The formula of ionic compound made from sodium (Na) and iodine (l) is
NaI
<em><u>Explanation</u></em>
- An ionic compound is are compounds made up of ions.
- The ions are atoms or ions that loses electron to form cation or gain electron to form anion.
- Metal for this case loses electron while non metal gain electrons
- In Nal Na <em>( a Metal</em>) loses one electron while I <em>(non metal) </em>gain one electron to form an ionic compound.
5/8 because it is slightly larger then the button being attached.
Answer:
P = 58.52 atm
Explanation:
Given data:
Mass of sample = 32.0 g
Pressure of sample = ?
Volume of gas = 850 cm³
Temperature of gas = 30°C
Solution:
Number of moles of gas:
Number of moles = mass/molar mass
Number of moles = 32.0 g/ 16 g/mol
Number of moles = 2 mol
Pressure of gas:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
30+273 = 303 K
850 cm³ × 1L /1000 cm³ = 0.85 L
by putting values,
P× 0.85 L = 2 mol × 0.0821 atm.L/ mol.K × 303 K
P = 49.75 atm.L/ 0.85 L
P = 58.52 atm
Answer:
There are three possible chemical equations for the combustion of sulfur:
- 2S (s) + O₂ (g) → 2SO (g)
- 2S (s) + 3O₂ (g) → 2SO₃ (g)
Explanation:
<em>Combustion</em> is a reaction with oxygen. The products of the reaction are oxides, and energy is released in the form of heat and light.
<em>Sulfur</em> iis a nonmetal, so the oxide formed is a nonmetal oxide.
The most common oxidation numbers of sulfur are -2, + 2, + 4, and + 6.
The combination of sulfur with oxygen may be only with the positive oxidation numbers (+2, + 4, and +6).
Then you have three different equations for sulfur combustion:
<u>1) Oxidation number +2:</u>
Which when balanced is: 2S(g) + O₂(g) → 2SO(g)
<u>2) Oxitation number +4:</u>
That equation is already balanced.
<u>3) Oxidation number +6:</u>
Which when balanced is: 2S(s) + 3O₂(g) → 2SO₃(g)