Earth is so hot inside that magma erupts and flies out. LOL hope this helps.
Bromine has the following electron configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5. categorize the electrons in each. Answer for video: The video player is loaded.
On the periodic chart, row 5, column 7, is where you can find a chemical element that was identified in 1811. It has a proton count of 53 and an atomic mass of 126.9. Iodine's atom, then, contains 53 electrons in the following configuration: 1s2, 2s2, 2p6, 3s2, 3d10, 4p6, 5s2, 4d10, 5p5 (Kr 4d10 5s2 5p5). Cu Z = 29 has an electrical arrangement of 1s2 2s2 2p6 3s2 3p6 3d10 4s1. Copper (Co) has the following electron configuration: 1s2 2s2 2p6 3s3 3p6 4s2 3d7. If a chemist were to refer to Copper by its subshell, they would abbreviate this notation to "3d7."
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Answer:
Mass of C₂H₄N₂ produced = 3.64 g
Explanation:
The balanced chemical equation for the reaction is given below:
3CH₄ (g) + 5CO₂ (g) + 8NH₃ (g) → 4C₂H₄N₂ (g) + 10H₂O (g)
From the equation, 3 moles of CH₄ reacts with 5 moles of CO₂ and 8 moles of NH₃ to produce 4 moles of C₂H₄N₂ and 10 moles of H₂O
Molar masses of the compounds are given below below:
CH₄ = 16 g/mol; CO₂ = 44 g/mol; NH3 = 17 g/mol; C₂H₄N₂ = 56 g/mol; H₂O g/mol
Comparing the mole ratios of the reacting masses;
CH₄ = 1.65/16 = 0.103
CO₂ = 13.5/44 = 0.307
NH₃ = 2.21/17 = 0.130
converting to whole number ratios by dividing with the smallest ratio
CH₄ = 0.103/0.103 = 1
CO₂ = 0.307/0.103 = 3
NH₃ = 0.130/0.103 = 1.3
Multiplying through with 5
CH₄ = 1 × 5 = 5
CO₂ = 3 × 5 = 15
NH₃ = 1.3 × 5 = 6.5
Therefore, the limiting reactant is NH₃
8 × 17 g (136 g) of NH₃ reacts to produce 4 × 56 g (224 g) of C₂H₄N₂
Therefore, 2.21 g of NH₃ will produce (2.21 × 224)/136 g of C₂H₄N₂ = 3.64 g of C₂H₄N₂
Mass of C₂H₄N₂ produced = 3.64 g
Iridium-192 is used in cancer treatment, a small cylindrical piece of 192 Ir, 0.6 mm in diameter (0.3mm radius) and 3.5 mm long, is surgically inserted into the tumor. if the density of iridium is 22.42 g/cm3, how many iridium atoms are present in the sample?
Let us start by computing for the volume of the cylinder. V = π(r^2)*h where r and h are the radius and height of the cylinder, respectively. Let's convert all given dimensions to cm first. Radius = 0.03 cm, height is 0.35cm long.
V = π * (0.03cm)^2 * 0.35 cm = 9.896*10^-4 cm^3
Now we have the volume of 192-Ir, let's use the density provided to get it's mass, and once we have the mass let's use the molar mass to get the amount of moles. After getting the amount of moles, we use Avogadro's number to convert moles into number of atoms. See the calculation below and see if all units "cancel":
9.896*10^-4 cm^3 * (22.42 g/cm3) * (1 mole / 191.963 g) * (6.022x10^23 atoms /mole)
= 6.96 x 10^19 atoms of Ir-122 are present.