To determine what elements are represented by the electron configuration given above, we need to know the sum of the exponents of each term or subshell involved in the configuration as this represent the atomic number of the element.
Atomic Number Element
<span>1s2 2s2 2p6: 2 + 2 + 6 = 10 neon
1s2 2s2 2p6 3s2 3p3: </span>2 + 2 + 6 + 2 + 3 = <span>15 phosphorus
1s2 2s2 2p6 3s2 3p6 4s1: </span>2 + 2 + 6 + 2 + 6+1 = <span>19 potassium
1s2 2s2 2p6 3s2 3p6 4s2 3d8: 20 + 8 = 28 nickel
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d3: 30 + 6 + 2 +3 = 41 niobium</span>
An r-selected species reproduces much faster than K-selected species.
r-selected species focuses on maturing and reproducing quickly. r-selected species will probably reproduce when the water supply is there for the short period of time; thus, increasing the chance of the r-selected species of surviving.
K-selected species, on the other hand, focus on raising their young and reproduce later. Since the K-selected species take long to mature before reproducing, water supply may run out before they have a chance of fully maturing; thus, K-selected species have a lower chance of survival.
Hope this helps.
If you need anything more, feel free to comment! Have an awesome day! :)
~Collinjun0827, Junior Moderator
Generally speaking, organic molecules tend to dissolve in solvents that have similar physical properties. A good rule of thumb is that "like dissolves like". Meaning, polar compounds can dissolve polar compounds and nonpolar compounds can dissolve nonpolar compounds.
To apply this to the current problem, we are told that the brushes are being cleaned with vegetable oil or mineral oil. In this case, the oils are used as solvents. In order for these solvents to be effective, the compounds they are trying to dissolve must be similar in structure and properties to other oils. Therefore, vegetable oil or mineral oil will be most effective in removing oil-based paints, as these will have the similar properties needed to dissolve in the oil solvents.
Predict what will be observed in each experiment below. Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in 500 mL of hot water (70 degree C). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature. He prepares batch B by dissolving sugar in 500 mL of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature
a. It is likely that more rock candy will be formed in batch A.
b. It is likely that less rock candy will be formed in batch A.
c. It is likely that no rock candy will be formed in either batch.
d. I need more information to predict which batch is more likely to form rock candy.
Answer: Option A
Explanation:
More rock candy will be formed in the batch A because it is dissolved in hot water and less rock candy will be formed in batch B because the water is not hot.
Formation of the candies require hot water as the solubility of sugar is more in hot water as compared to normal water.
The sugar will be dissolved in water until the time all the space is filled sugar molecules.
Hence, the correct answer is Option A.
1) Balanced chemical reaction:
2H2 + O2 -> 2H20
Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O
2) Reactant quantities converted to moles
H2: 5.00 g / 2 g/mol = 2.5 mol
O2: 50.0 g / 32 g/mol = 1.5625 mol
Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).
3) Products
H2 totally consumed -> 0 mol at the end
O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end
H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.
Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol
4) Pressure
Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K
p = nRT/V = 7.9 atm
