solute. With moles of solute we can find mass of the solute as follows:
<em>Moles KOH:</em>
15.2mL = 0.0152L * (2.26mol / L) = 0.0344moles
<em>Mass KOH:</em>
0.0344 moles * (56.11g/mol) = 1.93g of KOH
Squeezing just the juices out of the orange, like with your hand or whatever you use, is a physical change. yes :)
An orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy.
Explanation:
The only true statement from the given options is that "an orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy." Inner orbitals which are also known to contain core electrons feels the bulk of the nuclear pull on them compared to the outermost orbitals containing the valence electrons.
- The nuclear pull is the effect of the nucleus pulling and attracting the electrons in orbitals.
- This pull is stronger for inner orbitals and weak on the outer ones.
- The outer orbitals are said to be well shielded from the pull of the nuclear charge.
- Also, based on the quantum theory, electrons in the outer orbitals have higher energies because they occupy orbitals at having higher energy value.
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Answer:
114 kPa
Explanation:
By Bernoulli's equation when a fluid flows steadily through a pipe:
P + ρ*g*y + v² = constant in the pipe, where P is the pressure, ρ is the density of the fluid, g is the gravity acceleration (9.8 m/s²), y is the high, and v the velocity.
By the continuity equation, the liquid flow must be constant in the pipe, and then:
A1*v1 = A2*v2
Where A is the area, v is the velocity, 1 is the point 1, and 2 the point 2 in the pipe. The are is the circle area: π*(d/2)². So:
π*(0.105/2)²*9.91 = π*(0.167/2)²*v2
0.007v2 = 0.027
v2 = 3.9 m/s
Then:
P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²
ρ*g*y1 - ρ*g*y2 + v1² - v2² = P2 - P1
ρ*g*Δy + v1² - v2² = ΔP
ΔP = 1290*9.8*9.01 + 9.91² - 3.9²
ΔP = 113,987.42 Pa
ΔP = 114 kPa
