Answer:
by statistical analyses, especially by determining the p-value
Explanation:
In general, observations and results obtained from experimental procedures are subjected to a statistical test to check the robustness of the working hypothesis. The p-value is the most widely used statistical index in order to test such observations and results. The p-value is the statistical probability of obtaining extreme observed results when the null hypothesis is considered correct. A p-value lesser than 0.05 generally is considered statistically significant and then the null hypothesis can be rejected. In consequence, a very low p-value (which is obtained by statistical analysis of the observations and results), indicates that there is strong evidence in support of the alternative hypothesis.
Because their species is in danger of going extinct<span />
The more acidic the substance is, the more the iron nails will corrode (this obviously depends on what your experiment is but hope this helped in some way)
<h3>Answer:</h3>
Limiting reactant is Lithium
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of Lithium as 1.50 g
- Mass of nitrogen is 1.50 g
We are required to determine the rate limiting reagent.
- First, we write the balanced equation for the reaction
6Li(s) + N₂(g) → 2Li₃N
From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Second, we determine moles of Lithium and nitrogen given.
Moles = Mass ÷ Molar mass
Moles of Lithium
Molar mass of Li = 6.941 g/mol
Moles of Li = 1.50 g ÷ 6.941 g/mol
= 0.216 moles
Moles of nitrogen gas
Molar mass of Nitrogen gas is 28.0 g/mol
Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol
= 0.054 moles
- According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
- On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.
Thus, Lithium is the limiting reagent while nitrogen is in excess.
It would be solid because you said it dissolves...Hope this helped :)
