Answer: The noble gas preceding tin is Krypton [Kr].
Explanation: Atomic Number of tin is 50 , Atomic Number of neon is 10, Atomic Number of argon is 18, Atomic Number of krypton is 36 and Atomic Number of xenon is 54.
Atomic number tells us the number of electrons present in the atom.
Thus the electronic configuration of tin according to increasing energy of orbitals is written as
.
Physical Properties: Sodium bicarbonate is an odorless, white crystalline solid or fine powder. It has a slightly alkaline taste. Its density is 2.20 g mL-1 and it decomposes in temperatures above 50 ºC. The decomposition yields to sodium carbonate. It is highly soluble in water and poorly soluble in acetone and methanol. It is insoluble in ethanol.
Chemical Properties: Sodium bicarbonate is an amphoteric compounds, it means the compound has a character acids an basic at the same time. It is highly soluble in water, resulting in a slighty alkaline solution.
i hope this part helps i will continue to research on Toothpaste's properties
Answer:
The answer to your question is the letter C) 5648 kJ/mol
Explanation:
Data
C₁₂H₂₂O₁₁ + 12 O₂ ⇒ 12 CO₂ + 11 H₂O
H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol
H° O₂ = 0 kJ / mol
H° CO₂ = -393.5 kJ/mol
H° H₂O = -285.8 kJ/mol
Formula
ΔH° = ∑H° products - ∑H° reactants
Substitution
ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)
ΔH° = -4722 - 3143.8 + 2221.8
Result
ΔH° = -5644 kJ/mol
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:
1.57 mol NaN₃
Explanation:
- 2 NaN₃ (s) → 2 Na (s) + 3 N₂ (g)
First we <u>use PV=nRT to calculate the number of N₂ moles that need to be produced</u>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 23.7 °C ⇒ 23.7 + 273.16 = 296.86 K
<u>Inputing the data</u>:
- 1.07 atm * 53.4 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.86
And <u>solving for n</u>:
Finally we <u>convert N₂ moles into NaN₃ moles</u>, using <em>the stoichiometric coefficients of the balanced reaction</em>:
- 2.35 mol N₂ *
= 1.57 mol NaN₃