Answer:
THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g
Explanation:
In an ideal condition
PV = nRT or PV = MRT/ MM where:
M = mass = unknown
MM =molar mass = 28 g/mol
P = pressure = 2 atm
V = volume = 25 mL = 0.025 L
R = gas constant = 0.082 L atm/mol K
T = temperature = 290 K
n = number of moles
The gas in the question is nitrogen gas
Molar mass of nitrogen gas = 14 * 2 = 28 g/mol
Then equating the variables and solving for M, we have
M = PV MM/ RT
M = 2 * 0.025 * 28 / 0.082 * 290
M = 1.4 / 23.78
M = 0.0589 g
The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g
Answer:
A: Synonyms
Explanation:
Synonyms are two words that have same meaning but are different words (tell me if this works!)
Answer:
HF
H₂S
H₂CO₃
NH₄⁺
Explanation:
<em>Which acid in each of the following pairs has the stronger conjugate base?</em>
According to Bronsted-Lowry acid-base theory, <em>the weaker an acid, the stronger its conjugate acid</em>. Especially for weak acids, pKa gives information about the strength of such acid. <em>The higher the pKa, the weaker the acid.</em>
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- Of the acids HCl or HF, the one with the stronger conjugate base is HF because it is a weak acid.
- Of the acids H₂S or HNO₂, the one with the stronger conjugate base is H₂S because it is a weaker acid. pKa (H₂S) = 7.04 > pKa (HNO₂) = 3.39
- Of the acids H₂CO₃ or HClO₄, the one with the stronger conjugate base is H₂CO₃ because it is a weak acid.
- Of the acids HF or NH₄⁺, the one with the stronger conjugate base is NH₄⁺ because it is a weaker acid. pKa (HF) = 3.17 < pKa (NH₄⁺) = 9.25
Answer:
INTENSIVE PROPERTY OF MATTER
Explanation:
DENSITY IS AN INTENSIVE PROPERTY OF MATTER THAT ILLUSTRATES HOW MUCH MASS A SUBSTANCE HAS IN A GIVEN AMOUNT OF VALUE,
We assume that the method that made use of urea was able to recover all of the recoverable substance. The method in question is the method that makes use of water.
The total amount of substance is 43 mg/dl. The recovered amount is 25 mg/dl. The percent recovery is
(25 mg/dl / 43 mg/dl) * 100 = 58.14%