The answer is lithium chlorite.
Answer:
Solid Osmium transition metal reacts with Oxygen gas to produce solid Osmium tetroxide.
Os(s) + 2O₂(g) -> OsO₄(s)
Explanation:
Osmium tetroxide is another way of writing Osmium (VIII) oxide.
Leaving powdered osmium exposed to air in a room will slowly create osmium tetroxide at room temperature.
Similarly, osmium tetroxide vapor will readily be released from a liquid solution at room temperature.
Answer:
I think the insect would be the one in the second photo reading from left to right
Explanation:
the other insects:
the first of the first photo is a change of what would be the external surface of the insect, it is not the insect itself.
In the third photo, the insect has short legs with little grip, which indicates that it does not climb large areas on top, and finally, the 4 image is of an insect that lives in desert areas and is poorly coordinated in areas of high heat. and sand.
These are the reasons why I chose photo number two reading from left to right
The phenotype for the genotype bb would be Purple and white
Answer:
41.9(w/w) %
Explanation:
Based on the reaction:
Na₂C₂O₄(s) + 2HCl(aq) → H₂C₂O₄(aq) + 2NaCl(aq)
<em>Where 1 mole of sodium oxalate reacts with 2 moles of HCl</em>
Moles of HCl solution to reach end point are:
44.15mL = 0.04415L ₓ (0.250mol / L) = 0.01104 moles of HCl
As 2 moles of HCl reacts per mole of Na₂C₂O₄:
0.01104mol HCl ₓ (1 mol Na₂C₂O₄ / 2 mol HCl) = <em>5.519x10⁻³ moles Na₂C₂O₄</em> are in the sample.
Molar mass of Na₂C₂O₄ is 134g/mol; thus, mass of 5.519x10⁻³ moles Na₂C₂O₄ is:
5.519x10⁻³ moles Na₂C₂O₄ ₓ (134g / mol) = <em>0.740g of Na₂C₂O₄</em> in the sample.
Thus, percent by mass of sodium oxalate in the sample is:
0.740g of Na₂C₂O₄ / 1.766g ₓ 100 =
<h3>41.9(w/w) %</h3>