The Lewis structures for formate ion and carbon dioxide are shown in the attached pictures. As you can see, the formate ion consists of one C=O bond and one C-O bond, while carbon dioxide has two C=O bonds. Double bonds are weaker, hence, longer compared to single bonds. Therefore, the C-O bonds in the formate ion is shorter than those in carbon dioxide.
Answer is: <span>volume of oxygen is 14.7 liters.
</span>Balanced chemical
reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.<span>
m(</span>C₃H₈-propane) = 5.53 g.
n(C₃H₈) = m(C₃H₈) ÷ M(C₃H₈).
n(C₃H₈) = 5.53 g ÷ 44.1 g/mol.
n(C₃H₈) = 0.125 mol.
From chemical reaction: n(C₃H₈) : n(O₂) = 1 : 5.
n(O₂) = 0.625 mol.
T = 25° = 298.15K.
p = 1.04 atm.
<span>R = 0.08206
L·atm/mol·K.
</span>Ideal gas law: p·V = n·R·T.
V(O₂) = n·R·T / p.
V(O₂) = 0.625 mol · 0.08206 L·atm/mol·K · 298.15 K / 1.04 atm.
V(O₂) = 14.7 L.
It's oxidation number is related to its charge.
Since this atom's charge is very negative in the 1st molecule, it is more electronegative. Which means the other atom is less electronegative.
In the 2nd molecule, it has a positive charge, so THIS atom is less electronegative. Or, you can say the OTHER atom is more electronegative.
Answer:
mass of hydrogen collected is 0.016 gram
Explanation:
Given values:
For calculating mass we have to find the number of moles first
Ideal gas equation PV =nRT
Volume= 195 ml
Pressure: 753 torr =0.99 atm
Temperature: 25+273= 298 Kelvin
Ideal gas constant R= 0.0821 Latm/molK
Number of moles n= ?
So n= PV/RT
Adding the values
n= = (753 torr)(1 atm/760 torr)(195 mL)(1 L/1000 mL)/(0.0821 L·atm/mol·K)(273 K)
n = 0.007897 moles of H₂
Now mass of hydrogen collected = number of moles x Molar mass of H₂
= 0.007897 x 2
= 0.0157 g H₂
Now mass of hydrogen collected is 0.016 gram (rounding the amount)
We can use the ideal gas law equation to find the volume of the balloon.
PV = nRT
where
P - pressure - 0.992 atm x 101 325 Pa/atm = 100 514 Pa
V - volume
n - number of moles - 8.80 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 25 °C + 273 = 298 K
Substituting these values in the equation
100 514 Pa x V = 8.80 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
V = 217 L
volume of balloon is 217 L
