Aluminum has a density of 2.70 g/m. Calculate the mass (in grams) of a piece of aluminum having a volume of 382 mL

Which method is used to obtain petrol from petroleum?

kotykmax [81]

Explanation:

fractional distillation method is used to obtain petrol from petroleum...

hope it helps

6 0

3 years ago

Read 2 more answers

The decomposition of SOCl2 is first-order in SOCl2. If the half-life for the reaction is 4.1 hr, how long would it take for the

qaws [65]

4.1 h = 14760 s

t 1/2 = ln 2 / k 

k = rate reaction = 4.97 x 10^-5 

ln 0.045 / 0.36 = - 4.97 x 10^-5 t 

2.08 = 4.97 x 10^-5 t 

t = 41839.9 s = 11 h 37 min 19 s

3 0

3 years ago

If the pressure of a gas is 1.01 atm and the temperature is 25°C, then if the pressure is increased to 1.10, what is the new tem

vodka [1.7K]

Answer:

T2 = 51.6°C

Explanation:

Given:

P1 = 1.01 atm

T1 = 25°C + 273 = 298K

P2 = 1.10 atm

T2 = ?

P1/T1 = P2/T2

Solving for T2,

T2 = (P2/P1)T1

= (1.10 atm/1.01 atm)(298K)

= 324.6 K

= 51.6°C

where Tc = Tk - 273

3 0

3 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

ASAP worth 15 points also

Taya2010 [7]

The person above me is correct I took a test on this so it’s the right answer

7 0

3 years ago