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guajiro [1.7K]
3 years ago
5

Why are iodoalkanes very reactive

Chemistry
1 answer:
makvit [3.9K]3 years ago
5 0

Answer:

Iodoalkanes are very reactive because they have the weakest bond of C-I, thus they are most easily broken, making them most reactive. B. The carbon where the halogen is bonded is only attached to one other carbon.

Explanation:

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How many formula units of CaO are in 32.7 g of CaO?
max2010maxim [7]

Answer:

3.51× 10²³ formula units

Explanation:

Given data:

Mass of CaO = 32.7 g

Number of formula units = ?

Solution:

First of all we will calculate the number of moles.

Number of moles = mass/molar mass

Number of moles = 32.7 g/ 56.1 g/mol

Number of moles = 0.583 mol

Number of formula units:

1 mole = 6.022 × 10²³ formula units

0.583 mol ×  6.022 × 10²³ formula units / 1 mol

3.51× 10²³ formula units

The number 6.022 × 10²³ is called Avogadro number.

7 0
3 years ago
What sample size (grams) of Na3PO4 (FW 164.00) known to be 50.00% pure should be used to consume exactly 40.00 mL of 0.1000 M HC
Mkey [24]

Answer:

0.109 g.

Explanation:

Equation of the reaction:

Na3PO4 + 3HCl --> 3NaCl + H3PO4

Number of moles of HCl = molar concentration × volume

= 0.1 × 0.04

= 0.004 mol.

By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3

= 0.0013 mol

Mass of Na3PO4 = molar mass × number of moles

= 0.0013 × 164

= 0.219 g

Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance

= 0.219 × 50 g/100 g

= 0.109 g.

3 0
3 years ago
The alkali earth metal beryllium (Be) engages in a chemical reaction and loses all of its valence electrons.
denpristay [2]

The loss of electron from an results in the formation of cation represented by the positive charge on the element whereas gaining of electron results in the formation of anion represented by the negative charge on the element.

The alkali earth metal beryllium (Be) belongs to the second group of the periodic table. The ground state electronic configuration of Be is:1s^{2}2s^{2}

From the electronic configuration it is clear that it has 2 valence electrons in its valence shell (2s^{2}).

After losing all valence electrons that is 2 electrons from 2s orbital. The electronic configuration will be:

1s^{2}2s^{0}

Since, lose of electron is represented by positive charge on the element symbol. So, the beryllium will have +2 charge on its symbol as Be^{2+}.

Hence, beryllium will have 2+ charge on it after losing all its valence electrons in the chemical reaction.

6 0
3 years ago
Read 2 more answers
Why is it important for scientists to use the scientific method?
MakcuM [25]

Explanation:

It <em><u>provides an objective, standardized approach to conducting experiments</u></em> and, in doing so, improves their results. By using a standardized approach in their investigations, scientists can feel confident that they will stick to the facts and limit the influence of personal, preconceived notions.

I hope this helps you out!

4 0
3 years ago
Read 2 more answers
Given the following reaction: 2D(g) + 3E(g) + F(g) \longrightarrow⟶ 2G(g) + H(g) When the concentration of D is decreasing by 0.
Zepler [3.9K]

Answer:

Rate of reaction = -d[D] / 2dt  = -d[E]/ 3dt = -d[F]/dt  = d[G]/2dt = d[H]/dt

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

E decreseas 3/2 as fast as G increases = 0.30 M/s

Explanation:

Rate of reaction = -d[D] / 2dt  = -d[E]/ 3dt = -d[F]/dt  = d[G]/2dt = d[H]/dt

When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing:

Given data = d[D]/dt = 0.10 M/s

-d[D] / 2dt  = d[H]/dt

d[H]/dt = 0.05 M/s

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing:

d[G] / 2dt  = -d[H]/3dt

E decreseas 3/2 as fast as G increases = 0.30 M/s

5 0
3 years ago
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