126,720 inches are in 2.0 miles
Answer:
115.2 °C since melting point is an intensive property
Step-by-step explanation:
The melting point of a substance does not depend on how much you have.
For example, the melting point of water is 0 °C, whether it is an ice cube from the refrigerator or in the frozen pond outside.
The freezing point of a substance is an <em>intensive property</em>.
Thus, the melting point of 100 g of sulfur is 115.2 °C because melting point in an intensive property.
Answer:
(a) 7.11x10⁻⁴ M/s
(b) 2.56 mol.L⁻¹.h⁻¹
Explanation:
(a) The reaction is:
O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)
The reaction rate of equation (1) is given by:
(2)
<u>We have:</u>
k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹
[O₃]₀ = 2.35x10⁻⁶ M
[NO]₀ = 7.74x10⁻⁵ M
Hence, to find the inital reacion rate we will use equation (2):
Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s
(b) The number of moles of NO₂(g) produced per hour per liter of air is:
t = 1 h
V = 1 L
![\frac{\Delta[NO_{2}]}{\Delta t} = rate](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20rate)
![\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%2A%5Cfrac%7B3600%20s%7D%7B1%20h%7D%20%3D%202.56%20mol.L%5E%7B-1%7D.h%7B-1%7D)
Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹
I hope it helps you!
Answer:
56.28 g
Explanation:
First change the grams of oxygen to moles.
(50.00 g)/(32.00 g/mol) = 1.5625 mol O₂
You have to use stoichiometry for the next part. Looking at the equation, you can see that for every 2 moles of H₂O, 1 mole of O₂ is produced. Convert from moles of O₂ to moles of H₂O using this relation.
(1.5625 mol O₂) × (2 mol H₂O/1 mol O₂) = 3.125 mol H₂O
Now convert moles of H₂O to grams.
(3.125 mol) × (18.01 g/mol) = 56.28125 g
Convert to significant figures.
56.28125 ≈ 56.28
Explanation:
While atomic radii DECREASE across a Period, a row of the Period Table, from left to right as we face the Table, atomic radii INCREASE down a Group, a column of the Periodic Table, due to the shielding of the valence electrons by the closed valence shell(s) that intervenes between the nucleus and the valence, outermost electrons.