Answer:
.081 g of O2
Explanation:
4Cr + 3O2 -----> 2Cr2O3
.175 g Cr x [1 mole / 52.0 g] x [2 moles Cr2O3 / 4 moles Cr] x [152 g / 1 mole] = .256 g of Cr2O3
.175 g Cr x [1 mole / 52.0 g] x [3 moles O2 / 4 moles Cr] x [32 g / 1 mole] = .081 g of O2
Answer : The ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2
Explanation :
The relation between the equilibrium constant and standard Gibbs free energy is:
![\Delta G^o=-RT\times \ln Q\\\\\Delta G^o=-RT\times \ln (\frac{[A]_{inside}}{[A]_{outside}})](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-RT%5Ctimes%20%5Cln%20Q%5C%5C%5C%5C%5CDelta%20G%5Eo%3D-RT%5Ctimes%20%5Cln%20%28%5Cfrac%7B%5BA%5D_%7Binside%7D%7D%7B%5BA%5D_%7Boutside%7D%7D%29)
where,
= standard Gibbs free energy = -14.1 kJ/mol
R = gas constant = 8.314 J/K.mol
T = temperature = 
Q = reaction quotient
![[A]_{inside}](https://tex.z-dn.net/?f=%5BA%5D_%7Binside%7D)
= concentration inside the cell
![[A]_{outside}](https://tex.z-dn.net/?f=%5BA%5D_%7Boutside%7D)
= concentration outside the cell
Now put all the given values in the above formula, we get:
![-14.1\times 10^3J/mol =-(8.314J/K.mol)\times (298K)\times \ln (\frac{[A]_{inside}}{[A]_{outside}})](https://tex.z-dn.net/?f=-14.1%5Ctimes%2010%5E3J%2Fmol%20%3D-%288.314J%2FK.mol%29%5Ctimes%20%28298K%29%5Ctimes%20%5Cln%20%28%5Cfrac%7B%5BA%5D_%7Binside%7D%7D%7B%5BA%5D_%7Boutside%7D%7D%29)
![\frac{[A]_{inside}}{[A]_{outside}}=296.2](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5D_%7Binside%7D%7D%7B%5BA%5D_%7Boutside%7D%7D%3D296.2)
Thus, the ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2
Answer:
Boron
Explanation:
You can find this by looking at the number of protons in Boron, 5.
Then calculate how many electrons you are given, in this case the 2 core plus the 3 valence equal 5 total electrons
Neutral elements have the same number of protons and electrons, so your answer would be the element with 5 electrons, Boron.
You can also know this by using electron configuration. Since you kow there are 5 electrons then you can use EC to find out where your element is. In this case it is: 1s2 2s2 2p1
<em>The number of protons and electrons in a neutral atom of the element are;</em>
D. 29 protons and 29 electrons
<u>The atomic number of an element is equal to the number of protons in the nucleus of each atom of that element.</u>
<u>Thus copper has an atomic number of 29, all atoms of copper will have 29 protons.</u>
