For a Forty miles above the earth's surface, the temperature at 250k and the pressure at only 0.20 torr, the density of air at this altitude is mathematically given as
d=3.732*10{-4}g/l
<h3>What is the density of air at this altitude?</h3>
Generally, the equation for the ideal gas is mathematically given as
PV=(m/mw)RT
Therefore
d=m/v
d=26,664*20/83.14*250
d=0.37152
d=3.732*10{-4}g/l
In conclusion, the density
d=3.732*10{-4}g/l
Read more about Temperature
brainly.com/question/13439286
Answer:
(a) a carbon–carbon triple bond is shorter than a carbon–carbon single bond.
The given statement is true.This because bond order in of (C-C) bond in triple bond is 3 whereas the bond order of (C-C) bond in single bond is 1. Higher the bond order shorter will be the bond length.
(b) There are exactly six bonding electrons in the oxygen molecule.
The given statement is false. This is because bond order present in oxygen molecule is 2 which means there are 4 bonding electrons in the oxygen molecule.
(c) the (C-O) bond in carbon monoxide is longer than the (C-O) bond in carbon dioxide.
The given statement is false.This because bond order in of (C-O) bond in CO is 3 whereas in the bond order of (C-O) bond is 2. Higher the bond order shorter will be the bond length.
(d) The (O-O) bond in ozone is shorter than the (O-O) bond in oxygen.
The given statement is false. This is because bond order in of (O-O) bond in ozone molecule is 1.5 whereas in the bond order of (O-O) bond is 2. Higher the bond order shorter will be the bond length.
(e) The more electro-negative the atom, the more bonds it makes to other atoms.
The given statement is false. This is because atom with more electro-negative character. Since, they are highly electron rich atoms so they will tend to form less number of bonds with other atom.
Answer:
190.92 J
Explanation:
By using the formula Q = mcΔT
Q = 1.5g × 4.18 J/g°C × 30.45°C
= 190.92 J
<span>U(0), which is the internal energy referenced tothe value, U(0) at absolute zero (T = 0 K).</span>
The volume of CO₂ at STP = 2.8 L
<h3>Further explanation</h3>
Given
21 gram of sodium hydrogen carbonate-NaHCO₃
Required
Volume of CO₂
Solution
The decomposition of sodium bicarbonate into sodium carbonate, carbon dioxide, and water :
<em>
2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)</em>
mol of NaHCO₃ :
= mass : MW NaHCO₃
= 21 g : 84 g/mol
= 0.25
From the equation, mol ratio of NaHCO₃(s) :CO₂(g) = 2 : 1, so mol CO₂ :
= 1/2 x mol NaHCO₃
= 1/2 x 0.25
= 0.125
At STP, 1 mol gas = 22.4 L, so for 0.125 mol :
= 0.125 x 22.4 L
= 2.8 L