Answer:
A. m C5H12 = 108.23 g
B. m F2 = 547.142 g
C. m Ca(CN)2 = 71.85 g
Explanation:
- mass (m) = mol (n) × molecular weigth (Mw)
∴ Mw C5H12 = ((12.011)(5)) + ((1.008)(12)) = 72.151 g/mol C5H12
∴ Mw F2 = (18.998)(2) = 37.996 g/mol F2
∴ Mw = Ca(CN)2 = 40.078+((12.011+14.007)(2)) = 92.114 g/mol Ca(CN)2
A. m C5H12 = ( 1.50 mol)×(72.151 g/mol) = 108.23 g C5H12
B. m F2 = (14.4 mol)×(37.996 g/mol) = 547.142 g F2
C. m Ca(CN)2 = (0.780 mol)×(92.114 g/mol) = 71.85 g Ca(CN)2
Answer:
E = 3.77×10⁻¹⁹ J
Explanation:
Given data:
Wavelength of absorption line = 527 nm (527×10⁻⁹m)
Energy of absorption line = ?
Solution:
Formula:
E = hc/λ
h = planck's constant = 6.63×10⁻³⁴ Js
c = speed of wave = 3×10⁸ m/s
by putting values,
E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 527×10⁻⁹m
E = 19.89×10⁻²⁶ Jm /527×10⁻⁹m
E = 0.0377×10⁻¹⁷ J
E = 3.77×10⁻¹⁹ J
When using ion-selective electrodes, to compensate for a complex or unknown matrix, the standard addition method can be used to determine the analyte concentration. Option D
<h3>What are ion-selective electrodes?</h3>
Analytical chemistry is a science that deal with the measurement and detection of the accurate amount of a substance. Analytical chemistry plays a large role in environmental management as it helps in the determination of the levels of contaminants in a sample.
An ion selective electrode is used in analytical chemistry to measure the amount of a target ion by converting its activity into a measurable electrical signal.
Hence, when using ion-selective electrodes, to compensate for a complex or unknown matrix, the standard addition method can be used to determine the analyte concentration.
Learn more about ion-selective electrodes:brainly.com/question/14987024
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35 b.
37 h
39 i
41 d
hope that helps
Answer:
88.46%
Explanation:
Percentage yield is actual/theoretical * 100
138/156 * 100 = 88.4615385