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strojnjashka [21]
3 years ago
9

An example of hydrogen bonding would be the attractions between outer hydrogen atoms and the central nitrogen atom within the NH

3 molecule.
TRUE

FALSE
Chemistry
2 answers:
marshall27 [118]3 years ago
7 0

Answer:

The answer to your question is: false.

Explanation:

Hydrogen bonding is tha attraction of hydrogen atoms to the oxygen or nitrogen of a different molecule not between the atoms of the same molecule.

So this description of hydrogen bonding is incorrect.

White raven [17]3 years ago
4 0
Completely false, hydrogen bonding occurs only between hydrogen and oxygen atoms.
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What is a substance that cannot be broken down further by chemical means
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3 years ago
A sample of ideal gas is in a sealed container. The pressure of the gas is 485 torr , and the temperature is 40 ∘C . If the temp
Fantom [35]

Answer:

537.68 torr.

Explanation:

  • We can use the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n and V are constant, and have different values of P and T:

<em>(P₁T₂) = (P₂T₁).</em>

P₁ = 485 torr, T₁ = 40°C + 273 = 313 K,

P₂ = ??? torr, ​T₂ = 74°C + 273 = 347 K.

∴ P₂ = (P₁T₂)/(P₁) = (485 torr)(347 K)/(313 K) = 537.68 torr.

3 0
2 years ago
If 0.0106 g of a gas dissolves in 0.792 L of water at 0.321 atm, what quantity of this gas (in grams) will dissolve at 5.73 atm?
Alex73 [517]

Answer:

0.189 g.

Explanation:

  • This problem is an application on <em>Henry's law.</em>
  • Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
  • Solubility of the gas ∝ partial pressure
  • If we have different solubility at different pressures, we can express Henry's law as:

<em>S₁/P₁ = S₂/P₂,</em>

S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm

S₂ = ??? g/L and P₂ = 5.73 atm

  • So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.

<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>

<em></em>

8 0
3 years ago
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