Answer:
As the electrons flow through the wire, <em>electric current </em>is generated around the wire.
Explanation:
The rate of flow of charged particles in a given time is termed as the flow of current. Mostly the charge carriers are termed as electrons in a conductor. So the flow of electrons or movement of charged particles in a given time is the generation of electric current in that current. As the ratio of charge to time at which the charge is moving from one point to another is termed as the current flow in that time.
Thus, in the present case of electrons flowing in that wire will lead to generating of electric current in the opposite direction around the wire.
In the follow
<span>1)N<span>H4</span>OH,N<span>H4</span>Cl<span>O3</span>,(N<span>H4</span><span>)2</span>S<span>O3</span>,(N<span>H4</span><span>)2</span>HP<span>O4</span></span>
<span>2)Al(OH<span>)3</span>,Al(Cl<span>O3</span><span>)3</span>,A<span>l2</span>(S<span>O3</span><span>)3</span>,A<span>l2</span>(HP<span>O4</span><span>)3</span></span>
<span><span>3)Pb(OH<span>)4</span>,Pb(Cl<span>O3</span><span>)4</span>,Pb(S<span>O3</span><span>)2</span>,Pb(HP<span>O4</span><span>)2</span></span></span>
Answer: noble gasses
Explanation: zoom on on the red section and look at the key for what red means.
First, recognize that this is an elimination reaction in which hydroxide must leave and a double bond must form in its place. It is likely an E2 reaction. Here is an efficient mechanism:
1) Pre-reaction: Protonate the -OH to make it a good leaving group, water. H2SO4 or any strong H+ donor works. The water is positively charged but still connected to the compound.
2) E2: Use a sterically hindered base, such as tert-butoxide (tButO-) to abstract the hydrogen from the secondary carbon. [You want a sterically hindered base because a strong, non-sterically hindered base could also abstract a hydrogen from one of the two methyl groups on the tertiary carbon, and that leads to unwanted products, which is not efficient]. As the proton of hydrogen is abstracted, water leaves at the same time, creating an intermediate tertiary carbocation, and the 2 electrons in the C-H bond immediately are used to make a double bond towards the partial positive charge.
In the products we see the major product and water, as expected. Even though you have an intermediate, remember that an E2 mechanism technically happens in one step after -OH protonation.
Answer:
4.14 x 10²⁴ molecules CO₂
Explanation:
2 C₄H₁₀ + 13 O₂ --> 8 CO₂ + 10 H₂O
To find the number of CO₂ molecules, you need to start with 100 grams of butane (C₄H₁₀), convert to moles (using the molar mass), convert to moles of CO₂ (using coefficients from equation), then convert to molecules (using Avagadro's number). The molar mass of C₄H₁₀ is calculated using the quantity of each element (subscript) multiplied by the number on the periodic table. The ratios should be arranged in a way that allows for units to be cancelled.
4(12.011g/mol) + 10(1.008 g/mol) = 58.124 g/mol C₄H₁₀
100 grams C₄H₁₀ 1 mol C₄H₁₀ 8 mol CO₂
-------------------------- x ---------------------- x ---------------------
58.124 g 2 mol C₄H₁₀
6.022 x 10²³ molecules
x ------------------------------------ = 4.14 x 10²⁴ molecules CO₂
1 mol CO₂
