Answer:
8.28 MPa
Explanation:
From the question given above, the following data were obtained:
Radius (r) = 2×10¯³ m
Force applied (F) = 104 N
Pressure (P) =?
Next, we shall determine the area of the nail (i.e circle). This can be obtained as follow:
Radius (r) = 2×10¯³ m
Area (A) of circle =?
Pi (π) = 3.14
A = πr²
A = 3.14 × (2×10¯³)²
A = 3.14 × 4×10¯⁶
A = 1.256×10¯⁵ m²
Next, we shall determine the pressure. This can be obtained as follow:
Force applied (F) = 104 N
Area (A) = 1.256×10¯⁵ m²
Pressure (P) =?
P = F / A
P = 104 / 1.256×10¯⁵
P = 8280254.78 Nm¯²
Finally, we shall convert 8280254.78 Nm¯² to MPa. This can be obtained as follow:
1 Nm¯² = 1×10¯⁶ MPa
Therefore,
8280254.78 Nm¯² = 8280254.78 Nm¯² × 1×10¯⁶ MPa / 1 Nm¯²
8280254.78 Nm¯² = 8.28 MPa
Thus, the pressure exerted on the wall is 8.28 MPa
Answer:
ΔG = 18KJ/mol
Explanation:
Given data:
ΔS = 0.09 Kj/mol.K
ΔH = 27 KJ/mol
Temperature = 100 K
ΔG = ?
Solution:
Formula:
ΔG = ΔH - TΔS
ΔH = enthalpy
ΔS = entropy
by putting values,
ΔG = 27 KJ/mol - 100K(0.09 Kj/mol.K)
ΔG = 27 KJ/mol - 9 KJ/mol
ΔG = 18KJ/mol
Answer:
The atom that loses the electrons becomes a positively charged ion, while the one that gains them becomes a negatively charged ion
Maybe to not get rained on.
Hahhahahaha I ain't sure tho
Answer:
- 0.99 °C ≅ - 1.0 °C.
Explanation:
- We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.
<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>