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2 years ago

Find the acceleration of the 2 kg block in the following diagram. ( Hint find the net force!)

Chemistry

1 answer:

mario62 [17]2 years ago

7 0

Answer:
2m/s^2
Explanation:
Fnet = F2-F1
ma = 8-4
2a = 4
Dividing both sides by 2
a = 2
Therefore (a) thus acceleration is 2m/s^2

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What are duplet and octet elements ?Differentiate.

Finger [1]

Octet means presence of a total of 8 electrons in its valence shell while in case of duplet only 2 electrons are present in valence shell.

4 0

3 years ago

A student sees an absorbance a=1.140 for his solution that has a concentration of c=1.50*10-4 m using 0.50 cm cuvette. what is t

777dan777 [17]

The molar extinction coefficient is 15,200 $M^{-1} cm^{-1}$ .

The formula to be used to calculate molar extinction coefficient is -

A = ξcl, where A represents absorption, ξ refers molar extinction coefficient, c refers to concentration and l represents length.

The given values are in required units, hence, there is no need to convert them. Directly keeping the values in formula to find the value of molar extinction coefficient.

Rewriting the formula as per molar extinction coefficient -

ξ = $\frac{A}{cl}$

ξ = $\frac{1.140}{1.5*10^{-4}*0.5 }$

Performing multiplication in denominator to find the value of molar extinction coefficient

ξ = $\frac{1.140}{0.000075}$

Performing division to find the value of molar extinction coefficient

ξ = 15,200 $M^{-1} cm^{-1}$

Hence, the molar extinction coefficient is 15,200 $M^{-1} cm^{-1}$ .

Learn more about molar extinction coefficient -

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1 year 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 $(\Delta S^o)$ .

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

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