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Rina8888 [55]
3 years ago
15

Describe the formation of covalent bond in methane (5 marks) ​

Chemistry
2 answers:
Lady bird [3.3K]3 years ago
8 0

Answer:

Explanation:

The structure of the methane, CH4, molecule exhibits single covalent bonds. Covalent bonding involves the sharing of electrons. In the methane molecule, the four hydrogen atom share one electron each with the carbon atom

bija089 [108]3 years ago
5 0

Answer: Covalent bonding is when atoms of different elements  share electrons

Explanation: Taking Methane CH₄

<h3>it contains Hydrogen and Carbon atoms pairing electrons. the electronic configuration for carbon is 2,4 that is to say it has 4 electrons in its valence shell (outer shell) and hydrogen has 1 so carbon shares its four electrons with four hydrogen atoms thus forming a covalent bond. so they now have a stable arrangement of electrons in their outer shell ,thus giving CH₄</h3>

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The whole cloves are broken up to help release the eugenol during distillation.
GREYUIT [131]

The statement is False.

No, whole cloves are not broken up to help release the eugenol during distillation.

Steam distillation:

  • Live steam is used in the co-distillation technique of steam distillation to separate mixture components.
  • It works well to extract essential oil constituents with high boiling points, such as those with boiling points of 200°C. However, the oil vapors themselves are warmer—around 100°C—helping to maintain the compounds' structural integrity.
  • It enables distillation to be carried out at lower temperatures than the boiling points of the constituent parts.
  • The high-boiling essential oils are vaporized by steam, and after passing through a cooling system, the hot vapors that were formed from them condense back into a liquid along with water.
  • A two-phase distillate, consisting of a water layer and an oil layer, is created because the oils are immiscible in water.

Learn more about the Steam distillation with the help of the given link:

brainly.com/question/14864901

#SPJ4

4 0
1 year ago
I RLLY NEED HELP. Use Image B (picture above) and calculate the density of a ring that has a mass of 32 grams. Read Page 11 "Mea
dimaraw [331]

Answer:

Mass of ring = 32 g

Volume of ring = 4 mL

Density of ring = 8 g/mL

Explanation:

From the question given above, the following data were obtained:

Mass of ring = 32 g

Volume of water = 64 mL

Volume of water + ring = 68 mL

Density of ring =?

Next, we shall determine the volume of the ring. This can be obtained as follow:

Volume of water = 64 mL

Volume of water + ring = 68 mL

Volume of ring =?

Volume of ring= (Volume of water + ring) – (Volume of water)

Volume of ring = 68 – 64

Volume of ring = 4 mL

Finally, we shall determine the density of the ring. This can be obtained as follow:

Mass of ring = 32 g

Volume of ring = 4 mL

Density of ring =?

Density = mass / volume

Density of ring = 32 / 4

Density of ring = 8 g/mL

8 0
3 years ago
Which statement is true about the gravity?
REY [17]

Answer:

B.

Explanation:

Gravity acts on all masses equally, even though the effects on both masses.

Hope it helps you! ^^



8 0
3 years ago
How many moles are contained in 3.131 × 1024 particles?
AleksAgata [21]

<u>Given:</u>

The number of particles = 3.131 * 10²⁴

<u>To determine:</u>

The number of moles corresponding to the given particles

<u>Explanation:</u>

1 mole corresponds to Avogadro's number of particles = 6.023 * 10²³ particles

Therefore, the given particles would correspond to:

3.131* 10²⁴ particles * 1 moles/6.023 * 10²³ particles = 5.199 moles

Ans: A) 5.199 moles are contained in 3.131* 10²⁴ particles



3 0
3 years ago
Read 2 more answers
A sample of a gas in a rigid cylinder with a movable piston has a volume of a lamp then .2 L at STP. What is the volume of this
Gnoma [55]

<u>Answer:</u> The new volume of the gas is 0.11 L

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

At STP:

The temperature at this condition is taken as 273 K and the pressure at this condition is taken as 1 atm or 101.3 kPa.

We are given:

P_1=101.3kPa\\V_1=0.2L\\T_1=273K\\P_2=202.6kPa\\V_2=?\\T_2=300K

Putting values in above equation, we get:

\frac{101.3kPa\times 0.2L}{273K}=\frac{202.6kPa\times V_2}{300K}\\\\V_2=\frac{101.3\times 0.2\times 300}{273\times 202.6}=0.11L

Hence, the new volume of the gas is 0.11 L

3 0
3 years ago
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