Answer:
3.8 g/mL
Explanation:
From the question given above, the following data were obtained:
Volume of Water = 60 mL
Volume of Water + Object = 73.5 mL
Mass of object = 51.3 g
Density of object =?
Next, we shall determine the volume of the object. This can be obtained as follow:
Volume of Water = 60 mL
Volume of Water + Object = 73.5 mL
Volume of object =?
Volume of object = (Volume of Water + Object) – (Volume of Water)
Volume of object = 73.5 – 60
Volume of object = 13.5 mL
Finally, we shall determine the density of the object as illustrated below:
Volume of object = 13.5 mL
Mass of object = 51.3 g
Density of object =?
Density = mass /volume
Density of object = 51.3 /13.5
Density of object = 3.8 g/mL
Thus, the density of the object is 3.8 g/mL
Answer: 4 x 10 ∧-2 moles of nitrogen.
Explanation:
The chemical formular for quinine is ; C20 H24 N2 O2
As can be seen from the chemical formular;
1 mole of quinine contains 2 moles of Nitrogen
Thus; 2.0 x 10 ∧-2 moles of quinine would contain
2.0 x 10 ∧-2 x 2 = 4 x 10 ∧-2 moles of nitrogen.
Therefore 4 x 10 ∧-2 moles of nitrogen are in 2.0×10−2mole of quinine
Diamond. It can only be cut with another diamond.
Given 3.72 g of P and 21.28g of Cl, converting these to moles: ( 3.72 g P)(mol P/30.97 g P) = 0.12 mol P (21.28 g Cl)(mol Cl/35.45 g Cl) = 0.60 mol Cl P:Cl = 0.12/0.60, therefore P: Cl =1/5 Therefore, PCl5 hope it helps
First, let's compute the number of moles in the system assuming ideal gas behavior.
PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
n = 2.176 moles
At standard conditions, the standard molar volume is 22.4 L/mol. Thus,
Standard volume = 22.4 L/mol * 2.176 mol =<em> 48.74 L</em>
