<span>Answer:
Some metals have the ability to form differently charged ions. For example, iron can form
2
+
or
3
+
ions. If you simply gave the name iron chloride, you would not know which charge the iron ion possessed.
A Roman numeral is to indicate the charge of the iron.
Iron (
II
) means the iron has a
2
+
charge
Iron (
III
) means that the iron has a
3
+
charge
So, iron (
II
) oxide would have a chemical formula of
FeO
.
(The oxide ion has a
2
â’
charge to balance the
2
+
of the iron to form a neutral compound.)
Iron (
III
) oxide would have a chemical formula of
Fe
2
O
3
(Here you need to find the common multiple of 6, so two iron ions with a
3
+
charge will balance the charge of three oxide ions with a
2
+
charge.)</span>
<span><span>Yes.
An element that is highly electronegative pulls more on the electrons
in a bond, such as oxygen in H20. This creates a polar bond, where
there is a small negative charge on the oxygen, and a small positive
charge in between the hydrogens.
</span>Credit goes to "Erin M" answered on yahoo answers a decade ago.
</span>
20600Cal
Explanation:
Given parameters:
Mass of water = 319.5g
Initial temperature = 35.7°C
Final temperature = 100°C
Unknown:
Calories needed to heat the water = ?
Solution:
The calories is the amount of heat added to the water. This can be determined using;
H = m c Ф
c = specific heat capacity of water = 4.186J/g°C
H is the amount of heat
Ф is the change in temperature
H = m c (Ф₂ - Ф₁)
H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J
Now;
1kilocalorie = 4184J
85996.56J to kCal; 
= 20.6kCal = 20600Cal
learn more:
Specific heat brainly.com/question/3032746
#learnwithBrainly
Explanation:
you look like a mohammed. im good, hbu?
In preparing diluted solutions from concentrated solutions we can use the following formula
c1v1 = c2v2
c1 and v1 are the concentration and volume of the concentrated solution respectively
c2 and v2 are the concentrations and volume of the diluted solution respectively
Substituting these values ,
20 mL x 1.0 M = C x 60 mL
C = 0.33 M
The concentration of the resulting diluted solutions is 0.33 M
