Answer:
84.8 mL
Explanation:
From the question given above, the following data were obtained:
Mass of CuNO₃ = 3.53 g
Molarity of CuNO₃ = 0.330 M
Volume of solution =?
Next, we shall determine the number of mole in 3.53 g of CuNO₃. This can be obtained as follow:
Mass of CuNO₃ = 3.53 g
Molar mass of CuNO₃ = 63.5 + 14 + (16×3)
= 63.5 + 14 + 48
= 125.5 g/mol
Mole of CuNO₃ =?
Mole = mass / Molar mass
Mole of CuNO₃ = 3.53 / 125.5
Mole of CuNO₃ = 0.028 moles
Next, we shall determine the volume of the solution. This can be obtained as follow:
Molarity of CuNO₃ = 0.330 M
Mole of CuNO₃ = 0.028 moles
Volume of solution =?
Molarity = mole /Volume
0.330 = 0.028 / Volume
Cross multiply
0.330 × Volume = 0.028
Divide both side by 0.330
Volume = 0.028 / 0.330
Volume = 0.0848 L
Finally, we shall convert 0.0848 L to millilitres (mL). This can be obtained as follow:
1 L = 1000 mL
Therefore,
0.0848 L = 0.0848 L × 1000 mL / 1 L
0.0848 L = 84.8 mL
Therefore, the volume of the solution is 84.8 mL.
Answer:
Electronegativity in group 1 decreases as we go from Lithium to Francium.
Explanation:
Electronegativity is defined as the tendency of an element to attract an electron pair towards itself.
In a group generally this tendency decreases from top to bottom as the size of the atom increases and hence the positive nucleus get far from the outer orbital.
In the same way group 1 elements i.e. from Lithium to Francium electronegativity decreases.
Answer: B its most likely
In the complete combustion of 1.60 moles of benzene, C6H6, 12 moles of oxygen, O2, is consumed.
Combustion is defined as the process of burning something. In chemistry, combustion refers to the chemical process between a fuel and an oxidant, usually oxygen to produce heat and light in the form of flame.
In a complete combustion, oxygen is sufficient to react with any hydrocarbons to produce carbon dioxide and water.
Balancing the combustion reaction of benzene, we have:
2C6H6 + 15 O2 = 12CO2 + 6H2O
Based on the balanced combustion reaction above, 2 moles of benzene requires 15 moles of oxygen to have a complete combustion.
If we have 1.60 moles C6H6,
moles O2 = mole ratio x mole of benzene
moles O2 = (15 moles O2/2 moles C6H6) x 1.60 moles C6H6
moles O2 = 12
To learn more about combustion: brainly.com/question/9913173
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