First you need to know the molecular weight of sugar (C6H12O6) which is 180.156g/mol
You have half a mole so you have 90.078g
If you wanted to make 1L of a 1.2M solution of glucose you would need 180.156*1.2=216.1872g
But you only have 90.078g
So you need to figure out how much this 90.078g will make if the solution must be 1.2M:
90.078g/216.1872g=xL/1L
solve for the X and you get 0.416666666...
so 416.7ml or 0.417L
Answer:
The mass of 2.35 moles of (NH4)3PO4 is 411.156 or 411.156 grams/g.
Explanation:
By calculating the mass of the formula, (NH4)3PO4, which equals up to 174.96 grams you divide 2.35 moles by 174.96 grams to equal 411.156 grams. The way to find out the mass is to multiple the molar mass of each chemical to the number beside it such as H multiplied by 4, P multiplied by 3, and so on.
Answer:
![[A]_0=0.400M](https://tex.z-dn.net/?f=%5BA%5D_0%3D0.400M)
Explanation:
Hello.
In this case, since the first-order reaction is said to be linearly related to the rate of reaction:
![r=-k[A]](https://tex.z-dn.net/?f=r%3D-k%5BA%5D)
Whereas [A] is the concentration of hydrogen peroxide, when writing it as a differential equation we have:
![\frac{d[A]}{dt} =-k[A]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3D-k%5BA%5D)
Which integrated is:
![ln(\frac{[A]}{[A]_0} )=-kt](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B%5BA%5D%7D%7B%5BA%5D_0%7D%20%29%3D-kt)
And we can calculate the initial concentration of the hydrogen peroxide as follows:
![[A]_0=\frac{[A]}{exp(-kt)}](https://tex.z-dn.net/?f=%5BA%5D_0%3D%5Cfrac%7B%5BA%5D%7D%7Bexp%28-kt%29%7D)
Thus, for the given data, we obtain:
![[A]_0=\frac{0.321M}{exp(-2.54x10^{-4}s^{-1}*855s)}](https://tex.z-dn.net/?f=%5BA%5D_0%3D%5Cfrac%7B0.321M%7D%7Bexp%28-2.54x10%5E%7B-4%7Ds%5E%7B-1%7D%2A855s%29%7D)
Best regards!
