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4 years ago

What happens when magnesium reacts with iodine to make magnesium iodide?

Chemistry

1 answer:

I am Lyosha [343]4 years ago

7 0

Explanation:

Atomic number of magnesium is 12 and its electronic distribution is 2, 8, 2. On the other hand, atomic number of iodine is 53 and its electronic configuration is $[Kr]4d^{10}5s^{2}5p^{5}$ .

Hence, there are 7 valence electrons in an iodine atom and there are 2 valence electrons in a magnesium atom.

So, one atom of iodine requires one electron from a donor atom to complete its octet. But one magnesium atom contains two valence electrons.

Therefore, one magnesium atom will combine with two iodine atoms to result in the formation of magnesium iodide as follows.

$Mg^{2+} + 2I^{-} \rightarrow MgI_{2}$

Therefore, an ionic bond will be formed when magnesium reacts with iodine to make magnesium iodide.

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The most abundant class of phospholipid in animal cell membranes, with a hydrophobic tail composed of two fatty acyl chains, est

juin [17]

Answer:

Phosphatidylcholine also known as lecithin

Explanation:

The most common phospholipids in order of abundance are

1). phosphatidylcholine, lecithin

2). phosphatidylethanolamine, cephalins 3).phosphatidylinositol, and 4).phosphatidylserine.

They have the common characteristics of esterified fatty acids to the 1 and 2 positions of the glycerol structure with the esterified phosphate group to the 3 position

7 0

4 years ago

Can some one help whith these

MA_775_DIABLO [31]

Answer:

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3 years ago

Read 2 more answers

A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug

OverLord2011 [107]

Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Henderson-Hasselbalch formula to find pH in a buffer is:

pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Replacing with known values:

5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

0.260 = log₁₀ [CH₃COO⁻] / [CH₃COOH]

1.820 = [CH₃COO⁻] / [CH₃COOH] (1)

As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M (2)

Replacing (2) in (1):

1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

[CH₃COOH] = 0.100M / 2.820

[CH₃COOH] = 0.035M

Thus: [CH₃COO⁻] = 0.100M - 0.035M = 0.065M

5.40 mL of a 0.490 M HCl are:

0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = 7.429x10⁻³ moles of CH₃COO⁻

Moles of CH₃COOH before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles + 2.646x10⁻³ moles = 8.071x10⁻³ moles of CH₃COOH. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

pH = 4.704

As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = 0.296

4 0

3 years ago

Convert 32.56 km/hr into ft/hr

Gekata [30.6K]

Note that
1 m = 3.2808 ft

Therefore
1 km = 3280.8 ft
and
$32.56 \, \frac{km}{h} = (32.56 \, \frac{km}{h})*(3280.8 \, \frac{ft}{km}) = 1.0682 \, \times 10^{5} \, \frac{ft}{h}$

Answer: 1.0682 x 10⁵ ft/hr

8 0

3 years ago

What happens when hydrogen gas is passed over hot ferric oxide

Roman55 [17]

Answer:

When hydrogen is passed over hot ferric oxide (FeO) hydrogen reacts with oxygen present in the compound and forms water (H2O) and pure Iron

Explanation:

7 0

3 years ago

Read 2 more answers