All categories

3 years ago

When does the electron shown release the greatest amount of energy as it moves from one level to another?

Chemistry

1 answer:

ELEN [110]3 years ago

8 0

S1 to GS as the amount of energy required to remove the electron is greatest when it is closest to the nucleus. think about two magents. does it take more energy to spereate them when they are far apart or when they are very close together? it's the same idea with electrons. the more energy required to remove, the more energy will be emmitted.

You might be interested in

Empirical formula can i have help

EastWind [94]

Answer:

what kind of class u in like dam

Explanation:

3 0

3 years ago

What is the molarity of a solution composed

Neporo4naja [7]

Answer:

0.0845 M

Explanation:

First we convert 4.27 grams of potassium iodide into moles, using its molar mass:

Molar Mass of KI = 166 g/mol

4.27 g ÷ 166 g/mol = 0.0257 mol

Now we calculate the molarity of the solution, using the number of moles and the given volume:

Molarity = moles / liters

Molarity = 0.0257 mol / 0.304 L = 0.0845 M

7 0

3 years ago

Please help me with my 8th grade chemistry, I'm really confused

saveliy_v [14]

Answer:

The answer would be C, in both chemical and physical change, because everything is made up of matter. So if you have a chemical or physical change, that doesent change the total mass of your product ever.

Explanation:

7 0

3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

Two solids of identical mass, A and B, are analyzed using identical calorimeters. Each calorimeter contains the same amount of w

DIA [1.3K]

Answer:

Specific heat of solid A is greater than specific heat of solid B.

Explanation:

In the calorimeter, as the temperature is increasing, the vibrational kinetic energy will increase and this means that additional amount of energy will be needed to increase the temperature by the same value. Therefore, we can conclude that specific heat increases as temperature increases.

Now, we are told that the final temperature of solid A's calorimeter is higher than that of B.

This means from our definition earlier, Solid A will have a higher specific heat that solid B.

5 0

2 years ago