Answer:
Option E. Zirconium
Explanation:
From the question given above, the following data were obtained:
Length of side (L) of cube = 0.2 cm
Mass (m) of cube = 52 mg
Name of the unknown metal =?
Next, we shall determine the volume of the cube. This can be obtained as follow:
Length of side (L) of cube = 0.2 cm
Volume (V) of the cube =?
V = L³
V = 0.2³
V = 0.008 cm³
Next, we shall convert 52 mg to g. This can be obtained as follow:
1000 mg = 1 g
Therefore,
52 mg = 52 mg × 1 g / 1000 mg
52 mg = 0.052 g
Thus, 52 mg is equivalent to 0.052 g.
Next, we shall determine the density of the unknown metal. This can be obtained as follow:
Mass = 0.052 g.
Volume = 0.008 cm³
Density =?
Density = mass / volume
Density = 0.052 / 0.008
Density of the unknown metal = 6.5 g/cm³
Comparing the density of the unknown metal i.e 6.5 g/cm³ with those given in table in the above, we can conclude that the unknown metal is zirconium
Answer:
The mole ratio of the cation and the anion in a precipitate is a simple fraction. ( im sorry if this dosent help a lot.)
Explanation:
Answer:
The value of the carbon bond angles are 109.5 °
Explanation:
CH3CH2CH2OH = propanol . This is an alcohol.
All bonds here are single bonds.
Single bonds are sp³- hybdridization type. To be sp3 hybridized, it has an s orbital and three p orbitals : sp³. This refers to the mixing character of one 2s-orbital and three 2p-orbitals. This will create four hybrid orbitals with similar characteristics.
Sp3- types have angles of 109.5 ° between the carbon - atoms.
This means that the value of the carbon bond angles are 109.5 °
Answer: 94.07%
Explanation:
Percentage yield can be calculated by the formula
%yield = Experimental yield/Theoretical yield x100
Experimental yield = 7.93g
Theoretical yield = 8.43
%yield = Experimental yield/Theoretical yield x100
%yield = 7.93/8.43 x 100 = 94.07%
