Chemistry, 15.07.2020 20:01 rgilliam3002

The rate at which a reaction is proceeding at a specific point in time is called the:

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Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

Answer from: Quest

i think the answer is terrae but not 100 percent sure. ▓▬▬ §

Answer from: Quest

Radon-222 or rn-222 has a half-life of 3.82 days. to determine the fraction of an original sample of rn-222 that remains unchanged after 7.646 days, we calculate the amount of rn-222 remaining after 7.646 days and divide it with the original amount. using 1 gram as a basis and using half-life calculation methods, we can determine that 0.249 or 0.25 grams of rn-222 will remain, and thus the fraction remaining is 0.25 or 1/4.

Answer from: Quest

Answer is: value of q is 5·10⁻⁶m.chemical reaction: mg²⁺(aq) + co₃²⁻(aq) → mgco₃(s).c(mg²⁺) = 2.5·10⁻³ m; concentration of magnesium cations.c(co₃²⁻) = 2.00·10⁻³ m; concentration of carbonate anions.q = c(mg²⁺) · c(co₃²⁻).q = 2.5·10⁻³ m · 2.00·10⁻³ m.q = 5·10⁻⁶m.q = 0.000005 m.

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