PH is a simplified observation of the acidity of a solution, defined as the -log [H+] (negative logarithm of the hydrogen concentration).
-log[10^-4] = 4
The pH of this solution is 4.
You are given
200 grams of H2O(s) at an initial temperature of 0°C. you are also given the
final temperature of water after heating at 65°C. You are required to get the
total amount of heat to melt the sample. The specific heat capacity, cp, of
water is 4.186 J/g-°C. Let us say that T1 = 0°C and T2 = 65°C. The equation for
heat, Q, is
Q = m(cp)(T2-T1)
Q = 200g(4.186
J/g-°C )(65°C - 0°C)
<u>Q =
54,418J</u>
A fish is a living thing and a rock is not
According to the kinetic theory, the mean free path is the average distance a single atom or molecule of an element or compound travels with respect with the other atoms during a collision. The greater the mean free path, the more ideal the behavior of a gas molecule is because intermolecular forces are minimum. To understand which factors affect the mean free path, the equation is written below.
l = μ/P * √(πkT/2m), where
l is the mean free path
μ is the viscosity of the fluid
P is the pressure
k is the Boltzmann's constant
T is the absolute temperature
m is the molar mass
So, here are the general effects of the factors on the mean free path:
Mean free path increases when:
1. The fluid is viscous (↑μ)
2. At low pressures (↓P)
3. At high temperatures (↑T)
4. Very light masses (↓m)
The opposite is also true for when the mean free path decreases. Factors that are not found here have little or no effect.
Answer:
47.1 L.
Explanation:
- To solve these problems, we can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 234.0 kPa/101.325 ≅ 2.31 atm) .
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 3.9 mol).
R is the general gas constant (R = 0.082 L.atm/mol.K),
T is the temperature of the gas in K (T = 67.0 °C + 273.15 = 340.15 K).
The volume of the gas = nRT/P.
<em>∴ V = nRT/P </em>= (3.9 mol)(0.082 L.atm/mol.K)(340.15 K)/(2.31 atm) = <em>47.1 L.</em>
