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3 years ago

74 g Potassium Chloride converted to moles

Chemistry

1 answer:

kramer3 years ago

5 0

Answer:

Number of moles = 0.99 mol

Explanation:

Given data:

Mass of potassium chloride = 74 g

Number of moles of potassium chloride = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of KCl = 74.6 g/mol

now we will put the values in formula:

Number of moles = mass/molar mass

Number of moles = 74 g/ 74.6 g/mol

Number of moles = 0.99 mol

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Question 3:

Oliga [24]

E. Molarity of vinegar

i hope u get this i tried to figure it out hopefully im right

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5 0

3 years ago

29. What is E for a system which has the following two steps:

jasenka [17]

Answer:

Zero

Explanation:

Recall that;

E = q + w

Where;

q = heat, w = work done

When heat is absorbed by the system q is positive

When heat is evolved by the system q is negative

When the system does work, w is negative

When work is done on the system w is positive

Step 1

ΔE1= 60 KJ + 40 KJ = 100KJ

Step 2

ΔE2= (-30 KJ) + (-70 KJ) = (-100) KJ

ΔE1 + ΔE2= 100KJ + (-100) KJ = 0KJ

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3 years ago

Which corresponds to a pressure of 1.23 atm??

Vadim26 [7]

Atmospheric pressure, sometimes also called barometric pressure, is the pressure exerted by the weight of air in the atmosphere of Earth (or that of another planet)

1 atm is equivalent to = 101325 Pa

= 760 mmHg

= 760 torr

= 1.01325 bar

So 1.23 atm is equal to

= 124629.8 Pa

= 934.8 mmHg

= 934.8 torr

= 1.2462 bar

6 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$