One mole of a substance contains 6.02 × 10∧23 particles. Thus we first convert 89.2 g to moles. 1 mole of sodium contains 23 g
Hence 89.2 g = 89.2 / 23 g = 3.878 moles
Therefore, 3.878 × 6.02×10∧23 particles= 23.346 × 10∧23 particles
Hence 89.2 g of sodium contains 2.335 ×10∧24 particles
Answer:
41.9 g
Explanation:
We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity
m: mass
ΔT: change in temperature
If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.
According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.
Qw + Qs = 0
Qw = -Qs
cw × mw × ΔTw = -cs × ms × ΔTs
(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)
ms = 41.9 g
Ok it’s 38 so I hope that helps ok
Answer:
The energy released will be -94.56 kJ or -94.6 kJ.
Explanation:
The molar mass of methane is 16g/mol
The given reaction is:
the enthalpy of reaction is given as ΔH = -890.0 kJ
This means that when one mole of methane undergoes combustion it gives this much of energy.
Now as given that the amount of methane combusted = 1.70g
The energy released will be:
Answer:
The correct answer is option b, that is, 2.1 M Na₃PO₄.
Explanation:
The solution with the largest concentration of ions will possess the highest conductivity.
a) 3.0 M NaCl
NaCl ⇔ Na⁺ + Cl⁻
Here the total number of ions is 2, therefore, the concentration of ions is 3.0 × 2 = 6.0 M
b) 2.1 M Na₃PO₄
Na₃PO₄ ⇔ 3 Na⁺ + PO₄³⁻
Here the total number of ions i 4. Therefore, the concentration of ions is
2.1 × 4 = 8.4 M.
c) 2.4 M CaCl₂
CaCl₂ ⇔ Ca²⁺ + 2Cl⁻
The total number of ions is 3. Therefore, the concentration of ions is
2.4 × 3 = 7.2 M
d) 3.2 M NH₄NO₃
NH₄NO₃ ⇔ NH₄⁺ + NO₃⁻
The total number of ions is 2. The concentration of ions will be,
3.2 × 2 = 6.4 M
Hence, the highest conductivity will be of 2.1 M Na₃PO₄.
