CuCl2+F2—>CuF2+Cl2.
This is a single replacement because there is one compound and one element. Picture Cu as ‘A’ Cl2 as ‘B’ and F2 as ‘C.’ So AB+C—>AC+B. A and B “broke up” and that resulted to A going with C to create the compound CuF2 leaving Cl2 alone.
Answer:
u just have to multiply 20x12 because its tells ya how fast the cart will be moving at 12 sec.
Explanation:because its tells ya how fast the cart will be moving at 12 sec.
OP already did it - CONGRATS!!
here are the steps 2 get the same ans:
(NH4)2 CO3 has 2x N, 8x H, 1x C and 3x O per molecule
so its molecular mass = 2x14 + 8x1 + 1x12 + 3x16
=28+8+12+48
=96g
of that 96g, 8x1=8g is due to Hydrogen
so by ratio n proportion, 1.00g will have 1x8/96 = 1/12g = 0.083g of H
greater the particles move on average because as the higher the temperature the more energy the particles have to the more they move .
The image provided shows the diene that is used for this question. We are told that the major product of the reaction is the 1,4-addition product. The result is the addition of a bromine atom to the first carbon and the fourth carbon of the diene. However, Br₂ can also add to just one alkene of the diene in a 1,2-addition to get the other product shown in the image.
As the first bromine atom adds to one of the alkenes, it adds to the first carbon which leads to the formation of a carbocation. The carbocation can be a stable tertiary center at the 2 carbon of the diene, or the less stable secondary center of the 4 carbon. To addition to the 4-carbon has a higher activation barrier which, but the product has a lower energy than the 1,2-product. Therefore, the 1,4-addition is the thermodynamic product and will form at higher temperatures. The 1,2-product is the kinetic product that will form at lower temperatures.