Since there is one carbon with 4 Fluorines attached to it, and both compounds are no metals, we use the covalent method for naming,
Here we ignore the prefix for the first element if it is 1. Mono. Then pay attention to the second one, it would be tetra, because tetra means 4. Here there are 4 fluorines.
Drop ine and place ide
CF4 = carbon tetrafluoride.
Answer:
The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg
Explanation:
Heat gain by ice = Heat lost by water
Thus,
Heat of fusion + 
Where, negative sign signifies heat loss
Or,
Heat of fusion + 
Heat of fusion = 334 J/g
Heat of fusion of ice with mass x = 334x J/g
For ice:
Mass = x g
Initial temperature = 0 °C
Final temperature = 6 °C
Specific heat of ice = 1.996 J/g°C
For water:
Volume = 353 mL
Density of water = 1.0 g/mL
So, mass of water = 353 g
Initial temperature = 26 °C
Final temperature = 6 °C
Specific heat of water = 4.186 J/g°C
So,


345.976x = 29553.16
x = 85.4197 kg
Thus,
<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>
Answer:
Put the water out in the sun and the water will evaporate.
Explanation:
Option B: 2, 8, 18, 32.
2 in the K shell, 8 in L shell, 18 in M shell and 32 in N shell.
Mass of Hydrogen gas required to react : 0.936 g
<h3>Further explanation</h3>
Reaction on Nitrogen gas and Hydrogen gas to produce Ammonia gas
N₂ (g) + 3 H₂ (g) ⇒ 2 NH₃ (g)
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol
so mol Nitrogen for 3.5 L at STP :

From the equation, mol ratio of N₂ : H₂ = 1 : 3, so mol H₂ :

then mass of Hydrogen(MW= 2 g/mol) :
