W = AB x F x Cos < AB, F
or just W= AB x F for short
Answer:
The specific heat of sodium is 1,23J/g°C
Explanation:
Using the atomic weight of sodium (23g/mol) and the atomic weight definition, we have that each mole of the substance has 23 grams of sodium.
starting from this, we use the atomic weight of sodium to convert the units from J / mol ° C to J / g ° C
Answer:
No, it is not sufficient
Please find the workings below
Explanation:
Using E = hf
Where;
E = energy of a photon (J)
h = Planck's constant (6.626 × 10^-34 J/s)
f = frequency
However, λ = v/f
f = v/λ
Where; λ = wavelength of light = 325nm = 325 × 10^-9m
v = speed of light (3 × 10^8 m/s)
Hence, E = hv/λ
E = 6.626 × 10^-34 × 3 × 10^8 ÷ 325 × 10^-9
E = 19.878 × 10^-26 ÷ 325 × 10^-9
E = 19.878/325 × 10^ (-26+9)
E = 0.061 × 10^-17
E = 6.1 × 10^-19J
Next, we work out the energy required to dissociate 1 mole of N=N. Since the bond energy is 418 kJ/mol.
E = 418 × 10³ ÷ 6.022 × 10^23
E = 69.412 × 10^(3-23)
E = 69.412 × 10^-20
E = 6.9412 × 10^-19J
6.9412 × 10^-19J is required to break one mole of N=N bond.
Based on the workings above, the photon, which has an energy of 6.1 × 10^-19J is not sufficient to break a N=N bond that has an energy of 6.9412 × 10^-19J
First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
