Answer:
the difference in electronegativity is so large (2.04) that the bonding electrons spend almost all their time on the nitrogen atom.
Explanation:
Because calcium loses 2 electrons to become Ca2+, and nitrogen gains 3 electrons to become N3−, you need two calcium atoms and three nitrogen atoms in order to form a neutral compound.
Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and
respectively.
Explanation :
Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).
As we are given that:

Fuel value = 
Molar mass of pentane = 72 g/mol
Fuel value = 
Fuel value = 49.09 kJ/g
Now we have to calculate the fuel density of pentane.
Fuel density = Fuel value × Density
Fuel density = (49.09 kJ/g) × (0.626g/mL)
Fuel density = 30.73 kJ/mL = 
Thus, the fuel density of pentane is 
The [OH⁻] of the solution is 5.37×10⁵ M.
<h3 /><h3>What is pOH?</h3>
This is the negative logarithm to base 10 of hydroxy ion [OH⁻] concentration.
To calculate the hydroxy ion [OH⁻] concentration we use the formula below.
Note:
- pOH = 14-pH
- pOH = 14-9.77
- pOH = 4.27
Formula:
- [OH⁻] = 1/
................. Equation 1
Given:
Substitute the value into equation 1
- [OH⁻] = 1/

- [OH⁻] = 5.37×10⁵
Hence, The [OH⁻] of the solution is 5.37×10⁵ M.
Learn more about hydroxy ion concentration here: brainly.com/question/17090407
Answer:
I think cold front if not than its c
Answer:
41.9 g
Explanation:
We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity
m: mass
ΔT: change in temperature
If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.
According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.
Qw + Qs = 0
Qw = -Qs
cw × mw × ΔTw = -cs × ms × ΔTs
(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)
ms = 41.9 g