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2 years ago

For the reaction:

Chemistry

1 answer:

vekshin12 years ago

3 0

Answer:

0-77.3kj/mol

Explanation:

add all of the mole

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Answer:

Because boys are overrated

Explanation:

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3 years ago

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What is the concentration of 60 mL of H3PO4 if it is neutralized by 225 mL of 2 M Ba(OH)2?

goldfiish [28.3K]

7.5 M is the concentration of 60 ml of H3PO4 if it is neutralized by 225 ml of 2 M Ba(OH)2.

Explanation:

Data given:

volume of phosphoric acid, Vacid =60 ml

volume of barium hydroxide, Vbase = 225 ml

molarity of barium hydroxide, Mbase = 2M

Molarity of phosphoric acid, Macid =?

the formula for titration is used as:

Macid x Vacid = Mbase x Vbase

rearranging the equation to get Macid

Macid = $\frac{Mbase x Vbase}{Vacid}$

Macid = $\frac{225 X 2}{60}$

Macid = 7.5 M

the concentration of the phosphoric acid is 7.5 M and the volume is 60 ml. Thus 7.5 M solution of phosphoric acid is used to neutralize the barium hydroxide solution of 2M.

8 0

3 years ago

Read 2 more answers

Please help!!! ASAP

bekas [8.4K]

Na 2Co3

1*23=23. 2*12=24. 6*16=96

23+24+96=143

(23*100)/143. (24*100)/143. (96*100)/143

=16%. =16.7%. =67.1%

5 0

3 years ago

What volume will 2.04g of helium gas occupy at 100C and 1.13 atm?

Igoryamba

Answer:

1.13 atm

Explanation:

6 0

3 years ago

I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

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

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0

3 years ago