Give the abbreviated electron configuration for silver, 47Ag, and underline the valence electrons.

Give the abbreviated electron configuration for silver, 47Ag, and underline the valence electrons. The Correct Answer and Explanation is: Abbreviated Electron Configuration for Silver (47Ag):[Kr] 4d10 5s1‾\text{[Kr]} \, 4d^{10} \, \underline{5s^1}[Kr]4d105s1​ Explanation Silver (Ag), with atomic number 47, is a transition metal located in group 11 of the periodic table. To understand its electron configuration, we begin

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Involved in the transport of substances within the neuron 7

Involved in the transport of substances within the neuron 7. essentially rough endoplasmic reticulum, important metabolically 8. impulse generator and transmitter Column B a. axon b. axon terminal c. axon hillock d. cell body e. chromatophilic substance f. dendrite g. myelin sheath h. neurofibril 5. Draw a “typical” multipolar neuron in the space below. Include

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Factor -6 out of 18z – 15.

Factor -6 out of 18z – 15. The Correct Answer and Explanation is: Factoring –6 out of the Expression: Given expression:18z – 15 We are asked to factor –6 out of this expression. Step-by-step Factoring: We factor –6 out of both terms:18z−15=−6(−3z+156)18z – 15 = -6(-3z + \frac{15}{6})18z−15=−6(−3z+615​) But we can simplify further. First, divide

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A map is drawn using the scale 2 cm:100 mi. On the map, Town B is 3.5 centimeters from Town A, and Town C is 2 centimeters past

A map is drawn using the scale 2 cm:100 mi. On the map, Town B is 3.5 centimeters from Town A, and Town C is 2 centimeters past Town B. How many miles apart are Town A and Town C? A.) 275 miles B.) 200 miles C.) 550 miles D.) 1,100 miles The Correct Answer

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