Answer:
Potassium nitrate react with aluminum to produce potassium aluminate, aluminum oxide and nitrogen. This reaction takes place at a temperature near 400°C.
Explanation:
6KNO3 + 10Al → 6KAlO2 + 2Al2O3 + 3N2
Gold has a heavy enough nucleus that its electrons must travel at speeds nearing the speed of light to prevent them from falling into the nucleus. This relativistic effect applies to those orbitals that have appreciable density at the nucleus, such as s and p orbitals. These relativistic electrons gain mass and as a consequence, their orbits contract. As these s and (to some degree) p orbits are contracted, the other electrons in d and f orbitals are better screened from the nucleus and their orbitals actually expand.
Since the 6s orbital with one electron is contracted, this electron is more tightly bound to the nucleus and less available for bonding with other atoms. The 4f and 5d orbitals expand, but can't be involved in bond formation since they are completely filled. This is why gold is relatively unreactive.
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Answer:
Explanation:
Most of this ancient space rubble can be found orbiting the Sun between Mars and Jupiter within the main asteroid belt. Asteroids range in size from Vesta – the largest at about 329 miles (530 kilometers) in diameter – to bodies that are less than 33 feet (10 meters) across.
The reaction, Fe2O3 + 3CO------> 2Fe + 3CO2 is an oxidation-reduction reaction.
An oxidation-reduction reaction is a reaction in which there is a change in oxidation number from left to right in the reaction. This is because, a specie is oxidized and another specie is reduced.
In the reaction; Fe2O3 + 3CO------> 2Fe + 3CO2, we can see that the oxidation number of iron decreased from +3 on the left hand side to zero on the right hand side. The oxidation number of carbon was increased from + 2 to +4.
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25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.
The equation of the reaction is;
NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)
We can use the titration formula;
CAVA/CBVB = NA/NB
CA= concentration of acid
VA = volume of acid
CB = concentration of base
VB = volume of base
NA = number of moles of acid
NB = number of moles of base
CB = 0.010 M
VB = 50.0 ml
CA = 0.50 M
VA = ?
NA = 1
NB = 1
Substituting values;
CAVANB = CBVBNA
VA = 0.010 × 50.0 × 1/ 0.50 × 1
VA = 1 ml
Since the total volume of acid used is 1 ml and each drop contains 0.040 ml
The number of drops required is 1ml/0.040 ml = 25 drops
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