Answer:
3.1 x 10⁻²¹ Nm
Explanation:
When placed in an external electric filed, an electric dipole experiences a torque. and this torque is represented mathematically with the equation:
torque (τ) = dipole moment vector (P) x electric field vector (E)
τ = P. E . sin θ
where θ is the angle between the water molecule and the electric field, which in this case is 90° (because this is where the torque is maximum)
τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . sin90
τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . 1
solve for τ
τ = 3.1 x 10⁻²¹ Nm
the maximum possible torque on the water molecule is therefore 3.1 x 10⁻²¹ Nm
Answer:
0.0303 Liters
Explanation:
Given:
Mass of the potassium hydrogen phosphate = 0.2352
Molarity of the HNO₃ Solution = 0.08892 M
Now,
From the reaction it can be observed that 1 mol of potassium hydrogen phosphate reacts with 2 mol of HNO₃
The number of moles of 0.2352 g of potassium hydrogen phosphate
= Mass / Molar mass
also,
Molar mass of potassium hydrogen phosphate
= 2 × (39.09) + 1 + 30.97 + 4 × 16 = 174.15 g / mol
Number of moles = 0.2352 / 174.15 = 0.00135 moles
thus,
The number of moles of HNO₃ required for 0.00135 moles
= 2 × 0.00135 mol of HNO₃
= 0.0027 mol of HNO₃
Now,
Molarity = Number of Moles / Volume
thus,
for 0.0027 mol of HNO₃, we have
0.08892 = 0.0027 / Volume
or
Volume = 0.0303 Liters
Answer:
248 mL
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by water (Qw) and the heat released by the coffee (Qc) is zero.
Qw + Qc = 0
Qw = -Qc [1]
We can calculate each heat using the following expression.
Q = c × m × ΔT
where,
- ΔT: change in the temperature
163 mL of coffee with a density of 0.997 g/mL have a mass of:
163 mL × 0.997 g/mL = 163 g
From [1]
Qw = -Qc
cw × mw × ΔTw = -cc × mc × ΔTc
mw × ΔTw = -mc × ΔTc
mw × (54.0°C-25.0°C) = -163 g × (54.0°C-97.9°C)
mw × 29.0°C = 163 g × 43.9°C
mw = 247 g
The volume corresponding to 247 g of water is:
247 g × (1 mL/0.997 g) = 248 mL
16 grams I think it might be it
