Answer:
1. DNA copying mistakes made during cell division, exposure to ionizing radiation, exposure to chemicals called mutagens, or infection by viruses.
2. both
Explanation: Since errors can occur when DNA replicates itself, mutations can occur in both asexually and sexually reproducing organisms.
atoms consist of protons and neutrons in the nucleus , surrounded by electrons that reside in orbitals . Orbitals are classified according to the four quantum number s that represent anyone particular orbitals energy , shape , orientation and the spin of the occupying electron.
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Answer:
dinitrogen pentoxide
Explanation:
Based on the research, dinitrogen pentoxide is a chemical compound containing only nitrogen and oxygen which is both nonmetal. When two nonmetals or two negatively charged atoms try to form bonds, they always form covalent bonds. Hence, Dinitrogen Pentoxide is not ionic - it is covalent.
Answer:
A. To study the chemical composition of the surface of a planet
Explanation:
Strictly speaking, a spectrometer is any instrument used to view and analyze a range (or a spectrum) of a given characteristic for a substance (for example, a range of mass-to-charge values as in mass spectrometry), or a range of wavelengths as in absorption spectrometry like nuclear magnetic radiation spectroscopy
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Answer:
pH = 4.543
Explanation:
- CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
- pKa = - Log Ka
∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
∴ pKa = 4.87
⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
added 300 mL 0f 0.02 M NaOH:
⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)
⇒ <em>C</em> CH3CH2COOH = 0.048 M
⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M
mass balance:
⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)
charge balance:
⇒ [H3O+] + [Na+] = [CH3CH2COO-]
∴ [Na+] = 0.02 M
⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)
(2) in (1):
⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]
replacing in Ka:
⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])
⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]
⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]
⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0
⇒ [H3O+ ] = 2.867 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = 4.543