Answer:Active immunity involves both cell mediated and humoral immunity. Passive immunity is due to the presence of ready-made antibodies. Active immunity is durable. Active immunity offers effective protection against microbes
Explanation:
Answer:

Explanation:
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In this case, the undergoing chemical reaction is:

Thus, in terms of pressures, the rate becomes:

Thus, the rate of change for the partial pressure of ammonia turns out:
![r_{NH_3}=2*(-r_{N_2H_4})\\r_{NH_3}=2*[-(-70torr/h)]\\r_{NH_3}=140torr/h](https://tex.z-dn.net/?f=r_%7BNH_3%7D%3D2%2A%28-r_%7BN_2H_4%7D%29%5C%5Cr_%7BNH_3%7D%3D2%2A%5B-%28-70torr%2Fh%29%5D%5C%5Cr_%7BNH_3%7D%3D140torr%2Fh)
The rate of decrease of partial pressure of urea is taken negative as it is a reactant whereas ammonia a product which has 2 as its stoichiometric coefficient.
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<u>Answer: </u>The molecular weight of the dibasic acid is 89.6 g/mol
<u>Explanation:</u>
Normality is defined as the amount of solute expressed in the number of gram equivalents present per liter of solution. The units of normality are eq/L. The formula used to calculate normality:
....(1)
We are given:
Normality of solution = 
Given mass of solute = 0.56 g
Volume of solution = 250 mL
Putting values in equation 1, we get:

Equivalent weight of an acid is calculated by using the equation:
.....(2)
Equivalent weight of acid = 44.8 g/eq
Basicity of an acid = 2 eq/mol
Putting values in equation 2, we get:

Hence, the molecular weight of the dibasic acid is 89.6 g/mol
Answer:
Heat required = 1.23×10⁵J
Explanation:
Given:
Mass (m) = 500 gram
Specific heat = 6,090 J/g
heat of fusion = 247 J/g.
Find:
Heat required
Computation:
Heat required = 247 J/g× 500 g
Heat required = 1.23×10⁵J