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kondaur [170]
4 years ago
15

Calculate the oxidation number of “C” in CH3COCH3 please

Chemistry
1 answer:
Triss [41]4 years ago
8 0

Answer:

The oxidation number of C (carbon) is +4

Explanation:

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How can you tell from the name the types of bonds present in a hydrocarbon?
Aneli [31]

Answer:

Of bond is formed by sharing of electrons - covalent

If bond is formed by donation and by accepting electrons - ionic bond

Bond formed between metals - metallic bond.

7 0
3 years ago
Read 2 more answers
A 5.5 g sample of a substance contains only carbon and oxygen. Carbon makes up 35% of the mass of the substance. The rest is mad
Umnica [9.8K]

We have been given the condition that carbon makes up 35% of the mass of the substance and the rest is made up of oxygen. With this, it can be concluded that 65% of the substance is made up of oxygen. If we let x be the mass of oxygen in the substance, the operation that would best represent the scenario is,

<span>                                       x = (0.65)(5.5 g)</span>

<span>                                       <em> </em><span><em>x = 3.575 g</em></span></span>

8 0
3 years ago
A 0.43g samle of KHP required 24.11cm of NaOH for neutralization. Calculate the molarity of NaOH
expeople1 [14]

Answer:

0.083 M

Explanation:

We'll begin by calculating the number of mole in 0.43 g of KHP (potassium hydrogen phthalate, C₈H₅O₄K). This is can be obtained as follow:

Mass of C₈H₅O₄K = 0.43 g

Molar mass of C₈H₅O₄K = (8×12) + (5×1) + (16× 4) + 39

= 96 + 5 + 64 + 39 = 204 g/mol

Mole of C₈H₅O₄K =?

Mole = mass / molar mass

Mole of C₈H₅O₄K = 0.43 / 204

Mole of C₈H₅O₄K = 0.002 mole

Next, we shall determine the number of mole of NaOH required to react with 0.43 g (i.e 0.002 mole) of KHP. This can be obtained as follow:

C₈H₅O₄K + NaOH → C₈H₄O₄KNa + H₂O

From the balanced equation above,

1 mole of KHP reacted with 1 mole of NaOH.

Therefore, 0.002 mole of KHP will also react with 0.002 mole of NaOH.

Next, we shall convert 24.11 cm³ to L. This can be obtained as follow:

1000 cm³ = 1 L

Therefore,

24.11 cm³ = 24.11 cm³ × 1 L / 1000 cm³

24.11 cm³ = 0.02411 L

Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.00 2 mole

Volume = 0.02411 L

Molarity of NaOH =?

Molarity = mole /Volume

Molarity of NaOH = 0.002 / 0.02411

Molarity of NaOH = 0.083 M

3 0
3 years ago
Suppose 2.8 moles of methane are allowed to react with 5 moles of oxygen.
Ronch [10]

Answer : The limiting reagent is O_2

Solution : Given,

Moles of methane = 2.8 moles

Moles of O_2 = 5 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 mole of O_2 react with 1 mole of CH_4

So, 5 moles of O_2 react with \frac{5}{2}=2.5 moles of CH_4

From this we conclude that, CH_4 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is O_2

8 0
3 years ago
When the amount of oxygen is limited, carbon and oxygen react to form carbon monoxide. How many grams of CO can be formed from 3
Alinara [238K]

<u>61.25 grams</u> of CO can be formed from 35 grams of oxygen.

The molecular mass of oxygen is <u>16 gmol⁻¹</u>

The molecular mass of carbon monoxide is<u> 28 gmol⁻¹</u>

Explanation:

The molar mass of carbon monoxide is molar mass of C added to that of O;

12 + 16 = 28

= 28g/mol

The molar mass of oxygen is 16 g/mol while that of oxygen gas (O₂) is 32 g/mol

Since the ration oxygen to carbon monoxide is 1: 2 moles, we begin to find out how many moles of carbon monoxide are formed by 35 g of oxygen;

35/32 * 2

= 70/32 moles

Then multiply by the molar mass of carbon monoxide;

70/32 * 28

= 61.25 g

4 0
3 years ago
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