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seropon [69]
3 years ago
7

Alcano con 1 carbono?

Chemistry
1 answer:
never [62]3 years ago
5 0

creo esto te va a ayudar

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What volume will 12.0 g of oxygen gas occupy at 25 c and a pressure of 52.7 kpa?
Stolb23 [73]
We can use the ideal gas law equation to find the volume occupied by oxygen gas
PV = nRT
where ;
P - pressure - 52.7 kPa
V - volume 
n - number of oxygen moles - 12.0 g / 32 g/mol = 0.375 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25 °C + 273 = 298 K
substituting the values in the equation 
52 700 Pa x V = 0.375 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
 V = 17.6 L
volume of the gas is 17.6 L
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The chemical bonding in sodium phosphate, Na3PO4, is classified as:
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<span>Ionic bonding between sodium and phosphate ions.</span>
7 0
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Someone please help me sort these out for either if they are Heterogenous, solution, suspension, or colloid. Tuna casserole, gre
Sphinxa [80]
Heterogeneous-a mixture of two or more things
solution-a substance that dissolves
suspension-mixture where solid particles do not dissolve
colloid-does not settle and cannot be seperated

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How are you guys today? o.0
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You have a 25.2 L sample of gas at 1.25 atm and 25.0 degrees Celsius. How many moles are present in this gas. For your answer, p
Elenna [48]

Answer:

  • <u>1.29 mol</u>

Explanation:

This is a direct application of the equation for ideal gases.

  • PV=nRT

Where:

  • P = pressure = 1.25 atm
  • V = volume = 25.2 liter
  • R = Universal constant of gases = 0.08206 atm-liter/K-mol
  • T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
  • n = number of moles

Solving for n:

  • n=\frac{PV}{RT}

Substituting:

n=\frac{1.25atm\times 25.2liter}{0.08206atm-liter/K-mol\times298.15K }\\\\n=1.29mol

8 0
3 years ago
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