I think it is C, because a covalent bond is a distribution of 2 atoms to 1 electron, meaning they are sharing and not exchanging, and the electronegravity would be above 1.7
Answer:
20 amu
Explanation:
An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.
All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.
Atomic mass = Number of protons + number of neutrons
Atomic number = Number of electrons or number of protons.
In given question it is stated that atom has 11 electrons and -1 charge it means this atom has 12 electrons in neutral state.
Thus it has 12 protons because number of electrons and protons are always equal.
Atomic mass of given atom:
Atomic mass = Number of protons + number of neutrons
Atomic mass = 12 + 8 = 20 amu
Answer:
Limiting reactant: O2
grams NO2 produced = 230.276 g NO2
grams of NO unused = 26.67 gNO
Explanation:
2NO + O2 --> 2NO2
Step 1: Determine the molar ratio NO:O2
molar ratio NO:O2 = 5.895: 2.503 = 2.35
stoichiometric molar ratio NO:O2 = 2:1
So, O2 is the limiting reactant.
Step2: Determine the grams of NO2:
?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2
Step 3: Determine the amount of excess reagent unreacted
moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted
moles NO unreacted = total moles NO - moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted
mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted
Answer:
Q = 233.42 J
Explanation:
Given data:
Mass of lead = 175 g
Initial temperature = 125.0°C
Final temperature = 22.0°C
Specific heat capacity of lead = 0.01295 J/g.°C
Heat absorbed by water = ?
Solution:
Heat absorbed by water is actually the heat lost by the metal.
Thus, we will calculate the heat lost by metal.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 22.0°C - 125.0°C
ΔT = -103°C
Q = 175 g × 0.01295 J/g.°C×-103°C
Q = -233.42 J
Heat absorbed by the water is 233.42 J.
Answer : The excess reactant in the combustion of methane in opem atmosphere is 
molecule.
Solution : Given,
Mass of methane = 23 g
Molar mass of methane = 16.04 g/mole
The Net balanced chemical reaction for combustion of methane is,
First we have to calculate the moles of methane.
= 
= 1.434 moles
From the above chemical reaction, we conclude that
1 mole of methane react with the 2 moles of oxygen
and 1.434 moles of methane react to give 
moles of oxygen
The Moles of oxygen = 2.868 moles
Now we conclude that the moles of oxygen are more than the moles of methane.
Therefore, the excess reactant in the combustion of methane in open atmosphere is molecule.
