Answer:
c) parallel to the wire
Explanation:
The concept is based on right-hand thumb rule that states that the thumb points in the direction of the current and the fingers will represent the direction of the lines of magnetic force.
As the current is moving in the east-west direction. Using the right-hand thumb rule, the direction of the magnetic field at a point below the wire is from north to south. So, when viewed from the west end, the magnetic field at a point will be anti- clockwise.
Hence, the correct answer is "b) anti- clockwise".
Answer:
Four substitution products are obtained. The carbocation that forms can react with either nucleophile (H2O or CH3OH) from either the top or bottom side of the molecule
Explanation:
An SN1 reaction usually involves the formation of a carbocation in the slow rate determining step. This carbocation is now attacked by a nucleophile in a subsequent fast step to give the desired product.
However, the product is obtained as a racemic mixture because the nucleophile may attack from the top or bottom of the carbocation hence both attacks are equally probable.
The attacking nucleophile in this case may be water or CH3OH
The missing part of the equation is found to be 4/2He. Option A
<h3>What are nuclear equations?</h3>
The term nuclear equations have to do with the type of equation in which one type of nucleus is transformed into another sometimes by the bombardment or loss of a particle.
Now the full equation ought to be written as 7/3Li + 1/1H -----> 4/2He + 4/2He. This is because the total mass on the left is 8 and the total charge on the left is 4.
Learn more about nuclear equations:brainly.com/question/19752321
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Hehehehwgwgw. Be the hardest thing ever for a long day and I have a windows
The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.
The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.
Now, we will consider each reaction individually to decode whether or not a precipitate is possible.
- In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
- In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
- In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.
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