Answer:
n₂ =1.4 mol
Explanation:
Given data:
Mass of nitrogen = 2 g
Initial Volume occupy by nitrogen = 1.25 L
Final volume occupy by nitrogen = 25.0 L
Final number of moles = ?
Solution;
Formula:
V₁ / n₁ = V₂ / n₂
Number of moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 2 g/ 28 g/mol
Number of moles = 0.07 mol
Now we will put the values in formula:
V₁ / n₁ = V₂ / n₂
n₂ = V₂× n₁ /V₁
n₂ = 25 L × 0.07 mol / 1.25 L
n₂ = 1.75 L. mol / 1.25 L
n₂ =1.4 mol
Answer:
The lymphatic system produces white blood cells, known as lymphocytes. There are two types of lymphocyte, T cells and B cells. They both travel through the lymphatic system. As they reach the lymph nodes, they are filtered and become activated by contact with viruses, bacteria, foreign particles, and so on in the lymph fluid.
Explanation:
Answer:
The correct answer is - the number of days the rainwater stands in the jars.
Explanation:
The independent variable or the manipulating variable is a variable in a research or study is the variable which affects the dependent variable or the variable that is going to measure in the study.
In the study, variables that recieves the treatment and changed in every subject are independent variables and in this study number of days the rainwater standing in various jars that might affect the growth of the bacteria in rainwater.
Answer:
Explanation:
In S₈ , there are 8 single bonds which breaks up first . Energy absorbed
= 8 x 240 = 1920 kJ
In S₈ four double bonds of S₂ are formed . Let bond energy be x . In this process energy will be released . energy released in four S₂ molecules formed = 4 x
Given
1920 + 4x = 239
4x = 239 - 1920
x = - 420.25 kJ .
So bond energy of S₂ = 420.25 kJ .
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
