The percentage yield of the product if the actual yield was measured as 2.80 g and the theoretical yield is measured as 3.12 g is 11.43%.
<h3>How to calculate percentage yield?</h3>
The percentage yield of a substance can be calculated by dividing the difference between the actual and theoretical yield by the actual yield, then multiplied by 100.
Percentage yield = (theoretical yield - actual yield)/actual yield × 100
According to this question, the actual yield of a product in a reaction was measured as 2.80 and the theoretical yield of the product for the reaction is 3.12 g.
Percent yield = (3.12g - 2.80g)/2.80 × 100
Percent yield = 11.43%.
Therefore, percentage yield of the product if the actual yield was measured as 2.80 g and the theoretical yield is measured as 3.12 g is 11.43%.
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Answer:
See explanation
Explanation:
The third and fourth elements in the first transition series are vanadium and chromium. They contain 23 and 24 electrons respectively.
In universe L, the electron configuration of vanadium is;
[X] 3d3 4s2
In universe L, the electron configuration of chromium is;
[X] 3d4 4s2
The electron configuration of chromium has to change since there are four instead of five d orbitals in universe L.
The wavelength of the photon required to excite this molecule from its ground state, to its first excited state is 1240 nm.
This is given by the equation:
wavelength = hc/(E_homo - E_lumo)
where h is Planck's constant =6.626070 * 10^-34 J.m , c is the speed of light = 3.0 x 10^8 m/s^2, and E_homo and E_lumo are the energies of the highest occupied molecular orbital and the lowest unoccupied molecular orbital, respectively.
In this particular case, the wavelength of the required photon would be:
wavelength = hc/(-2.42 hartree - 0.65 hartree)
= 6.626070 * 10^-34 X 3.0 x 10^8 / (-3.07)
= 1240 nm
Hence , The wavelength of the photon required to excite this molecule from its ground state, to its first excited state is 1240 nm.
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Answer:
Requires two Mg²⁺ cofactors and as such demonstrates metal-ion catalysis
Explanation:
Electrostatic catalysis or metal ion catalysis is a catalytic mechanism that makes use of metalloenzymes, such as enolase along with a metal ion which is bound tightly, including, Mo⁶⁺, Ni³⁺, Co³⁺, Mn²⁺, Zn²⁺, Cu²⁺, and Fe²⁺, to undertake a catalysis
For maximal activity, enolase requires the presence of 2 equivalent metal ions in each site which is active
Therefore, the correct option is; requires two Mg²⁺ cofactors and as such demonstrates metal-ion catalysis
Explanation: i stuck at this Answer:Sometimes atoms share more than one pair of electrons, for example a double covalent bond can occur, as in the carbon dioxide molecule where carbon shares 2 pairs of electrons with each oxygen atom: O=C=O. (where = represents a double bond). Another kind of covalent bonding is called co-ordinate or dative covalent.
