For the given question above, I think the bond between the carbon atoms is a double bond. <span>2 carbon atoms and 4 hydrogen atoms will give an ethene, which has a C=C bond between carbon atoms. I hope this is the answer you are looking for.</span>
Answer:
Initial rate of the reaction when concentration of hydrogen gas is doubled will be 
.
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
Initial rate of the reaction = R = 
![R = k\times [N_2][H_2]^3](https://tex.z-dn.net/?f=R%20%3D%20k%5Ctimes%20%5BN_2%5D%5BH_2%5D%5E3)
![4.0\times 10^5 M/s=k\times [N_2][H_2]^3](https://tex.z-dn.net/?f=4.0%5Ctimes%2010%5E5%20M%2Fs%3Dk%5Ctimes%20%5BN_2%5D%5BH_2%5D%5E3)
The initial rate of the reaction when concentration of hydrogen gas is doubled : R'
![[H_2]'=2[H_2]](https://tex.z-dn.net/?f=%5BH_2%5D%27%3D2%5BH_2%5D)
![R'=k\times [N_2][H_2]'^3=k\times [N_2][2H_2]^3](https://tex.z-dn.net/?f=R%27%3Dk%5Ctimes%20%5BN_2%5D%5BH_2%5D%27%5E3%3Dk%5Ctimes%20%5BN_2%5D%5B2H_2%5D%5E3)
![R'=8\times k\times [N_2][H_2]^3](https://tex.z-dn.net/?f=R%27%3D8%5Ctimes%20k%5Ctimes%20%5BN_2%5D%5BH_2%5D%5E3)
Initial rate of the reaction when concentration of hydrogen gas is doubled will be .
Answer:A compound is formed as a result of chemical reaction, between the constituent elements
Explanation:
A compound is formed as a result of chemical reaction, between the constituent elements. The properties of compound are different from the properties of the elements from which it is formed. Ex. Compounds can be further divided into three classes : acids, bases and salts, on the basis of their properties.
Molarity= moles/liter, so you would need 3mol KBr/1 liter
(0.3M)(1L)= 0.3mol KBr
<span>B)<span>C2H6O<span>2
</span></span></span>
First, convert each percentage to grams: 38.7g, 9.70g, and 51.6g.
Next, calculate the number of moles of each element, based on the number of grams given.
C = 3.23 mol
H = 8.91 mol
O = 3.23 mol
Set up the ratio of moles of each element:
C3.34H9.70O3.23. Convert the decimals to whole numbers by dividing by the smallest subscript, 3.23.
The empirical formula is CH3O.
Now, compute the formula mass, which is 31. Finally, divide the molecular mass by the formula mass, 62/31 = 2. Multiple the subscripts by 2 to get the molecular formula.
