Answer: The density of silver metal will be ![10.50g/ml[\tex]Explanation:Density is defined as the mass contained per unit volume.[tex]Density=\frac{mass}{Volume}](https://tex.z-dn.net/?f=10.50g%2Fml%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3EExplanation%3A%3C%2Fp%3E%3Cp%3EDensity%20is%20defined%20as%20the%20mass%20contained%20per%20unit%20volume.%3C%2Fp%3E%3Cp%3E%5Btex%5DDensity%3D%5Cfrac%7Bmass%7D%7BVolume%7D)
Given : Mass of silver = 194.3 grams
Volume of silver= volume of water displaced= ![260.5-242.0=18.5mltex]Putting in the values we get:[tex]Density=\frac{194.3g}{18.5ml}=10.50g/ml/tex]Thus density of silver metal will be [tex]10.50g/ml](https://tex.z-dn.net/?f=260.5-242.0%3D18.5mltex%5D%3C%2Fp%3E%3Cp%3EPutting%20in%20the%20values%20we%20get%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DDensity%3D%5Cfrac%7B194.3g%7D%7B18.5ml%7D%3D10.50g%2Fml%2Ftex%5D%3C%2Fp%3E%3Cp%3EThus%20density%20of%20silver%20metal%20will%20be%20%5Btex%5D10.50g%2Fml)
Answer: An ion with fewer electrons than its neutral atom is called a(n) cation. The charge of an ion with more electrons than its neutral atom is. negative.Name an ionic compound by the cation followed by the anion. ... Covalent compounds are formed when two or more nonmetal atoms bond by sharing valence electrons. Valence electrons are the outermost electrons of an atom.
Explanation:
The reaction to form NH3 is : N2 + 3H2-> 2NH3 12,33g NH3 is 12,33/17,03=0,3 =0,724 moles of NH3 moles NH3. So you need 1,5*0,724 = 1,086 moles H2 1,086*2,016 = 2,189 g of H2 is needed ro form 12,33 g NH3
Answer:

Explanation:
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In this case, when looking over the chemical formula of a compound formed when two elements interact, we need to realize about their location in the periodic table; thus, since lithium is a metal and oxygen a nonmetal, we infer lithium turns out the cation and oxygen the anion. Moreover, as the oxidation state of lithium is 1+ and that of oxygen is 2-, we can set up the chemical formula as follows:

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Answer:

Explanation:
First you should calculate the volume of a big sphere,so:



Then you calculate the volume of a small spehre, so:



Finally you subtract the two quantities:

