The number of Na atoms in the reactants = 2
<h3>Further explanation</h3>
Every chemical reaction involves compounds that can act as reactants or products
The reactants will react to produce products
The reactants are usually written on the left side of the reaction, while the products on the right side of the reaction
Reaction
2 NaCl + H₂SO₄ → 2 HCl + Na₂SO₄
The element with atomic number 11 = Na(Natrium)
The number of atoms of the element that reacts or is produced can be seen from the reaction coefficient and the number of atoms in the compound
Na in NaCl = 1 atom
From equation, cofficient for NaCl = 2, so number of atoms of Na=2 x 1 = 2
23.9 11.00 would be the wrong answer because I guessed
You will need the equation PV = nRT
P = Pressure in kPa
V = Volume in L
n = moles
R = 8.314 (constant)
T = Temperature in Kelvin
First convert 2.5 atm into kPa:
2.5 X 101.3 = 253.25 kPa
Convert 125 Celsius into Kelvin:
125 + 273 = 398 K
Convert Gallons to Litres:
1.25 X 3.79 = 4.74 L
Plug your values into the equation to solve for n:
(253.25)(4.74) = n(8.314)(398)
n = (253.25)(4.74)/(8.314)(398)
n = 0.362 moles
Now use M = m/n to solve for the mass of O2
M = Molar Mass
M = mass
n= moles
32 = m/(0.362)
m = (32)(0.362)
m = 11.58g
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
