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Vinvika [58]
3 years ago
10

How many moles of MgS2O3 are in 171 g of the compound?

Chemistry
1 answer:
Alexxx [7]3 years ago
8 0
Divide the 181g by the mol mass of the compound. Just add up the masses of the various elements.
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Given that ΔH = −571.6 kJ/mol for the reaction 2 H2(g) + O2(g) → 2 H2O(l), calculate ΔH for these reactions. (a) 2 H2O(l) → 2 H2
pashok25 [27]

Answer : The value of \Delta H for the reaction is +571.6 kJ/mole.

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

2H_2(g)+O_2(g)\rightarrow 2H_2O(l)     \Delta H_1=-571.6kJ/mole

Now we have to determine the value of \Delta H for the following reaction i.e,

2H_2O(l)\rightarrow 2H_2(g)+O_2(g)    \Delta H_2=?

According to the Hess’s law, if we reverse the reaction then the sign of \Delta H change.

So, the value \Delta H_2 for the reaction will be:

\Delta H_2=-(-571.6kJ/mole)

\Delta H_2=+571.6kJ/mole

Hence, the value of \Delta H for the reaction is +571.6 kJ/mole.

6 0
3 years ago
A buffer solution contains 0.496 M hydrocyanic acid and 0.399 M sodium cyanide . If 0.0461 moles of sodium hydroxide are added t
pochemuha

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = pK_a=9.31

First we have to calculate the moles of HCN and NaCN.

\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole

and,

\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole

The balanced chemical reaction is:

                          HCN+NaOH\rightarrow NaCN+H_2O

Initial moles     0.1116       0.0461     0.08978

At eqm.       (0.1116-0.0461)    0       (0.08978+0.0461)

                        0.0655                       0.1359

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

pH=9.31+\log (\frac{0.1359}{0.0655})

pH=9.63

Therefore, the pH of the solution is, 9.63

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