A scale and a ruler. The scale to measure the mass, and a ruler to measure the volume.
According to this formula:
Q = M*C*ΔT
when we have M ( the mass of water) = 200 g
and C ( specific heat capacity ) of water = 4.18 J/gC
ΔT (the difference in temperature) = Tf - Ti
= 100 - 24
= 76°C
So by substitution:
Q = 200 g * 4.18 J/gC * 76 °C
= 63536 J
∴ the amount of heat which be added and absorbed to raise the temp from 24°C to 100°C is = 63536 J
Tin and Hydrofluoric Acid reacts as shown,
<span> Sn + 2 HF </span>→ SnF₂<span> + H</span>₂
According to Equation,
40.02 g (2 Moles) HF Required = <span>1 Mole of Sn for complete Reaction
So,
40 g of HF will require = X Moles of Sn
Solving for X,
X = (40 g </span>× 1 Mole) ÷ 40.02 g
X = 0.999 Moles ≈ 1 Mole
Result:
40 g of HF requires 1 Mole of Tin (Sn) for complete Reaction to produce SnF₂ and H₂.
Answer: This is because, the stem symbolises the bronchioles of the lung through which oxygen is transfered to the lungs. The grabes represent the lungs, left and right respectively.
✩✩✩
Answer:
54.5%
Explanation:
The percentage composition of oxygen in C₆H₈O₆ can be obtained as follow:
Molar mass of C₆H₈O₆ = (12×6) + (8×1) + (16×6)
= 72 + 8 + 96
= 176 g/mol
Next, there are 6 oxygen atoms in C₆H₈O₆. Therefore the mass of oxygen in C₆H₈O₆ is:
Mass of Oxygen = 16 × 6 = 96 g
Finally, we shall determine the percentage composition of oxygen in C₆H₈O₆ as follow:
Percentage of oxygen =
Mass of Oxygen/mass of C₆H₈O₆ × 100
Percentage of oxygen = 96 / 176 × 100
Percentage of oxygen = 54.5%
Thus, the percentage composition of oxygen in C₆H₈O₆ is 54.5%.
