A. 599 mg x 1g/1000mg x 1mole/174.24g = 0.003438 moles
<span>b. 0.003438 moles x 1000 mmoles/mole = 3.438 moles </span>
<span>c. 0.003438 moles/0.741 L = 0.0046 M </span>
<span>d. [K+] = 2 x 0.0046 = 0.0092 M because each mole K2 SO4 yields 2 moles K+ </span>
<span>e. [SO42-] = 0.0046 M </span>
<span>f. 599 mg/0.741 L = 808.4 mg/L = 808 ppm </span>
<span>g. 0.599g/741 ml x 100 ml = 0.08 g/100 ml = 0.08% w/v </span>
<span>h. pK+ = -log [K+] = -log 0.0092 = 2.04 </span>
<span>i. pSO42- = -log [SO42-] = -log 0.0046 = 2.34</span>
Below are I think the data for this problem:
Given the following data:
<span>Ca (s) + 2 C (graphite) → CaC2 (s) ∆H = -62.8 kJ </span>
<span>Ca (s) + ½ O¬2 (g) → CaO (s) ∆H = -635.5 kJ </span>
<span>CaO (s) + H2O (l) → Ca(OH)2 (aq) ∆H = -653.1 kJ </span>
<span>C2H2 (g) + 5/2 O¬2 (g) → 2 CO2 (g) + H2O (l) ∆H = -1300 kJ </span>
<span>C (graphite) + O¬2 (g) → CO2 (g) ∆H = -393.51 kJ
</span>
Below is the answer:
CaC2 (s) + 2 H2O (l) → Ca(OH)2 (aq) + C2H2 (g)
<span>So what you do is: </span>
<span>Times the first equation by -1 Second by 1 Third By 1 Fourth by -1 and Fifth by 2 </span>
<span>So This gives us: </span>
<span>1.CaC2--> Ca+2C </span>
<span>2.Ca+1/2O2-->CaO </span>
<span>3.CaO+H2O-->Ca(OH)2 </span>
<span>4.2CO2+H2O-->C2H2+5/2O2 </span>
<span>5.2C+202-->2CO2 </span>
<span>Now you cancel out like terms on either sides of the equation and you end up with </span>
<span>CaC2 (s) + 2 H2O (l) → Ca(OH)2 (aq) + C2H2 (g) Just what you wanted </span>
<span>So to calculate ∆H: </span>
<span>62.8-635.5-653.1+1300-787.02= -712.82</span>
Answer : The normality of the solution is, 30.006 N
Explanation :
Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.
Mathematical expression of normality is:
or,
First we have to calculate the equivalent weight of solute.
Molar mass of solute 
= 94.97 g/mole
Now we have to calculate the normality of solution.
Therefore, the normality of the solution is, 30.006 N
Answer:
hmm
Explanation:
sorry for that have a great day
