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ExtremeBDS [4]
2 years ago
13

What is the name of the "invisible force" that holds you to Earth's surface

Chemistry
2 answers:
timofeeve [1]2 years ago
8 0

Answer: gravity

Explanation: gravity has is an “invisible force” that holds matter down instead of floating off into space.

Alex2 years ago
7 0

Answer:

Gravity

Explanation:

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Alex17521 [72]

Answer:

3

Explanation:

It is based on empirical evidence

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3 years ago
How can you write tht in mg ??
inessss [21]
Rris is how u right it Mg=? Because it ask u a question and then you put the equal sign and then you put the question mark because you don't know it yet
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3 years ago
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Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.
zubka84 [21]

<u>Answer:</u> The value of \Delta G^o of the reaction is 28.38 kJ/mol

<u>Explanation:</u>

For the given chemical reaction:

SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)

  • The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]

We are given:

\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol

  • The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]

We are given:

\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}

where,

\Delta H^o_{rxn} = standard enthalpy change of the reaction =-67200 J/mol

\Delta S^o_{rxn} = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol

Hence, the value of \Delta G^o of the reaction is 28.38 kJ/mol

7 0
3 years ago
Please help! Which Zeros are significant in the measurement .0506010 kg?
Vitek1552 [10]
It would be a,b,c as the answer. hope this helps!
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2 years ago
I WILL MARK BRAINLIEST!!!!! !!<br><br><br><br> (picture included with one question ^^^^^^^)
o-na [289]

Answer:

type D

balance form .b

Explanation:

5 0
2 years ago
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