A.) The amount of sunlight reflected onto the moon from the sun
Just like the Earth, half of the Moon is lit by the Sun while the other half is in darkness. The phases we see result from the angle the Moon makes with the Sun as viewed from Earth. The lit side does not always face the Earth, as the Moon circles the Earth, the amount of the lit side we see changes.
<em>≈ 0,27 g/cm³</em>
<em>36.5 l = 36.5 dm³ = 36500 cm³</em>
<em>10 kg = 10000 g</em>
<em>d = m/V</em>
<em>= 10000g/36500cm³</em>
<em>≈ 0,27 g/cm³</em>
Chemical is the answer to the question, but nuclear could also be a valid one since it is nearly impossible to reverse that.
Answer:
<h2>Because energy is conserved, the kinetic energy of a block at the bottom of a frictionless ramp is equal to the gravitational potential energy of the block at the top of the ramp. ... Therefore, the block's final velocity depends on the height of the ramp but not the steepness.</h2>
<span>Answer: 0.00649M
The question is incomplete,
</span>
<span>You are told that the first ionization of the sulfuric acid is complete and the second ionization of the sulfuric acid has a constant Ka₂ = 0.012
</span>
<span>
With that you can solve the question following these steps"
</span>
<span>1) First ionization:
</span>
<span>
H₂SO₄(aq) --> H⁺ (aq) + HSO₄⁻ (aq)
Under the fully ionization assumption the concentration of HSO4- is the same of the acid = 0.01 M
2) Second ionization
</span>
<span>HSO₄⁻ (aq) ⇄ H⁺ + SO₄²⁻ with a Ka₂ = 0.012
</span>
<span>Do the mass balance:
</span>
<span><span> HSO₄⁻ (aq) H⁺ SO₄²⁻</span>
</span>
<span /><span /><span> 0.01 M - x x x
</span><span>Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]</span>
<span /><span>
=> Ka₂ = (x²) / (0.01 - x) = 0.012
</span><span />
<span>3) Solve the equation:
</span><span>x² = 0.012(0.01 - x) = 0.00012 - 0.012x</span>
<span /><span>
x² + 0.012x - 0.0012 = 0
</span><span />
<span>Using the quadratic formula: x = 0.00649
</span><span />
<span>So, the requested concentratioN is [SO₄²⁻] = 0.00649M</span>
