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Answer:
Molarity = 0.08 M
Explanation:
Given data:
Mass of sodium carbonate = 10.6 g
Volume of water = 1.25 L
Molarity of solution = ?
Solution:
First of all we will calculate the moles of solute.
Number of moles = mass/molar mass
Number of moles = 10.6 g/ 106 g/mol
Number of moles = 0.1 mol
Formula:
Molarity = moles of solute / volume of solution in L
Now we will put the values in formula.
Molarity = 0.1 mol / 1.25 L
Molarity = 0.08 M
Answer:
15.3 %
Explanation:
Step 1: Given data
- Mass of the sample (ms): 230 g
- Mass of carbon (mC); 136.6 g
- Mass of hydrogen (mH): 26.4 g
- Mass of nitrogen (mN): 31.8 g
Step 2: Calculate the mass of oxygen (mO)
The mass of the sample is equal to the sum of the masses of all the elements.
ms = mC + mH + mN + mO
mO = ms - mC - mH - mN
mO = 230 g - 136.6 g - 26.4 g - 31.8 g
mO = 35.2 g
Step 3: Calculate the mass percent of oxygen
%O = (mO / ms) × 100% = (35.2 g / 230 g) × 100% = 15.3 %
This question can be simply solved by using heat formula,
Q = mCΔT
Q = heat energy (J)
m = Mass (kg)
C = Specific heat capacity (J / kg K)
ΔT = Temperature change (K)
when water freezes, it produces ice at 0°C (273 K)
hence the temperature change is 25 K (298 K - 273 K)
C for water is 4186 J / kg K or 4.186 J / g K
By applying the equation,
Q = 456 g x 4.186 J / g K x 25 K
= 47720.4 J
= 47.72 kJ
hence 47.72 kJ of heat energy should be removed.
Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp =
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp =
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp =
0.0104 =
0.0104 + - 200 = 0
Resolving the second degree equation:
=
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are = 98.7 MPa and P(N₂O₄) = 101.3 MPa