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gavmur [86]
3 years ago
10

What kind of bond does these have?

Chemistry
1 answer:
RUDIKE [14]3 years ago
7 0
<span>1. Tap water has a small concentration of H+ & OH- ions as well as water molecules, hence there would be permanent dipole-permanent dipole (p.d.-p.d.) forces of attraction between the water molecules (aka H-bonds) as well as ionic bonds between the H+ & OH- ions.

2. Distilled water does not have H+ & OH- ions, hence only H-bonds exist between the water molecules.

3. There are covalent bonds between the individual sugar molecules.
 
4. There are ionic bonds between the Na+ & Cl- ions in NaCl.
 
5. There are p.d.-p.d. forces of attraction between the Na+ ions and the O2- partial ions of the water molecules as well as between the Cl- ions and the H+ partial ions of the water molecules. There are also H-bonds between the individual water molecules and ionic bonds between the Na+ & Cl- ions (although these are in much lower abundance than in unsolvated solid NaCl).

6. There are i.d.-i.d. as well as p.d.-p.d. forces of attraction between the sugar molecules and the water molecules. There are also H-bonds between the individual water molecules and covalent bonds within the sugar molecules.</span>
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How much time would it take for 336 mg of copper to be plated at a current of 5.6 A ? Express your answer using two significant
schepotkina [342]

Answer:

1.8 × 10² s

Explanation:

Let's consider the reduction that occurs upon the electroplating of copper.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We will establish the following relationships:

  • 1 g = 1,000 mg
  • The molar mass of Cu is 63.55 g/mol
  • When 1 mole of Cu is deposited, 2 moles of electrons circulate.
  • The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
  • 1 A = 1 C/s

The time  that it would take for 336 mg of copper to be plated at a current of 5.6 A is:

336mgCu \times \frac{1gCu}{1,000mgCu} \times \frac{1molCu}{63.55gCu} \times \frac{2mole^{-} }{1molCu} \times \frac{94,486C}{1mole^{-}} \times \frac{1s}{5.6C} = 1.8 \times 10^{2} s

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Combustion vapor-air mixtures are flammable over a limited range of concentrations. The minimum volume % of vapor that gives a c
DochEvi [55]
Combustion equation of n-hexane:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles

LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1% 

b)
1.1 volume percent required for LFL

1.1% x 1
= 0.0011 m³ of n-hexane required
6 0
2 years ago
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Differentiate the 3 types of evolution in Science
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Answer:

Brainliest pls

Explanation:

shows the three main types of evolution: divergent, convergent, and parallel evolution.

Divergent: tending to be different or develop in different directions.

Convergent: coming closer together, especially in characteristics or ideas.

Parallel evolution:  implies that two or more lineages have changed in similar ways

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What is the correct electron configuration for magnesium? ( use periodic table)
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Option c. 1s²2s²2p63s²
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Calculate the enthalpy of the reaction: 4 B(s) + 3 O2(g) --&gt; 2 B2O3(s) given the following pertinent information: 1. B2O3 (s)
Dennis_Churaev [7]

Answer:

ΔH = -2552kJ

Explanation:

Using Hess's law, it is possible to obtain ΔH of a reaction by the sum of similar reactions, thus:

1. B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g), ΔHoA = +2035 kJ

2. 2 B (s) + 3 H2 (g) → B2H6 (g), ΔHoB = +36 kJ

3. H2 (g) + 1/2 O2 (g) → H2O (l), ΔHoC = −285 kJ

4. H2O (l) → H2O (g), ΔHoD = +44 kJ

Subtracting of 2*(2) - 2*(1):

4 B (s) + 6 H2 (g) +  6 O2 (g) → 2 B2O3 (s) + 6 H2O (g)

ΔH' = 2*36kJ - 2*2035kJ = -3998kJ

Now, this reaction - 6*(3):

4 B (s) + 3 O2(g) + 6 H2O(l) → 2 B2O3 (s) + 6 H2O (g)

ΔH'' = -3998kJ - (6*-285kJ)

ΔH'' = -2288kJ

The last reaction - 6*(4):

4 B (s) + 3 O2(g) → 2 B2O3 (s)

ΔH = -2288kJ - 6*44kJ

<h3>ΔH = -2552kJ</h3>
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