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____ [38]
4 years ago
15

Which compound is an alkyne? (1) C2H2 (3) C4H8 (2) C2H4 (4) C4H10

Chemistry
2 answers:
Romashka [77]4 years ago
6 0

Answer: C_2H_2

Explanation:

Saturated hydrocarbons are defined as the hydrocarbons in which a single bond is present between carbon and carbon atoms. The general formula for these hydrocarbons is C_nH_{2n+2} and are called as alkanes.

Unsaturated hydrocarbons are defined as the hydrocarbons which have double or triple covalent C-C bonds. They are known as alkenes and alkynes respectively. The general formula for these hydrocarbons is C_nH_{2n} and C_nH_{2n-2} respectively.

C_4H_8 and C_2H_4 are alkenes.

C_4H_{10} is an alkane.

Thus C_2H_2 with general formula C_nH_{2n-2} is an alkyne.

olga2289 [7]4 years ago
5 0
The answer is C2H2 because an alkyne compound consists of a series of unsaturated hydrocarbons with a triple bond. 
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A 1.44-g sample of an unknown pure gas occupies a volume of 0.335 L at a pressure of and a temperature of 100.0°C. The unknown g
Zepler [3.9K]

Answer:

Xenon

Explanation:

Step 1: Given data

  • Mass (m): 1.44 g
  • Volume (V): 0.335 L
  • Pressure (P): 1.00 atm (I looked it up)
  • Temperature (T): 100.0°C

Step 2: Convert the temperature to Kelvin

K = °C + 273.15 = 100.0°C + 273.15 = 373.2 K

Step 3: Calculate the number of moles (n)

We will use the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.00 atm × 0.335 L / (0.0821 atm.L/mol.K) × 373.2 K

n = 0.0109 mol

Step 4: Calculate the molar mass of the gas

M = 1.44 g / 0.0109 mol = 132 g/mol

Step 5: Identify the gas

The gas with a molar mass of about 132 g/mol is xenon.

8 0
4 years ago
Which of the following elements has the smallest atomic radius?
HACTEHA [7]
Fluorine has the smallest atomic radius at a 42 
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How are different natural disasters related? What different causes and effects of natural disasters do you know about?
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A photon of light possesses 5 x 10^-19 J of energy. Calculate its frequency
saveliy_v [14]

Answer:

The frequency of photon is 0.75×10¹⁵ s⁻¹.

Explanation:

Given data:

Energy of photon = 5×10⁻¹⁹ J

Frequency of photon = ?

Solution:

Formula;

E = hf

h = planck's constant = 6.63×10⁻³⁴ Js

5×10⁻¹⁹ J =  6.63×10⁻³⁴ Js ×f

f =  5×10⁻¹⁹ J / 6.63×10⁻³⁴ Js

f = 0.75×10¹⁵ s⁻¹

The frequency of photon is 0.75×10¹⁵ s⁻¹.

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