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lidiya [134]
3 years ago
15

Write the condensed electron configurations for the Ca atom. Express your answer in condensed form as a series of orbitals. For

example, [He]2s22p2 should be entered as [He]2s^22p^2.
Chemistry
1 answer:
umka21 [38]3 years ago
6 0

Answer:

[Ar] 4s²

Explanation:

Ca is the symbol for Calcium. It is the 20th element and it has 20 electrons.

The full electronic configuration for calcium is given as;

1s²2s²2p⁶3s²3p⁶4s²

The condensed electronic configuration is given as;

[Ar] 4s²

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Compute the boiling point of this solution:
ryzh [129]

Answer:

The boiling point of the solution is 101,42ºC

Explanation:

Formula for boiling point elevation is:

ΔT = Kb · molality . i

Water boils at 100ºC, at 1 atm.

NiSO4 ----> Ni2+  +  SO4-2  (assume 100% ionization, the value of i, is 2)

Tº boiling point solution - Tº boiling point = Ebulloscopic constant . molality . Van't Hoff factor.

Tº bp solution - 100ºC = 0,512 ºC/molal . molality . 2

Molality means moles of solute in 1kg of solvent (1000g)

Molar mass NiSO4 = 154,75 g/m

Mass/ Molar mass = Moles --> 21,6 g / 154,75 g/m = 0,139 moles

These moles are in 100 g of H2O, so we have to make a rule of three to find, the moles in 1000g

100 g _________0,139 moles

1000 g _______ 1,39 moles (molality)

Tº bp solution - 100ºC = 0,512 ºC/molal . 1,39molal . 2

Tº bp solution = (0,512 ºC/molal . 1,39molal . 2) + 100ºC

Tº bp solution = 101,42ºC

4 0
3 years ago
What was ernest rutherfords experiment
ohaa [14]

Answer: he shot tiny alpha particles throug a piece of gold foil.

Explanation:

6 0
3 years ago
Any help would be appreciated. Confused.
masya89 [10]

Answer:

q(problem 1) = 25,050 joules;  q(problem 2) = 4.52 x 10⁶ joules

Explanation:

To understand these type problems one needs to go through a simple set of calculations relating to the 'HEATING CURVE OF WATER'. That is, consider the following problem ...

=> Calculate the total amount of heat needed to convert 10g ice at -10°C to steam at 110°C. Given are the following constants:

Heat of fusion (ΔHₓ) = 80 cal/gram

Heat of vaporization (ΔHv) = 540 cal/gram

specific heat of ice [c(i)] = 0.50 cal/gram·°C

specific heat of water [c(w)] = 1.00 cal/gram·°C

specific heat of steam [c(s)] = 0.48 cal/gram·°C

Now, the problem calculates the heat flow in each of five (5) phase transition regions based on the heating curve of water (see attached graph below this post) ...   Note two types of regions (1) regions of increasing slopes use q = mcΔT and (2) regions of zero slopes use q = m·ΔH.

q(warming ice) =  m·c(i)·ΔT = (10g)(0.50 cal/g°C)(10°C) = 50 cal

q(melting) = m·ΔHₓ = (10g)(80cal/g) 800 cal

q(warming water) = m·c(w)·ΔT = (10g)(1.00 cal/g°C)(100°C) = 1000 cal

q(evaporation of water) =  m·ΔHv = (10g)(540cal/g) = 5400 cal

q(heating steam) = m·c(s)·ΔT = (10g)(0.48 cal/g°C)(10°C) = 48 cal

Q(total) = ∑q = (50 + 800 + 1000 + 5400 + 48) = 7298 cals. => to convert to joules, multiply by 4.184 j/cal => q = 7298 cals x 4.184 j/cal = 30,534 joules = 30.5 Kj.

Now, for the problems in your post ... they represent fragments of the above problem. All you need to do is decide if the problem contains a temperature change (use q = m·c·ΔT) or does NOT contain a temperature change (use q = m·ΔH).    

Problem 1: Given Heat of Fusion of Water = 334 j/g, determine heat needed to melt 75g ice.

Since this is a phase transition (melting), NO temperature change occurs; use q = m·ΔHₓ = (75g)(334 j/g) = 25,050 joules.

Problem 2: Given Heat of Vaporization = 2260 j/g; determine the amount of heat needed to boil to vapor 2 Liters water ( = 2000 grams water ).

Since this is a phase transition (boiling = evaporation), NO temperature change occurs; use q = m·ΔHf = (2000g)(2260 j/g) = 4,520,000 joules = 4.52 x 10⁶ joules.

Problems containing a temperature change:

NOTE: A specific temperature change will be evident in the context of problems containing temperature change => use q = m·c·ΔT. Such is associated with the increasing slope regions of the heating curve.  Good luck on your efforts. Doc :-)

5 0
3 years ago
Sabendo que A Densidade do Metal De Chumbo e 11,3/cm3
a_sh-v [17]
Sabendo que A Densidade do Metal De Chumbo e 11,3/cm3 means Knowing that the density of lead Metal and 11,3/cm3.
3 0
3 years ago
1. Given the hypotenuse of a 45°, 45°, 90° triangle is 8, find the length of the legs. Leave your answers in radical form if nee
Alex17521 [72]

Answer:

5.66

Explanation:

sin = opposite/hypotenuse

8 sin = opposite

8 sin 45 = 5.66

5 0
3 years ago
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