Answer:
it's an indicator that a chemical reaction occurred.
I think it’s D. Stars and planets.
a) The change here is that metallic iron is converted into ions and copper is deposited. This is called a displacement reaction.
b)
is oxidised in this reaction.
c)$ \mathrm{Fe_{(s)}+ CuSO_{4(aq)} \rightarrow FeSO_{4(aq)} + Cu_{(s)}}$
Answer:
The rate at which
is being produced is 0.0228 M/s.
The rate at which
is being consumed is 0.0912 M/s.
Explanation:

Rate of the reaction : R
![R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B-1%7D%7B4%7D%5Cfrac%7Bd%5BPH_3%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B1%7D%5Cfrac%7Bd%5BP_4%5D%7D%7Bdt%7D)
The rate at which hydrogen is being formed = ![\frac{d[H_2]}{dt}=0.137 M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D0.137%20M%2Fs)
![R=\frac{1}{6}\frac{d[H_2]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D)

The rate at which
is being produced:
![R=\frac{1}{1}\frac{d[P_4]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B1%7D%7B1%7D%5Cfrac%7Bd%5BP_4%5D%7D%7Bdt%7D)
![0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}](https://tex.z-dn.net/?f=0.0228%20M%2Fs%3D%5Cfrac%7B1%7D%7B1%7D%5Cfrac%7Bd%5BP_4%5D%7D%7Bdt%7D)
The rate at which
is being consumed :
![R=\frac{-1}{4}\frac{d[PH_3]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B-1%7D%7B4%7D%5Cfrac%7Bd%5BPH_3%5D%7D%7Bdt%7D)
![0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}](https://tex.z-dn.net/?f=0.0228%20M%2Fs%5Ctimes%204%3D%5Cfrac%7B-1%7D%7B1%7D%5Cfrac%7Bd%5BPH_3%5D%7D%7Bdt%7D)
![\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B1%7D%5Cfrac%7Bd%5BPH_3%5D%7D%7Bdt%7D%3D0.912%20M%2Fs)
Answer:
d = 9.5× 10⁻⁴ Kg/cm³
Explanation:
Given data:
Length of cube = 1.02 cm
Mass of cube = 0.001012 Kg
Density of cube = ?
Solution:
The length , height and width of cube is always same. Thus the volume of cube is,
Volume = length × height× width
Volume = 1.02 cm × 1.02 cm × 1.02 cm
Volume = 1.06 cm³
Density of cube:
d = 0.001012 Kg /1.06 cm³
d = 9.5× 10⁻⁴ Kg/cm³