Because each element has an exactly defined line emission spectrum, scientists are able to identify them by the color of flame they produce.
Answer:
[NH₃] = 14.7 mol/L
Explanation:
28 wt% is a type of concentration that indicates that 28 g of ammonia is contained in 100 g of solution.
Let's determine the amount of ammonia:
28 g . 1 mol / 17.03g = 1.64 moles of NH₃
You need to consider that, when you have density's data it is always referred to solution:
Mass of solution is 100 g, let's find out the volume
0.90 g/mL = 100 g /V
V = 100 g / 0.90mL/g → 111.1 mL
We convert the volume to L → 111.1 mL . 1 L/1000mL = 0.1111 L
mol/L = 1.64 mol/0.1111L → 14.7 M
mol/L = M → molarity a sort of concentration that indicates the moles of solute in 1L of solution
Answer:
(a) The rate of formation of K2O is 0.12 M/s.
The rate of formation of N2 is also 0.12 M/s
(b) The rate of decomposition of KNO3 is 0.24 M/s
Explanation:
(a) From the equation of reaction, the mole ratio of K2O to O2 is 2:5.
Rate of formation of O2 is 0.3 M/s
Therefore, rate of formation of K2O = (2×0.3/5) = 0.12 M/s
Also from the equation of reaction, mole ratio of N2 to O2 is 2:5.
Rate of formation of N2 = (2×0.3/5) = 0.12 M/s
(b) From the equation of reaction, mole ratio of KNO3 to O2 is 4:5.
Therefore, rate of decomposition of KNO3 = (4×0.3/5) = 0.24 M/s
Answer is: B. C(s) + 2S(s) + 89.4 kJ → CS2(l).
Missing question:
A. C(s) + 2S(s) → CS2(l) + 89.4 kJ.
B. C(s) + 2S(s) + 89.4 kJ → CS2(l).
C. C(s) + 2S(s) + 89.4 kJ → CS2(l) + 89.4 kJ.
D. C(s) + 2S(s) → CS2(l).
Because enthalpy of
the system is greater that zero, this is endothermic reaction (<span>chemical reaction that
absorbs more energy than it releases)</span>, heat is included as a reactant.
