Answer:
heating the reactant mix
Explanation:
Heat can result in to speed the reaction process.
hope it helps!
Answer:
Explanation:
The metric system is a system of measurement that uses the meter, liter, and gram as base units of length (distance), capacity (volume), and weight (mass) respectively.
To measure smaller or larger quantities, we use units derived from the metric units
metric-system
The given figure shows the arrangement of the metric units, which are smaller or bigger than the base unit.
The units to the right of the base unit are smaller than the base unit. As we move to the right, each unit is 10 times smaller or one-tenth of the unit to its left. So, a ‘deci’ means one-tenth of the base unit, ‘centi’ is one-tenth of ‘deci’ or one-hundredth of the base unit and ‘milli’ is one-tenth of ‘centi’ or one-thousandth of the base unit.
The units to the left of the base unit are bigger than the base unit. As we move to the left, each unit is 10 times greater than the unit to its right. So, a ‘deca’ means ten times of the base unit, ‘hecto’ is ten times of ‘deca’ or hundred times of the base unit and ‘killo’ is ten times of ‘hecto’ or thousand times of the base unit.
The molar concentration of the nitric acid solution was 0.6666 mol/L.
<em>Balanced equation</em>: KOH + HNO_3 → KNO_3 + H_2O
<em>Moles of KOH</em>: 32.33 mL KOH × (1.031 mmol KOH /1 mL KOH)
= 33.33 mmol KOH
<em>Moles of HNO_3</em>: 33.33 mmol KOH× (1 mmol HNO_3/1 mmol KOH)
= 33.33 mmol HNO_3
<em>Concentration of KOH</em>: <em>c </em>= "moles"/"litres" = 33.33 mmol/50.00 mL
= 0.6666 mol/L
Oxidizing agent is that which is reduced and the reducing agent is that which is oxidized. Reduced is when the charged is decreased and oxidized when the charge is increased.
(1) 2Na + 2H2O(l) --> 2NaOH(aq) + H2(g)
The charge of Na in the reactant is 0 and the charge of Na in the NaOH is +1. Na is oxidized. Hence, it is the reducing agent.
The charge of H in H2O is +1 while that in H2 is 0. H is reduced. Hence, it is the oxidizing agent.
(2) C(s) + O2(g) --> CO2(g)
The charge of C in the reactant side is 0 and that its charge in CO2 is +4. C is oxidized. Hence, it is the reducing agent.
The charge of O in O2 is 0 while in CO2, its charge is -2. O is reduced. Hence, it is the oxidizing agent.
(3) 2MnO⁻⁴ + SO2 + 2H2O --> 2Mn²⁺ + 5SO2⁻⁴ 4H⁺
The charge of Mn in MnO⁻⁴ is 4+ while its charge in Mn²⁺ is 2+. Mn is reduced. Hence, it is the oxidizing agent.
The charged of S in SO2 is -4 while its charge in SO₂⁻⁴ is 0. S is oxidized. Hence, it is the reducing agent.
The NADPH contributions Electrons to the cycle
