Answer:
0.877 mol
Step-by-step explanation:
We can use the<em> Ideal Gas Law </em>to solve this problem.
pV = nRT Divide both sides by RT
n = (pV)/(RT)
Data:
p = 646 torr
V = 25.0 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 22.0 °C
Calculations:
(a) <em>Convert the pressure to atmospheres
</em>
p = 646 torr × (1 atm/760 torr) = 0.8500 atm
(b) <em>Convert the temperature to kelvins
</em>
T = (22.0 + 273.15) K = 295.15 K
(c) <em>Calculate the number of moles
</em>
n = (0.8500 × 25.0)/(0.082 06 × 295.15)
= 0.877 mol
a. 1,4332 g
b. 7.54~g
<h3>Further explanation</h3>
Given
Reaction
MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)
20 cm of 2.5 mol/dm^3 of MgCl2
20 cm of 2.5 g/dm^3 of MgCl2
Required
the mass of silver chloride - AgCl
Solution
a. mol MgCl2 :
From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1
mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g
b. mol MgCl2 (MW=95.211 /mol):
From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526
mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g
This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:
Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰
Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻
Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.
In such a way, the correct choice is C.
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It could them know how it start when it started wats in the infectious disease and why
Answer:
b I my current pls let me known
