Answer:
Transition Element
Explanation:
Transition elements are defined as those elements which can form at least one stable ion and has partially filled d-orbitals. They are also characterized by forming complex compounds and having different oxidation states for a single metal element.
Transition metals are present between the metals and the non metals in the periodic table occupying groups from 3 to 12. There general electronic configuration is as follow,
(n-1)d
¹⁻¹⁰ns
¹⁻²
The general configuration shows that for a given metal, the d sublevel will be in lower energy level as compared to corresponding s sublevel. For example,
Scandium is present in fourth period hence, its s sublevel is present in 4rth energy level so its d sublevel will be present in 3rd energy level respectively.
Hence, we can conclude that for transition metals the electron are present in highest occupied s sublevel and a nearby d sublevel
.
Answer:
<h2>(40/111)×100</h2><h2>Here is a bonus Answer✍️✍️</h2><h2>Percent composition by element</h2><h2>Element Symbol Mass Percent</h2><h2>Chlorine Cl 63.888%</h2><h2>Calcium Ca 36.112</h2>
The correct option is B.
Single replacement reaction is a type of chemical reaction, in which one element is substituted for another element in a compound. There are usually two reactants, one is a pure element and the other one is an aqueous compound. The products formed is usually made up of another pure element and a new compound. In the option B, you will notice that, in the reactant part, Mg is a pure element while in the product side, H2 is a pure element.
NaCl= sodium chloride or salt
KBr= potassium bromide
Answer:
1. 15.71 g CO2
2. 38.19 % of efficiency
Explanation:
According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:
For CO:
C = 12.01 g/mol
O = 16 g/mol
CO = 28.01 g/mol
(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO
For CO2:
C = 12.01 g/mol
O = 16 x 2 = 32 g/mol
CO2 = 44.01 g/mol
We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:
(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2
We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:
(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2
Now for the efficiency question:
From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,
(6g / 15.71g) x 100 = 38.19 % of efficiency
