Cl2=3.17g/L
Ne=.901g/L
CO2=1.96g/l
therefore Cl2 is the densest gas under the given conditions.
Answer:
333.7 g.
Explanation:
- The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: <em>ΔTf = Kf.m.</em>
Where, ΔTf is the depression in water freezing point (ΔTf = 20.0°C).
Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).
m is the molality of the solution.
<em>∴ m = ΔTf/Kf</em> = (20.0°C)/(1.86 °C/m) = <em>10.75 m.</em>
molaity (m) is the no. of moles of solute per kg of the solvent.
∵ m = (no. of moles of antifreeze C₂H₄(OH)₂)/(mass of water (kg))
∴ no. of moles of antifreeze C₂H₄(OH)₂ = (m)(mass of water (kg)) = (10.75 m)(0.5 kg) = 5.376 mol.
∵ no. of moles = mass/molar mass.
<em>∴ mass of antifreeze C₂H₄(OH)₂ = no. of moles x molar mass </em>= (5.376 mol)(62.07 g/mol) =<em> 333.7 g.</em>
Full Question:
A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?
2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)
Answer:
13.1 g K2CO3 required to neutralize spill
Explanation:
2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)
Number of moles = Volume * Molar Concentration
moles HBr= 0.42L x .45 M= 0.189 moles HBr
From the stoichiometry of the reaction;
1 mole of K2CO3 reacts with 2 moles of HBr
1 mole = 2 mole
x mole = 0.189
x = 0.189 / 2 = 0.0945 moles
Mass = Number of moles * Molar mass
Mass = 0.0945 * 138.21 = 13.1 g
Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
Answer:
2.94x10²² atoms of Cu
Explanation:
We must work with NA to solve this, where NA is the number of Avogadro, number of particles of 1 mol of anything.
Molar mass Cu = 63.55 g/mol
Mass / Molar mass = Mol → 3.11 g / 63.55 g/m = 0.0489 moles
1 mol of Cu has 6.02x10²³ atoms of Cu
0.0489 moles of Cu, will have (0.0489 .NA)/ 1 = 2.94x10²² atoms of Cu
