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Vanyuwa [196]
2 years ago
5

4. At least one carbon-carbon bond in an alkyne is a triple covalent bond. Other bonds may be single or double carbon-carbon bon

ds and carbon-hydrogen bonds.
Chemistry
1 answer:
DanielleElmas [232]2 years ago
3 0

Answer:

hope it helps cuz i did not really understand

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Phosphorus p4 burns in oxygen to form diphosphorus pentoxide what is the balanced chemical equation representing this chemical r
ahrayia [7]

Answer:

P₄ + 5O₂ → 2P₂O₅

Explanation:

Phosphorus burn in the presence of air and produced diphosphorus pentoxide.

Chemical equation:

P₄ + O₂ → P₂O₅

Balanced chemical equation:

P₄ + 5O₂ → 2P₂O₅

Equation is balanced because there are four phosphorus atoms ans ten oxygen atoms in both side of equation.

Coefficient with reactant and product:

P₄         1

O₂        5

P₂O₅    2

7 0
3 years ago
How many grams of O2 are present in 44.1 L of O2 at STP?
ycow [4]

Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

First of all, the STP conditions refer to the standard temperature and pressure, where the values ​​used are: pressure at 1 atmosphere and temperature at 0°C. These values ​​are reference values ​​for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that occupies.
  • T is its temperature.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

Then, in this case:

  • P= 1 atm
  • V= 44.1 L
  • n= ?
  • R= 0.082 \frac{atmL}{molK}
  • T= 0°C =273 K

Replacing in the expression for the ideal gas law:

1 atm× 44.1 L= n× 0.082 \frac{atmL}{molK}× 273 K

Solving:

n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}

n=1.97 moles

Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:

1.97 molesx\frac{32 g}{1 mole}= 63.04 g ≈ <u><em>63 g</em></u>

Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

Learn more about the ideal gas law:

  • <u>brainly.com/question/4147359?referrer=searchResults</u>
7 0
3 years ago
HEELP PLEASE Any change that alters a substance without changing it into another substance is a(n)________change.
Trava [24]
Sounds like a physical change.
8 0
3 years ago
Please help will give brainliest
DochEvi [55]

Answer:

Number of moles = 1.57 mol

Explanation:

Given data:

Mass of propanol = 94.1 g

Molar mass of propanol = 60.1 g/mol

Number of moles of propanol = ?

Solution:

Formula:

Number of moles = mass/molar mass

by putting values,

Number of moles = 94.1 g/ 60.1 g/mol

Number of moles = 1.57 mol

5 0
4 years ago
Why isn't the mass of the electron included in the mass of an atom on the periodic table?
marta [7]
Its A because............
5 0
4 years ago
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