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
Answer:
Answer is option C
Explanation:
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Answer is: there are 3.011·10²³ atoms of calcium.
n(Ca) = 0.50 mol; amount of substance(calcium).
Na = 6.022·10²³ 1/mol; Avogadro's constant or number.
N(Ca) = n(Ca) · Na.
N(Ca) = 0.50 mol · 6.022·10²³ 1/mol.
N(Ca) = 3.011·10²³; number of calcium atoms.
The mole is an SI unit which measures the number of particles in substance. One mole is equal to <span><span>6.022</span></span>·<span><span><span>10</span></span></span>²³<span> atoms.</span>
Answer:
the answer is D. Because the 1 atm pressure of water is 40.65 or 40.7.
Answer:
CH4
Explanation:
The number of moles of carbon and hydrogen has been given as follows:
C = 0.300 mol
H = 1.20 mol
Next, we divide each mole value by the smallest (0.300)
C = 0.300 ÷ 0.300 = 1
H = 1.20 ÷ 0.300 = 4
The empirical ratio of Carbon and Hydrogen is 1:4, hence, the empirical formula is CH4
