What is the total electrons needed for an Octet of SiCL4?

What do these two changes have in common sauce burning on a stove and jewelry tarnishing

Andrews [41]

The changes that are common between sauce burning on a stove, and jewelry tarnishing, which is a chemical change.

How to define chemical and physical changes?

Chemical Change-

Any alteration that produces new chemical substances with distinct properties is considered a chemical change. Chemical reactions involve the rearrangement and recombination of elements and compounds to create new substances. Examples of chemical changes are listed below:

Burning
Digestion
chemicals changing colors
Tarnishing
compost rotting

Physical Change-

A substance is not destroyed or transformed into something new by physical changes. A substance can undergo physical changes that alter its shape, size, or phase. The constituents of an element or compound do not change during a physical change. Examples of physical changes are listed below:

Boiling water
Chopping, Cutting, Carving
Evaporation
Freezing, Melting, Condensation

To know more about chemical and physical changes, visit the given link:

brainly.com/question/20628019

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5 0

11 months 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

A sample of nitrogen gas has the temperature drop from 250.°c to 150.°c at constant pressure. what is the final volume if the in

Leviafan [203]

Charles law gives the relationship between temperature of gas and volume of gas.
It states that for a fixed amount of gas, temperature is directly proportional to volume of gas.
V / T = k
where V- volume , T - temperature and k - constant
$\frac{V1}{T1} = \frac{V2}{T2}$
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation.
T1 = 250 °C + 273 = 523 K
T2 = 150 °C + 273 = 423 K
Substituting the values in the equation,

$\frac{310 mL}{523 K} = \frac{V}{423K}$
V = 251 mL
the new volume is 251 mL

6 0

3 years ago

Consider the following data showing the initial rate of a reaction (A→products) at several different concentrations of A.

OlgaM077 [116]

Answer:

(1) order = 2

(2) R = K [A]²

Explanation:

Given the reaction:

A--------->Product

The rate constant relation for the reaction is given as:

R(i) = K [A]............(*)

Where R(I) is rate constant at different concentration of A.

Taking the rate constant as R1, R2 and R3 for the different concentrations respectively. Then the following equations results

0.011 = K [0.15] ⁿ.........(1)

0.044 = K [0.30]ⁿ .......(2)

0.177 = K [0.60]ⁿ .........(3)

Dividing (2) by (1) and (3) by (1)

Gives:

0.044/0.011 = [0.3/0.15]ⁿ

4 = 2ⁿ; 2² = 2ⁿ; n = 2

Similarly

0.177/0.011 = [0.60/0.15]ⁿ

16.09 = 4ⁿ

16.09 = 16 (approximately)

4² = 4ⁿ ; n = 2

Hence the order of the reaction is 2.

The rate law is R = K [A]²

4 0

3 years ago

How to balance this equation: Cu(s) + O2(g) --> CuO(s)

Lelu [443]

2Cu(s) + O2(g) --> 2CuO(s)

3 0

2 years ago