Answer:
semiconducting and tellurium
Explanation:
u did the test hope this helps babes
Answer:
Percentage yield = 85.2%
Explanation:
Given data:
Mass of Mg = 21.3 g
Actual yield of MgO = 30.2 g
Percentage yield = ?
Solution:
Chemical equation:
2Mg + O₂ → 2MgO
Number of moles of Mg = mass/molar mass
Number of moles of Mg = 21.3 g / 24.3 g/mol
Number of moles of Mg = 0.88 mol
Now we will compare the moles of MgO with Mg.
Mg : MgO
2 : 2
0.88 : 0.88
Mass of MgO:
Mass of MgO= moles × molar mass
Mass of MgO= 0.88 mol × 40.3g/mol
Mass of MgO = 35.46 g
Actual yield of MgO = 30.2 g
Percentage yield:
Percentage yield = Actual yield/theoretical yield × 100
Percentage yield = 30.2 g/ 35.46 g × 100
Percentage yield = 85.2%
Answer:Student 2
Explanation:
Student 2 repeated the experiment several times with different seeds to make sure the experiment would come out with the same answers and was reliable, using the same area would make sure the environment wouldn't interfere. The other students didn't do all the things that student 2 needed for the experiment.
Answer:
The Retention factor (rf) value is = 0.2
Explanation:
- Retention factor (Rf) is factor used substances that could be separated using Chromatography. Retention factor determines how fast the component can move on the chromatogram (stationary phase) after elution. Elution occurs when mobile phase (solvent) moves across the stationary phase when the solute has been spotted on the origin.
- Retention factor (Rf) ranges from value between 0 and 1. The closer the value to 1, the faster it can move upon elution. Rf can be calculated.
- Rf value = distance moved by the solute / distance moved by the solvent
= 0.40cm / 2.00cm
= 0.2
