A compound is known to have an empirical formula of CH and a molecular mass of 78.11 g/ mom what is he molecular formula

Mashcka [7]

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases. -google

its probably 40

8 0

2 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

Why do different materials have similar properties

Deffense [45]

Answer:

The difference in the number of protons and neutrons in atoms account for many of the different properties of elements.

7 0

2 years ago

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Which describes radioactive decay of a substance? More of the radioactivity is lost during the first half-life than in later hal

vovangra [49]

Answer:More of the radioactivity is lost during the first half-life than in later half-lives.

Explanation: Passed test with 98

5 0

2 years ago

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An investor has an account with stock from two different companies. Last year, his stock in Company A was worth $7200 and his st

djyliett [7]

Answer:

22.7%

Explanation:

Given that last year stock for company A was $7200

The stock for company B last year was worth $3510

Stock in company A decreased by 24%

This means the new value of stock for company A became;

(100-24)/100 *$7200

76/100*$7200

0.76*7200 =$5472

Stock in company B decreased by 20%

This means the new value of stock for company B became;

(100-20)/100 *$3510

80/100*$3510

0.8*$3510

$2808

Original investors stock value was = $7200+$3510 =$10710

New investors stock value is = $5472+$2808=$8280

Decrease in value of stock = $10710-$8280 =$2430

percentage decrease in stock value = decrease in stock/original value of stock *100%

=2430/10710 *100 =22.689

=22.7%

7 0

2 years ago