Answer:
Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M
Explanation:
4A + 3B ------> C + 2D
In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be
0.08 × 3 = 0.24 M
Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.
From the chemical reaction,
4 moles of A gives 1 mole of C
0.24 M of reacted A will form (0.24 × 1)/4 M of C
Amount of C formed at the end of the 3s interval = 0.06 M
If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M
The sets of characteristics describe what we know about
the inner planets is that have rocky
surfaces, no rings, and few or no moons. The answer is letter C. Also, they are
smaller in sizes compared to the outer planets. The rest of the choices are
characteristics of the outer planets.
The mass of water is equal to the combined mass of hydrogen and oxygen.
<h3>What is Mass?</h3>
This is defined as the quantity of matter in a physical body. The electrolysis reaction of water can be seen below:
2 H₂O ---> 2 H₂ + O₂
We can deduce that 36 grams of H₂O dissociated to give 4 grams of H₂ and 32 grams of O₂ which option D was chosen as the appropriate choice.
Read more about Mass here brainly.com/question/25121535
Answer:
972.3 Torr
Explanation:
P2=P1V1/V2
You can check this by knowing that P and V at constant T have an an inverse relationship. Hence, this is correct.
According to Henry's law, solubility of solution is directly proportional to partial pressure thus,
Solubility at pressure 3.08 atm is 72.5/100, solubility at pressure 8 atm should be calculated.
Putting the values in equation:
On rearranging,
Therefore, solubility will be 1.88 mg of 
gas in 1 g of water or, 188 mg of tex]N_{2}[/tex] gas in 100 g of water.
