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max2010maxim [7]
3 years ago
5

Give the following compound's base name.

Chemistry
1 answer:
Murljashka [212]3 years ago
6 0

Answer:

2-Butyne

Explanation:

The given compound name is 2-Butyne.

According to the IUPAC rule, the compound is consist of 4 carbon atoms with triple bond at second carbon atom, so its name will be 2-Butyne or but-2-yne. 2-Butyne is an alkyne which is produced artificially at standard temperature and pressure.

Hence, the correct answer is "2-Butyne".

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2Li + H2SO4=Li2SO4 + H2 How many liters of hydrogen gas, H2 at STP can be produced from 3.0 moles of Li? The molar volume of a g
Gemiola [76]

Answer:

volume of H_2=33.6 litre

Explanation:

Firstly balance the given chemical equation,

2Li + H_2SO_4=Li_2SO_4 + H_2

From the given balance equation it is clearly that,

2 mole of Li gives  1 mole of H2 gas

2 mole Li⇔1 mole H_2

1 mole Li⇔0.5 mole H_2

3 mole Li⇔1.5 mole H_2

hence

3 mole of Li will give 1.5 mole H2 gas

therefore volume of gas produced from 3 mole Li at STP = 1.5\times22.4 \frac{L}{mol}

volume of H2=33.6 litre

7 0
3 years ago
Determine the value of the equilibrium constant, Kgoal, for the reaction C(s)+12O2(g)+H2(g)⇌12CH3OH(g)+12CO(g), Kgoal=? by makin
Licemer1 [7]

Answer:

1.71x10²⁷

Explanation:

If we sum 1/2 of (3) + 1/2 of (1):

1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷  = 4.58x10²³

1/2 (1)   1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8

C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>

K' = 4.58x10²³ * 11.8 = 5.42x10²⁴

+1/2 (2):

<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2

C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)

K'' = 5.42x10²⁴* 316.2 =

<h3>1.71x10²⁷</h3>

5 0
4 years ago
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