A solution is the answer to a problem
Answer:
Explanation:
Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.
Hence, the rate law is:
![r=d[A]/dt=-k[A]](https://tex.z-dn.net/?f=r%3Dd%5BA%5D%2Fdt%3D-k%5BA%5D)
Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:
![[A]=[A]_0e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_0e%5E%7B-kt%7D)
You know [A]₀, k, and t, thus you can calculate [A].
![[A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}](https://tex.z-dn.net/?f=%5BA%5D%3D0.548M%5Ctimes%20e%5E%7B-3.6%5Ccdot%2010%5E%7B-4%7D%2Fs%5Ctimes99.2s%7D)
![[A]=0.529M](https://tex.z-dn.net/?f=%5BA%5D%3D0.529M)
First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2
It’s positive when you use energy for work
The atoms of hydrogen that are present in 7.63 g of ammonia(NH3)
find the moles of NH3 =mass/molar mass
7.63 g/ 17 g/mol = 0.449 moles
since there is 3 atoms of H in NH3 the moles of H = 0.449 x 3 = 1.347 moles
by use of 1 mole = 6.02 x10^23 atoms
what about 1.347 moles
= 1.347 moles/1 moles x 6.02 x10^23 atoms = 8.11 x10^23 atoms of Hydrogen
