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zzz [600]
3 years ago
9

How much heat (in kj) is released when 3.600 mol naoh(s) is dissolved in water? (the molar heat of solution of naoh is â445.1 kj

/mol.)?
Chemistry
2 answers:
Fantom [35]3 years ago
8 0
<span>-1602 kj

Sauce:me

Your welcome
</span>
pav-90 [236]3 years ago
6 0

Answer : The amount of heat released is, -1602.36 KJ

Explanation : Given,

Moles of NaOH = 3.600 mole

Molar heat of solution of NaOH = -445.1 KJ/mole

Now we have to calculate the amount of heat released.

As, 1 mole of NaOH is dissolved in water then heat released = - 445.1 KJ

So, 3.600 mole of NaOH is dissolved in water then heat released = -445.1\times 3.600=-1602.36KJ

Therefore, the amount of heat released is, -1602.36 KJ

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How many formula units of silver fluoride, AgF, are equal to 42.15 g f this substance?
s2008m [1.1K]

Answer:

1.99 x 10²³ formula units

Explanation:

Given parameters:

Mass of AgF  = 42.15g

Unknown:

The amount of formula units

Solution:

To solve this problem, we set out by find the number of moles in this compound from the given mass.

   Number of moles  = \frac{mass}{molar mass}

molar mass of AgF  = 107.9 + 19  = 126.9g/mol

  Number of moles  = \frac{42.15}{126.9}   = 0.33 moles

       1 mole of a substance =  6.02 x 10²³ formula units

         0.33 moles of AgF  = 0.33 x 6.02 x 10²³  = 1.99 x 10²³ formula units

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2 years ago
After ensuring the equipment is turned off and unplugged what is the next step in cleaning large equipment by hand
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8 0
3 years ago
Rxn
givi [52]

Answer: The enthalpy of formation of SO_3 is  -396 kJ/mol

Explanation:

Calculating the enthalpy of formation of SO_3

The chemical equation for the combustion of propane follows:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]

We are given:

\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-297kJ/mol\\\Delta H^o_{rxn}=-198kJ

Putting values in above equation, we get:

-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol

The enthalpy of formation of SO_3 is -396 kJ/mol

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3 years ago
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