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:
CH3COOH would be more concentrated
Explanation:
The higher the concentration value, the more concentrated it is.
The relationship between concentration, moles and volume is given by the equation;
Concentration = No of moles / Volume
5.0 grams of HCOOH dissolved in 189 mL of water
Number of moles = Mass / Molar mass = 5 / 46.03 = 0.1086 mol
Concentration = 0.1086 / 0.189 = 0.5746 mol/L
1.5 moles of CH3COOH dissolved in twice as much water
Volume = 2 * 189 = 378 ml = 0.378 L
Concentration = 1.5 / 0.378 = 3.9683 mol/L
Comparing both concentration values;
CH3COOH would be more concentrated
4 mol NH₃ → 5 mol N₂
x mol NH₃ → 0.824 mol N₂
x=0.824*4/5=0.6592 mol
The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles
<h3>Balanced equation </h3>
2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
<h3>How to determine the theoretical yield of NaBr</h3>
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
Therefore,
2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr
Therefore,
Thus, the theoretical yield of NaBr is 7.08 moles
Learn more about stoichiometry:
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I think it is c but I’m not sure