For this problem, we use Graham's Effusion Law to find out the rate of effusion of chlorine gas. The formula is as follows:
R₁/R₂ = √(M₂/M₁)
Let 1 be N₂ while 2 be Cl₂
255/R₂ = √(28/70.8)
Solving for R₂,
R₂ = 405.5 s
<em>Thus, it would take 405.5 s to effuse chlorine gas.</em>
The oxidation number of Mn is 0. Details about oxidation number can be found below.
<h3>What is oxidation number?</h3>
Oxidation number is the hypothetical charge of an atom within a molecule.
However, when an atom exists independently as an element, the oxidation number of such element is 0.
The oxidation number is assigned to an atom in a element/compound. It can either be +, -, or 0 and it tells us if electrons are lost or gained.
Therefore, according to this question, the oxidation number of Mn is 0.
Learn more about oxidation number at: brainly.com/question/15167411
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Answer:
Explanation: For example, pH 4 is ten times more acidic than pH 5 and 100 times (10 times 10) more acidic than pH 6. The same holds true for pH values above 7, each of which is ten times more alkaline (another way to say basic) than the next lower whole value.
HOPE THAT HELPS
All the solar energy is generated in the core by nuclear fusion. Around the core there is the radiative zone. ... It extends from a depth of 200,000 km up to the visible surface of the Sun. Energy is transported by convection in this region. The surface of the convection zone is where light (photons) is created.
Answer:
#1: 0.00144 mmolHCl/mg Sample
#2: 0.00155 mmolHCl/mg Sample
#3: 0.00153 mmolHCl/mg Sample
Explanation:
A antiacid (weak base) will react with the HCl thus:
Antiacid + HCl → Water + Salt.
In the titration of antiacid, the strong acid (HCl) is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.
Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:
Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl
Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl
Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl
The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.
Thus, mmoles HCl /mg OF SAMPLE<em> </em>for each trial is:
#1: 2.178mmol / 1515mg
#2: 2.253mmol / 1452mg
#3: 2.219mmol / 1443mg
<h3>#1: 0.00144 mmolHCl/mg Sample</h3><h3>#2: 0.00155 mmolHCl/mg Sample</h3><h3>#3: 0.00153 mmolHCl/mg Sample</h3>
