(1) False, lots of energy is actually produced from nuclear fuel, if we didn't get much then we probably wouldn't use it
(2) False, its burning coal that contributes to acid rain, since it contains sulfur
(3) False again, we can control the reaction with aptly named control rods, which are typically made of boron, to absorb some of the neutrons flying around in the chain reaction
(4) True, radioactive waste is very difficult to dispose of, and is also very dangerous. Sources of radiation can remain so for millions of years
I hope it helps you ❤️❤️❤️❤️
Answer:
Explanation: When solutions of potassium iodide and lead nitrate are combined?
The lead nitrate solution contains particles (ions) of lead, and the potassium iodide solution contains particles of iodide. When the solutions mix, the lead particles and iodide particles combine and create two new compounds, a yellow solid called lead iodide and a white solid called potassium nitrate. Chemical Equation Balancer Pb(NO3)2 + KI = KNO3 + PbI2. Potassium iodide and lead(II) nitrate are combined and undergo a double replacement reaction. Potassium iodide reacts with lead(II) nitrate and produces lead(II) iodide and potassium nitrate. Potassium nitrate is water soluble. The reaction is an example of a metathesis reaction, which involves the exchange of ions between the Pb(NO3)2 and KI. The Pb+2 ends up going after the I- resulting in the formation of PbI2, and the K+ ends up combining with the NO3- forming KNO3. NO3- All nitrates are soluble. ... (Many acid phosphates are soluble.)
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
Vol.250 before its to much pressure
