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Naddik [55]
3 years ago
8

Which kinds of bonds are present in a single water molecule?.

Chemistry
1 answer:
Elden [556K]3 years ago
4 0

The type of bonding present in water (H2O) is hydrogen bonding.

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Draw the lewis structure for the polyatomic trisulfide s−23 anion. be sure to include all resonance structures that satisfy the
mezya [45]

The following are the steps involved in drawing Lewis structure of the polyatomic trisulfide anion S_{3}^{2-}:

Total number of valence electrons = (3 * 6) + 2 = 20 electrons

Resonance is not possible in this ion. The molecular geometry of the ion will be bent as there are two lone pairs and two bond pairs on the central atom.



3 0
3 years ago
Chemistry! help! Please!
VMariaS [17]

Answer:

2. a. All three solutes are nonelectrolytes.

3. a. the solution of solute X

Explanation:

The freezing point depression (ΔTf) and <em>boiling point elevation</em> (ΔTb) are <em>colligative properties</em>: they depend on the <em>numbers of particles</em>.

The formula for ΔTf is

ΔTf = iKf·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For sucrose,

Sucrose (s) ⟶ sucrose (aq)

1 mole sucrose ⟶ 1 mol particles         i = 1

For NaCl

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

1 mol           1 mol     +   1 mol                 i = 2

For Ca(NO₃)₂

Ca(NO₃)₂(s) ⟶ Ca²⁺(aq) + 2NO₃⁻(aq)

1 mol                     1 mol       +  2 mol     i = 3

===============

2. <em>Freezing points </em>

For a nonelectrolyte, i = 1.

Kf = 1.86 °C·kg⁻¹mol⁻¹

b = 1 mol/kg          Calculate ΔTf

ΔTf = 1 × 1.86 × 1

ΔTf = 1.86 °C

Tf = Tf⁰ - ΔTf =0.00 °C – 1.86 °C = -1.86 °C

All the other solutions have lower freezing points, so the solutes must be <em>electrolytes</em>.

===============

3. <em>Boiling points</em>

The formula for ΔTb is

ΔTb = iKb·m

The solution with the <em>highest boiling point</em> will have the <em>highest value of i. </em>

In other words, the solution with the highest boiling point will be the one with the <em>lowest freezing point</em>.

That’s the solution of solute X.

5 0
3 years ago
10. The elements in group 1 are called
Alina [70]

Answer:

.

Explanation:

3 0
3 years ago
2. How many calories are needed to raise 50 g of iron from 55°C to 200°C? (c = 0.110 cal/g °C)
lord [1]

Answer:

Q = 797.5 cal

Explanation:

Given that,

Mass of iron, m = 50 g

The temperature rises from 55°C to 200°C.

We need to find the heat needed to raise the temperature. The heat raised is given by :

Q=mc\Delta T

Put all the values,

Q=50\times 0.11\times (200-55)\\\\Q=797.5\ cal

So, 797.5 calories of heat is needed.

3 0
3 years ago
Which statement correctly describes a reaction in dynamic equilibrium? At dynamic equilibrium, the reactions stop and the amount
harkovskaia [24]

Answer:

At dynamic equilibrium, the reactions continue but the amounts of reactants and products do not change.

Explanation:

At dynamic equilibrium, the reactions continue but according to the law of conservation of mass, the amounts of reactants and products don't change.

7 0
3 years ago
Read 2 more answers
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