Answer:
26.20% w/w of KBr in the sample
Explanation:
Mohr titration is a way to quantify Br⁻ and Cl⁻ ions in solution. The reaction is:
KBr(aq) + AgNO₃(aq) → KNO₃(aq) + AgBr(s)
<em>where 1 mole of KBr reacts per mole of AgNO₃</em>
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Thus, moles of AgNO₃ in (25.13mL-0.65mL = 24.48mL) of a 0.04614M solution are:
0.02448L × (0.04614mol / L) = 1.130x10⁻³ moles of AgNO₃ = moles of KBr.
Mass of KBr -Molar mass: 119g/mol- is:
1.130x10⁻³ moles of KBr × (119g / 1mol) = <em>0.1344g of KBr</em>
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Thus, %w/w of KBr in the sample is:
%w/w = 0.1344g / 0.5131g ×100 = <em>26.20% w/w of KBr in the sample</em>
<span>6.38x10^-2 moles
First, let's determine how many moles of gas particles are in the two-liter container. The molar volume for 1 mole at 25C and 1 atmosphere is 24.465 liters/mole. So
2 L / 24.465 L/mol = 0.081749438 mol
Now air doesn't just consist of nitrogen. It also has oxygen, carbon dioxide, argon, water vapor, etc. and the total number of moles includes all of those other gasses. So let's multiply by the percentage of nitrogen in the atmosphere which is 78%
0.081749438 mol * 0.78 = 0.063764562 mol.
Rounding to 3 significant figures gives 6.38x10^-2 moles</span>
Correct lol, Be nice they favourite ur class nad u bearly get homework
Answer:
1.2 L
Explanation:
Step 1: Write the balanced equation
Mg + 2 HCl ⇒ MgCl₂ + H₂
Step 2: Calculate the moles corresponding to 2.3 g of Mg
The molar mass of Mg is 24.31 g/mol.
2.3 g × 1 mol/24.31 g = 0.095 mol
Step 3: Calculate the moles of H₂ produced
0.095 mol Mg × 1 mol H₂/1 mol Mg = 0.095 mol H₂
Step 4: Calculate the volume occupied by the hydrogen
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 0.095 mol × (0.0821 atm.L/mol.K) × 298 K/2 atm = 1.2 L