V₂=0.894 L
<h3>Further explanation</h3>
Given
V₁=1.55 L
T₁=27 + 273 = 300 K
P₁=1 atm
T₂=-100+273 = 173 K
Required
The final volume(V₂)
Solution
Charles's Law
When the gas pressure is kept constant, the gas volume is proportional to the temperature
Input the value :
V₂=V₁T₂/T₁
V₂=1.55 x 173/300
V₂=0.894 L
<span>An Arrhenius base adds ions into water when added to the water hence it increases the concentration of ions in aqueous solution. Therefore, KOH is an example because the ions added to the water are called OH-ions.</span>
Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
Granite is an igneous type of rock.
1. The mass of 1.33×10²² mole of Sb is 1.62×10²⁴ g
2. The mass of 4.75×10¹⁴ mole of Pt is 9.26×10¹⁶ g
3. The mass of 1.22×10²³ mole of Ag is 1.32×10²⁵ g
4. The mass of 9.85×10²⁴ mole of Cr is 5.12×10²⁶ g
<h3>1. Determination of the mass of 1.33×10²² mole of Sb</h3>
- Mole of Sb = 1.33×10²² mole
- Molar mass of Sb = 122 g/mol
Mass = mole × molar mass
Mass of Sb = 1.33×10²² × 122
Mass of Sb = 1.62×10²⁴ g
<h3>2. Determination of the mass of 4.75×10¹⁴ mole of Pt</h3>
- Mole of Pt = 4.75×10¹⁴ mole
- Molar mass of Pt = 122 g/mol
Mass = mole × molar mass
Mass of Pt = 4.75×10¹⁴ × 195
Mass of Pt = 9.26×10¹⁶ g
<h3>3. Determination of the mass of 1.22×10²³ mole of Ag</h3>
- Mole of Ag = 1.22×10²³ mole
- Molar mass of Ag = 108 g/mol
Mass = mole × molar mass
Mass of Ag = 1.22×10²³ × 108
Mass of Ag = 1.32×10²⁵ g
<h3>4. Determination of the mass of 9.85×10²⁴ mole of Cr</h3>
- Mole of Cr = 9.85×10²⁴ mole
- Molar mass of Cr = 52 g/mol
Mass = mole × molar mass
Mass of Cr = 9.85×10²⁴ × 52
Mass of Cr = 5.12×10²⁶ g
Learn more about mole:
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