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exis [7]
3 years ago
11

Determine the concentration of a solution made by dissolving 44.0 g of calcium chloride (CaCl2) in 0.30 L of solution. SHOW YOUR

WORK for credit. Partial credit is awarded! Correct answers with no work will not receive credit. Reminder of the the steps: 1. Identify what you know and what you want, and the units. Known: 0.30 L, 44.0 g CaCl2 Want: molarity (mol /L ) 2. Find the amount of CaCl2 in moles. molar Mass of CaCl2 = 110.98 g/mol. Formula: Moles = grams / molar mass 3. Use dimensional analysis to set up and solve equation. Formula: M = mol solute / L solution
Chemistry
1 answer:
ycow [4]3 years ago
6 0

Answer:

[CaCl₂] = 1.32 M

Explanation:

We know the volume of solution → 0.30 L

We know the mass of solute → 44 g of CaCl₂

Let's convert the mass of solute to moles.

44 g . 1 mol / 110.98 g = 0.396 moles

Molarity (mol/L) → 0.396 mol / 0.3 L  = 1.32 M

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Which property do metalloids share with non metals?
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their ability to conduct heat and electricity.

Explanation:

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Which subatomic particle is located outside the nucleus
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Electrons is located outside the nucleus
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2 years ago
According to Hund's rule of maximum spin multiplicity, how many singly-occupied orbitals are there in the valence shells of the
leva [86]

Answer:

A) carbon  - 2

B) cobalt  - 3

C) sulfur   - 2

D) fluorine   - 1

E) titanium   - 2

F) germanium  - 2

Explanation:

Hund's rule of maximum multiplicity:-

Firstly, every orbital which is present in the sublevel is singly occupied and then the orbital is doubly occupied.  

(A) Carbon.

The electronic configuration is -  

1s^22s^22p^2

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, Carbon has 2 singly occupied orbitals.

(B) Cobalt.

The electronic configuration is -  

1s^22s^22p^63s^23p^63d^{7}4s^2

Thus, 4s orbital is fully filled and d orbital can singly filled 5 electrons. Thus, 4 electrons will be paired in 2 orbitals and 3 orbitals will be singly filled in cobalt.

(C) Sulfur.

The electronic configuration is -  

1s^22s^22p^63s^23p^4

Thus, 3s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, 2 electrons will be paired in 1 orbital and 2 orbitals will be singly filled in sulfur.

D) fluorine

The electronic configuration is -  

1s^22s^22p^5

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, 4 electrons will be paired in 2 orbitals and 1 orbital will be singly filled in fluorine.

E) Titanium

The electronic configuration is -  

1s^22s^22p^63s^23p^63d^{2}4s^2

Thus, 4s orbital is fully filled and d orbital can singly filled 5 electrons. Thus, 2 orbitals will be singly filled in titanium.

F) Germanium

The electronic configuration is -  

1s^22s^22p^63s^23p^63d^{10}4s^24p^2

Thus, 4s, 3d orbitals are fully filled and p orbital can singly filled 3 electrons. Thus, Germanium has 2 singly occupied orbitals.

4 0
3 years ago
Can someone help me please and please show work
FinnZ [79.3K]

Answer:

Average of the trial is: 288.50 C

Percent Error: 3.83%

Explanation:

(291 + 287 + 295 + 281) : 4 = 288.50 C

Average: 288.50 C

Percent Error: {(300 - 288.50) : 300} x 100%  = 3.83 %

                   

4 0
2 years ago
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