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jarptica [38.1K]
3 years ago
5

Atoms of iron (fe) and sulfur (s) form iron sulfide (fes) when the iron atom donates electrons to the sulfur atom. the type of b

ond formed would be a(n) __________.
Chemistry
1 answer:
marysya [2.9K]3 years ago
4 0

Answer:

               The type of bond formed in FeS would be <u>Polar Covalent Bond</u>.

Explanation:

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

               Less than 0.4 then it is Non Polar Covalent  Bond

               Between 0.4 and 1.7 then it is Polar Covalent  Bond

               Greater than 1.7 then it is Ionic  Bond

For Fe and S,

                   E.N of Sulfur           =   2.58

                   E.N of Iron              =   1.83

                                                  ________

                   E.N Difference            0.75          (Polar Covalent Bond)


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<h3>Answer:</h3>

2.04 mol CBr₄

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
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  4. Multiplication
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

675 g CBr₄

<u>Step 2: Identify Conversions</u>

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<u />\displaystyle 675 \ g \ CBr_4(\frac{1 \ mol \ CBr_4}{331.61 \ g \ CBr_4}) = 2.03552 \ mol \ CBr_4<u />

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2.03552 mol CBr₄ ≈ 2.04 mol CBr₄

7 0
3 years ago
Read 2 more answers
Suppose a salt and a glucose solution are separated by a membrane that is permeable to water but not to the solutes. the nacl so
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1)

<span>m(NaCl) = 1.95 g
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Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:

</span>V(H2O) = 250ml = 250g = 0.25 kg<span>

</span><span>molality of NaCl:
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n(NaCl)=m/M=1.95/58.5= 0.033 mole

</span>molality b(NaCl)=n(NaCl) / V (H2O)= 0.033/0.25 = 0.132 mol/kg
<span>
milimolality of NaOH = 0.132/0,001 = 132 mmole/kg
</span>
milliosmolality of NaOH = milimolality x N of ions formed in dissociation

Since NaCl dissociates into 2 ions in solution:
<span>                                        
</span>milliosmolality of NaOH = 132 x 2 = 264  osmol<span>es/kg
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2)

m(gl) = 9 g
V(H2O) = 250mL
M(NaCl) = 180 g/mole

Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:

V(H2O) = 250ml = 250g = 0.25 kg

molality of glucose:

n(gl)=m/M=9/180= 0.05 mole

molality b(gl)=n(gl) / V (H2O)= 0.05/0.25 = 0.2 mol/kg

milimolality of glucose = 0.132/0,001 = 200 mmole/kg

milliosmolality of glucose = milimolality x N of ions formed in dissociation

Since glucose does not dissociate, milimolality and milliosmolality are same:
                                        
milliosmolality of glucose = 200 osmoles/kg

3)

The osmosis represents the diffusion of solvent molecules through a semi-permeable membrane that allows passage solvent molecules but does not to the dissolved substance molecule. The osmosis occurs when the concentrations of the solution on both sides of the membrane are different. Since the semi-permeable membrane only permeates the solvent molecules, but not the particles of the dissolved substance, it occurs the solvent diffusion through the membrane, i.e. the solvent molecules pass through the membrane to equalize the concentration on both sides of the membrane. Solvents molecules move from the middle with a lower concentration in the middle with a higher concentration of dissolved substances.

In our case, osmosis will occur because the concentration of NaCl solution and the concentration of glucose solution do not have same values. Osmosis will occur in the direction of glucose solution because it has a lower concentration.

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When 55.0 grams of metal at 75.0°C is added to 100. grams of water at 15.0°C, the temperature of the water rises to 18.7°C. Assu
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Answer:

The specific heat of the metal is 0,50 J/gºC

Explanation:

Assume that no heat is lost to the surroundings

(Q = m . C . ΔT)metal + (Q = m . C . ΔT)water = 0

Let's replace our values.

55g . C . (18,7ºC - 75ºC) + 100g . 4,184 J/g·°C . (18,7ºC - 15ºC) = 0

55g . C . -56,3 ºC + 418,4J/·°C . 3,7ºC = 0

-3096,5 gºC . C + 1548,08 J = 0

1548,08 J = 3096,5 gºC . C

1548,08 J / 3096,5 gºC  = C = 0,50 J/gºC

8 0
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