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3 years ago

7

Which of the following best describes the conditions created by El Niño? When is that normally blowup to east get stronger water

s off the coast of north America cool down and upwelling increases in proving the nutrients in the water. Wednesday normally blood to the west get stronger waters off the coast of South America cool down and upwelling increases improving the nutrients in water. When is the normally blowup to the west slacken waters off the coast of South America walk up and upwelling decreases reducing the nutrients in the water

2 answers:

Whitepunk [10]3 years ago

7 0

Answer:

its C the one at the bottom

Explanation:

SVETLANKA909090 [29]3 years ago

3 0

The answer is c i had it on my test today

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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

Sodium react gently with water ture or false<br>

Maksim231197 [3]

false,

sodium has a vigorous reaction with water.

This can be observed by putting a piece of sodium in water,

it moves swiftly across the surface of water and reacts violently.

5 0

2 years ago

17 points help asap!! :) thank youuuu :) Jeff is in a stationary school bus. Which is the best reference point for him to use to

3241004551 [841]

Answer:

The answer would be A

Hope this helps!!!

5 0

3 years ago

An astronomer wants to study celestial objects with the shortest wavelengths. What kind of telescope should the astronomer use f

Leni [432]

Answer:

telescope

Explanation:

i took the test

8 0

3 years ago

12oz of water initially at 75oF is mixed with 20oz of water intiially at 140oF. What is the final temperature?

Kaylis [27]

Answer:

$115.625^{\circ}\text{F}$

Explanation:

$m_1$ = First mass of water = 12 oz

$m_2$ = Second mass of water = 20 oz

$\Delta T_1$ = Temperature difference of the solution with respect to the first mass of water = $(T-75)^{\circ}\text{F}$

$\Delta T_2$ = Temperature difference of the solution with respect to the second mass of water = $(T-75)^{\circ}\text{F}$

c = Specific heat of water

As heat gain and loss in the system is equal we have

$m_1c\Delta T_1=m_2c\Delta T_2\\\Rightarrow m_1\Delta T_1=m_2\Delta T_2\\\Rightarrow 12(T-75)=20(140-T)\\\Rightarrow 12T-900=2800-20T\\\Rightarrow 12T+20T=2800+900\\\Rightarrow 32T=3700\\\Rightarrow T=\dfrac{3700}{32}\\\Rightarrow T=115.625^{\circ}\text{F}$

The final temperature of the solution is $115.625^{\circ}\text{F}$ .

3 0

2 years ago