The concentration of ca 2 in blood plasma in equivalent if its concentration is 0.0080.
Concentration means the amount of or parts in terms of the alternative components. For acid–base reactions, the equivalent weight of an acid or base is the mass which materials or reacts with one mole of hydrogen
In chemistry, equal weight (additionally known as gram equal is the mass of 1 equivalent, that is the mass of a given substance as a way to integrate with or displace a fixed amount of any other substance. The equivalent weight of an detail is the mass which combines with or displaces 1.008 gram of hydrogen or eight. Zero grams of oxygen or 35.five grams of chlorine. those values correspond to the atomic weight divided through the usual valence for oxygen as example that is sixteen.
Charge on one Ca 2+ ion = + 2 .
Concentration of calcium ion = 0.0040M
⇒ Concentration of Ca 2 + in blood plasma = 0.0040 × 2
= 0.0080 equivalent Ca 2 answer
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Answser
i didnt do dis my mom has a answer key book
Explanation:
Answer:
A. C(s) + O2(g) → CO2(g) + 393.5 kJ
Explanation:
The formation of carbon IV oxide (CO2) is an exothermic process. n exothermic process is a chemical process in which heat is evolved. Speaking in lay man's terminology, heat is one of the 'products' of the reaction. This implies that heat is evolved by the process.
The formation of CO2 is always a combustion reaction where heat is evolved or released by the reaction system. Hence the reaction could be shown as;
C(s) + O2(g) → CO2(g) + 393.5 kJ
This implies that 393.5 kJ of energy is released in the reaction in the form of heat. Hence the answer given in the answer box.
Answer:
+523 kJ.
Explanation:
The following data will be used to calculate the average C-S bond energy in CS2(l).
S(s) ---> S(g)
ΔH = 223 kJ/mol
C(s) ---> C(g)
ΔH = 715 kJ/mol
Enthalpy of formation of CS2(l)
ΔH = 88 kJ/mol
CS2(l) ---> CS2(g)
ΔH = 27 kJ/mol
CS2(g) --> C(g) + 2S(g)
So we must construct it stepwise.
1: C(s) ---> C(g) ΔH = 715 kJ
2: 2S(s) ---> 2S(g) ΔH = 446 kJ
adding 1 + 2 = 3
ΔH = 715 + 446
= 1161 kJ
3: C(s) + 2S(s) --> C(g) + 2S(g) ΔH = 1161 kJ
4: C(s) + 2S(s) --> CS2(l) ΔH = 88 kJ
adding (reversed 3) from 4 = 5
ΔH = -1161 + 88
= -1073 kJ
5: C(g) + 2S(g) --> CS2(l) ΔH = -1073 kJ
6: CS2(l) ---> CS2(g) ΔH = 27 kJ
adding 5 + 6 = 7
ΔH = -1073 + 27
= -1046 kJ
7. C(g) + 2S(g) --> CS2(g) ΔH = -1046 kJ
Reverse and divide by 2 for C-S bond enthalpy
= -(-1046)/2
= +523 kJ.
It is the answer of 10 51
