Elements within the same group of the periodic table behave similarly because they have the same number of ?

1 answer:

4 0

The elements in each group have the same number of electrons in the outer orbital. Or also called valence electrons. Khan academy has a great video online explaining why this happens. (It only happens for main group elements). Here is a link (sorry you can’t click it in Brainly) https://www.khanacademy.org/science/chemistry/periodic-table/copy-of-periodic-table-of-elements/v/periodic-table-valence-electrons. Feel free to message me for a better explanation, I would explain now but I’m not sure how much you know about this. If you know how to write an electron configuration you can see how all the electron configurations for the same group (not the transitional metals only the main groups) have the same number of valence electrons. I hope that helped, sorry I was vague about the explanation :)

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What volume of a 0.35M solution of sodium chloride could be formed with 25 grams of NaCl?

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2 years ago

If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4

statuscvo [17]

Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

Step 1: Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

Step 2: Calculate moles of Ca^2+

Moles of Ca^2+ = Molarity Ca^2+ * volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

Step 3: Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Calculate Q

Q = [Ca2+] [SO42-]

[Ca2+]= 0.050 M [O42-]

Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

Step 7: Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

Step 8: Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

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3 years ago

When 1.14 g of octane (molar mass = 114 g/mol) reacts with excess oxygen in a constant volume calorimeter, the temperature of th

maxonik [38]

I can't answer this question without knowing what the specific heat capacity of the calorimeter is. Luckily, I found a similar problem from another website which is shown in the attached picture.

Q = nCpΔT
Q = (1.14 g)(1 mol/114 g)(6.97 kJ/kmol·°C)(10°C)(1000 mol/1 kmol)
Q = +6970 kJ