Molar mass of N2 = 28
Moles of N2 = 25 / 28 = 0.89
So, moles of NH3 produce = 2 x 0.89 = 1.78
Note: H2 is in excess. so no need to care about it.
Answer:
1(a) N = 3
(b) N = 0
(c) N = 5
(d) N = -2
(2) Molecular formula for benzene is C6H6
Explanation:
1(a) N02 1-
N + (2×-2) = -1
N-4 = -1
N = -1+4 = 3
(b) N2
2(N) = 0
N = 0/2 = 0
(c) NO2Cl
N + ( 2×-2) + (-1) = 0
N - 4 - 1 = 0
N - 5 = 0
N = 0+5 = 5
(d) N2H4
2(N) + (4×1) = 0
2N + 4 = 0
2N = 0 - 4 = -4
N = -4/2 = -2
(2) Molcular mass of benzene = 78g/mole = (6×12g of carbon) + (6×1g of hydrogen) = 72+6 = 78g/mole
Therefore, molecular formula for benzene is C6H6
Answer:
with the molecular formula C3H5(ONO2)3, has a high nitrogen content (18.5 percent) and contains sufficient oxygen atoms to oxidize the carbon and hydrogen atoms while nitrogen is being liberated, so that it is one of the most powerful explosives known.
Explanation:
NTG reduces preload via venous dilation, and achieves modest afterload reduction via arterial dilation. These effects result in decreased myocardial oxygen demand. In addition, NTG induces coronary vasodilation, thereby increasing oxygen delivery.
Answer:
3,964 years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of the element is 5,730 years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of the sample ([A₀] = 100%).
[A] is the remaining concentration of the sample ([A] = 61.9%).
∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.
