Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 3.16 L of chlorine were consumed at STP.
Be sure your answer has the correct number of significant digits.
Answer: Thus volume of carbon tetrachloride that would be produced is 0.788 L
Explanation:
According to ideal gas equation:
P = pressure of gas = 1 atm (at STP)
V = Volume of gas = 3.16 L
n = number of moles = ?
R = gas constant =
T =temperature =
According to stoichiometry:
4 moles of chlorine produces = 1 mole of carbon tetrachloride
Thus 0.141 moles of methane produces = 
moles of carbon tetrachloride
volume of carbon tetrachloride =
Thus volume of carbon tetrachloride that would be produced is 0.788 L
the oxygen atom
Explanation:
Water is a molecular compound consisting of polar molecules that have a bent shape. The oxygen atom acquires a partial negative charge while the hydrogen atom acquires a partial positive charge.
Answer:
3.47 ×10^-10
Explanation:
The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)
A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.
E°cell = E°cathode - E°anode
E°cathode = -0.41 V
E°anode = -0.13 V
E°cell = -0.41 -(-0.13) = -0.28 V
From
E°cell = 0.0592/n log K
n= 2, K= the unknown
-0.28 = 0.0592/2 log K
log K = -0.28/0.0296
log K = -9.4595
K = Antilog ( -9.4595)
K= 3.47 ×10^-10
Answer:
they are 10 but I'll be listing three
Explanation:
cirrus,cirrocumulus and cirrostratus
I hope this helps....good luck with the rest...again, I hope this helps
C₆H₆ is benzene which has a molar mass of 78 g/mol. When benzene is burned, the reaction is called combustion. The heat produced in this reaction is called the heat of combustion. For benzene, the heat of combustion is -3271 kJ/mol.
Heat of benzene = (8.7 g)(1 mol/78 g)(-3271 kJ/mol) = -364.84 kJ
By conservation of energy,
Heat of benzene = - Heat of water
where
Heat of Water = mCp(Tf - T₀)
where Cp for water is 4.187 kJ/kg·°C
Thus,
-364.84 kJ = -(5691 g)(1 kg/1000 g)(4.187 kJ/kg·°C)(Tf - 21)
<em>Tf = 36.31°C</em>
