Answer: -
6
Explanation: -
The given unbalanced chemical equation is As + NaOH -- > Na3AsO3 + H2
We see there 3 sodium on the right side from Na3AsO3.
But there are only 1 sodium on the left from NaOH.
So we multiply NaOH by 3.
As + 3 NaOH -- > Na3AsO3 + H2
Now we see the number of Hydrogen on the left is 3.
But the number of hydrogens is 2 on the left.
So, we multiply to get both sides 6 hydrogen.
As + 6NaOH -- > Na3AsO3 + 3 H2
Rebalancing for Na,
As + 6NaOH -- > 2Na3AsO3 + 3 H2.
Finally balancing As,
2 As + 6 NaOH -- > 2Na3AsO3 + 3H2
The coefficient of the NaOH molecule in the balanced reaction is thus 6
Answer:
1.
work out the mean mode median and range
Explanation:
Explanation:
Haemoglobin consists of heme unit which is comprised of an <u>
</u> and porphyrin ring. The ring has four pyrrole molecules which are linked to the iron ion. In oxyhaemoglobin, the iron has coordinates with four nitrogen atoms and one to the F8 histidine residue and the sixth one to the oxygen. In deoxyhaemoglobin, the ion is displaced out of the ring by 0.4 Å.
The prosthetic group of hemoglobin and myoglobin is - <u>Heme</u>
The organic ring component of heme is - <u>Porphyrin</u>
Under normal conditions, the central atom of heme is - <u></u>
In <u>deoxyhemoglobin</u> , the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system.
The central atom has <u>six</u> bonds: <u>four</u> to nitrogen atoms in the porphyrin, one to a <u>histidine</u> residue, and one to oxygen.
You can't usually just use a single spectrum line to confirm the identity of an element because there are cases that the emission line id not clearly defined. When the emission line is very weak compared to surrounding noise, in which case the more datapoints you have to build up confidence for the existence of a particular emission spectra, the better.
Answer:
-12
Explanation:
a negative times a positive is negative
