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meriva
3 years ago
5

Hello how to draw the electron dot structure of H2S and F2?​

Chemistry
1 answer:
lidiya [134]3 years ago
5 0

Explanation:

1. Attachment

Electron dot structure of H2S (hydrogen sulfide)

2. Attachment

Electron dot structure of F2 (Fluorine).

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Which subatomic particles have a mass of approximately 1 atomic mass unit each?
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Great question! The subatomic particles that have most of the mass in the particle are the positive proton, and the neutral neutron. Therefore, protons and neutrons are the two subatomic particles have approximately 1 atomic mass unit each.
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Who was J.J. Thomson
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Sir Joseph John Thomson OM PRS was a British physicist and Nobel Laureate in Physics, credited with the discovery of the electron, the first subatomic particle to be discovered.

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3 years ago
Read 2 more answers
4.07 x 10-17<br> (5.6 x 10") (5.8 x 105)
Andrej [43]
4.07 x 10-17 = 23.7
(5.6 x 10”) (5.8 x 105)= log(100)
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3 0
3 years ago
Help on science please!
erik [133]

Answer:

See explanation.

Explanation:

1. Forrest: From 425,000,000 to 300,000,000 is a change of 125,000,000. Divide the 125,000,000 by 425,000,000 to get the percent change:

125,000,000 / 425,000,000 = .294, which is a 29.4% change, a decrease.

2. Population: From 4,500 to 326,500,000 is a change of 326,495,500. Divide 326,495,500 by 4,500 to get the percent change.

326,495,500 / 4,500 = 72,554.55 which is a change of 7,255,455% change, an increase.

3. Conclusion: The growth in population required living area. People needed farms, houses, cities, and communities to live in. Forests were cleared to make room for the people to live. Forests are also used to provide building materials. The population increase contributed to the decrease in forest lands.

6 0
3 years ago
The bomb that destroyed tje Murrah Federal Office Building in Oklahoma City in April 1995 was constructed from ordinary material
Leto [7]

Answer:

ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of  NH₄NO₃(s) given) (exothermic)

Explanation:

           3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g)  +  17H₂O(g)   +  10CO₂(g)

ΔH⁰(f):  3(-365.6)Kj     1(-301)Kj    14(0)Kj       3(0)Kj    17(-241.8)Kj    10(-393.5)Kj

            = -1096.8Kj     = -301Kj     = 0Kj         = 0Kj      = -4110.6Kj    = -3930.5Kj

ΔHₙ°(rxn) = ∑ (ΔH˚(f)products) - ∑(ΔH˚(f)reactants)

= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]

= [-(8041.1) - (-1397.8)]Kj

= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)  

∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)

ΔH°(rxn for 14.11g  NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of  NH₄NO₃(s) given)

5 0
3 years ago
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