the heat of reaction for a chemical reaction
<h3>
Answer:</h3>
2.125 g
<h3>
Explanation:</h3>
We have;
- Mass of NaBr sample is 11.97 g
- % composition by mass of Na in the sample is 22.34%
We are required to determine the mass of 9.51 g of a NaBr sample.
- Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.
In this case,
- A sample of 11.97 g of NaBr contains 22.34% of Na by mass
A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass
% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100
Mass of the element = (% composition of an element × mass of the compound) ÷ 100
Therefore;
Mass of sodium = (22.34% × 9.51 g) ÷ 100
= 2.125 g
Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g
Ni oso sois ene e eisieknew. Jajaj
Lymphocytes and the other formed elements are developed from pluripotent stem cells. The pluripotent stem cells generate myeloid stem cells and lymphoid stem cells. Myeloid cells start and complete their development in red bone marrow and give rise to red blood cells, platelets, eosinophils, basophils, neutrophils, and monocytes. Lymphoid stem cells begin development in the red bone marrow, but some are completed in the lymphatic tissues, where they give rise to lymphocytes. The B cell lymphocytes begin and finish in the red bone marrow and the T cell lymphocytes begin in the red bone marrow, but they mature in the thymus.
Answer:
e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g)
Explanation:
All the following are oxidation–reduction reactions except:________
a. H₂(g) + F₂(g) → 2HF(g). Redox. H is oxidized and F is reduced.
b. Ca(s) + H₂(g) → CaH₂(s). Redox. Ca is oxidized and H is reduced.
c. 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g). Redox. K is oxidized and H is reduced.
d. 6Li(s) + N₂(g) → 2Li₃N(s). Redox. Li is oxidized and N is reduced.
e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g). Not redox. All the elements have the same oxidation number