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tino4ka555 [31]

2 years ago

10

The temperature of 15 grams of water is increased by 3.0 Celsius degrees. How much heat was absorbed by the water? (specific hea

t of water: 4.18 J/g deg C)

1 answer:

frozen [14]2 years ago

8 0

Answer:

188.1J

Explanation:

obtained from Q=MCT

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Fireworks change (blank) into (blank) and (blank) energy.

Nitella [24]

Fireworks changes chemical energy into light energy

5 0

3 years ago

Pls Help!!!!! ASAP!!!!

HACTEHA [7]

Answer:

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Explanation:

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7 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

Which of the following is an example of a compound?

VladimirAG [237]

Answer:

H2O is a compound because its a main constitute of earths hydrosphere

3 0

3 years ago

Rb2co3(aq)+fec2h302(aq)-->

nata0808 [166]

Answer:

Rb2CO3(aq)+Fe(C2H3O2)2(aq)--> 2Rb(C2H3O2)(aq) + FeCO3(s)

Explanation:

The reaction shown in the answer is the reaction of rubidium carbonate and iron II acetate. Rubidium is far more reducing than Fe II hence it can displace Fe II from its salt as shown.

The reducing property of metals depends on the value of their individual electrode potential values. For rubidium, its standard reduction potential is -2.98 V while that of Fe II is -0.44V. Hence rubidium can displace Fe II from its salt as shown above.

4 0

3 years ago