Answer:
The temperature associated with this radiation is 0.014K.
Explanation:
If we assume that the astronomical object behaves as a black body, the relation between its <em>wavelength</em> and <em>temperature</em> is given by Wien's displacement law.
where,
λmax is the wavelength at the peak of emission
b is Wien's displacement constant (2.89×10⁻³ m⋅K)
T is the absolute temperature
For a wavelength of 21 cm,
Answer:
3.925 mol.
Explanation:
- From the balanced equation:
<em>2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(s) + O₂(g)
,</em>
It is clear that 2 moles of Na₂O₂ react with 2 moles of H₂O to produce 4 moles of NaOH and 1 mole of O₂
.
<em>Using cross multiplication:</em>
4 moles of NaOH produced with → 1 mole of O₂
.
15.7 moles of NaOH produced with → ??? mole of O₂
.
<em>∴ The no. of moles of O₂ made =</em> (1 mole)(15.7 mole)/(4 mole) = <em>3.925 mol.</em>
Answer:
C.
Explanation:
Because it decreases from October trough december
when heat gained = heat lost
when AL is lost heat and water gain heat
∴ (M*C*ΔT)AL = (M*C*ΔT) water
when M(Al) is the mass of Al= 225g
C(Al) is the specific heat of Al = 0.9
ΔT(Al) = (125.5 - Tf)
and Mw is mass of water = 500g
Cw is the specific heat of water = 4.81
ΔT = (Tf - 22.5)
so by substitution:
∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)
∴Tf = 30.5 °C
