Answer:
612 K
Explanation:
From the question given above, the following data were obtained:
Initial temperature (T₁) = 306 K
Initial pressure (P₁) = 150 kPa
Final pressure (P₂) = 300 kPa
Volume = 4 L = constant
Final temperature (T₂) =?
Since the volume is constant, the final (i.e the new) temperature of the gas can be obtained as follow:
P₁ / T₁ = P₂ / T₂
150 / 306 = 300 / T₂
Cross multiply
150 × T₂ = 306 × 300
150 × T₂ = 91800
Divide both side by 150
T₂ = 91800 / 150
T₂ = 612 K
Thus, the new temperature of the gas is 612 K
Answer:
Grey precipitate implies the presence of silver ions
Yellow precipitate implies the presence of lead II ions
Explanation:
Qualitative analysis provides us a quick method of identifying ions present in a sample by chemical reactions involving simple reagents. Precipitates having a unique colour is formed. The identity of ions in the sample is deduced from the colour of precipitate obtained when particular reagents are added.
In the question, a precipitate containing silver ions upon standing turn into grey colour. Similarly, lead II ions give a yellow precipitate.
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Answer:
1.73 M
Explanation:
We must first obtain the concentration of the concentrated acid from the formula;
Co= 10pd/M
Where
Co= concentration of concentrated acid = (the unknown)
p= percentage concentration of concentrated acid= 37.3%
d= density of concentrated acid = 1.19 g/ml
M= Molar mass of the anhydrous acid
Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1
Substituting values;
Co= 10 × 37.3 × 1.19/36.5
Co= 443.87/36.6
Co= 12.16 M
We can now use the dilution formula
CoVo= CdVd
Where;
Co= concentration of concentrated acid= 12.16 M
Vo= volume of concentrated acid = 35.5 ml
Cd= concentration of dilute acid =(the unknown)
Vd= volume of dilute acid = 250ml
Substituting values and making Cd the subject of the formula;
Cd= CoVo/Vd
Cd= 12.16 × 35.5/250
Cd= 1.73 M
Answer:
Mass = 9.58 g
Explanation:
Given data:
Mass of Fe₂O₃ formed = ?
Mass of Fe = 6.7 g
Solution:
Chemical equation:
4Fe + 3O₂ → 2Fe₂O₃
Number of moles of Fe:
Number of moles = mass/molar mass
Number of moles = 6.7 g/ 55.8 g/mol
Number of moles = 0.12 mol
now we will compare the moles of Fe and Fe₂O₃.
Fe : Fe₂O₃
4 : 2
0.12 : 2/4×0.12 = 0.06 mol
Mass of Fe₂O₃:
Mass = number of moles × molar mass
Mass = 0.06 mol × 159.69 g/mol
Mass = 9.58 g