Answer:
1.The electonic configuration of elements and their position in the periodic table are related to each other, From the electronic configuration of the elements, we can determine the period and the group to which the element belongs
Let's consider, sodium with atomic number 11 and k, l, and M shells have 2,8,and 1 electrons. since, there are 3 principal energy levels so we concluded sodium belongs to third period M Shell(valance shell) has only 1 electrons. so sodium belongs to group 1.
2. Entire D-block elements are known as Transition Elements.
3. Group 17 is the halogen group.
4. Main group of elements are...... 1,2, and 13 through 18.
5. Group 18 are the noble gas elements .
12. a). Smaller
b). Increases
c). More reactive
d). Softer
7. a). k › Ca › Ge › Br › Kr
b). Ra › Ba › Sr › Ca › Mg › Be
9. a). Ca(calcium) ion is smaller.
b). Cl(chlorine) atom is smaller.
c). Mg(magnesium) atom is smaller.
10. a). F(fluorine)
b). Sr(strontium)
c). Pb(lead)
d). At(Astatine)
Sodium(Na) is the limiting reagent.
<h3>What is Limiting reagent?</h3>
The reactant that is totally consumed during a reaction, or the limiting reagent, decides when the process comes to an end. The precise quantity of reactant required to react with another element may be estimated from the reaction stoichiometry.
How do you identify a limiting reagent?
The limiting reactant is the one that is consumed first and sets a limit on the quantity of product(s) that can be produced. Calculate how many moles of each reactant are present and contrast this ratio with the mole ratio of the reactants in the balanced chemical equation to get the limiting reactant.
Start by writing the balanced chemical equation that describes this reaction
Notice that the reaction consumes 2 moles of sodium metal for every 1 mole of chlorine gas that takes part in the reaction and produces 2 moles of sodium chloride.
now we can see that we have 3 moles of sodium and 3 moles of chlorine, according to question. so, we can say that sodium is the limiting reagent in the given situation.
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Air is mainly composed of N2 (78%), O2 (21%) and other trace gases. Now, the total pressure of air is the sum of the partial pressures of the constituent gases. The partial pressure of each gas, for example say O2, can be expressed as:
p(O2) = mole fraction of O2 * P(total, air) ----(1)
Thus, the partial pressure is directly proportional to the total pressure. If we consider a sealed container then, as the temperature of air increases so will its pressure. Based on equation (1) an increase in the pressure of air should also increase the partial pressure of oxygen.
C + O2= CO2
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
