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3 years ago

Plss help me solve everything it’s for science

Chemistry

1 answer:

Maksim231197 [3]3 years ago

4 0

Answer:

b) heat from the lemonade move to the ice.

Explanation:

he he can only move one way and that is (hot to cold). because of this you know A and b is wrong.

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How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo

riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

$(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)$

The expression used will be:

$\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = heat released by the reaction = ?

m = mass of benzene = 125 g

$c_{p,g}$ = specific heat of gaseous benzene = $1.06J/g^oC$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC$

$\Delta H_{vap}$ = enthalpy change for vaporization = $33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g$

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

$\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]$

$\Delta H=-67682.5J=-67.7kJ$

Therefore, the energy removed must be, -67.7 kJ

4 0

3 years ago

What forces are important in the creation of magma?

kherson [118]

"the factors that mainly affect in the formation of magma can be summarized into three: temperature, pressure, and composition"

I hope this helps =^._.^=

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3 years ago

Rewrite each equation below with the delta h value included with either the reactants or he products , and identify the reaction

grin007 [14]

I will take a stab at it, but there are not equations, did you forget them?

5 0

3 years ago

Study the graph in Figure 6.32 and answer to these questions. (see attached image)

Dimas [21]

Hey there :)

We can see that the solubility of salt increases with increasing temperature. This happens with most substances.

To find out the maximum mass of copper sulfate that can be dissolved in water at these temperatures, just interpret the graph.

Considering Y-axis as g copper sulfate/100 g water and the X-axis as the temperature in °C:-

1)

a: 0 °C - 14 g of copper sulfate/100 g of water

b: 50 °C - 34 g of copper sulfate/100 g of water

c: 90 °C - 66 g of copper sulfate/100 g of water

2) From the graph, we can infer that temperature affects the solubility of the salt.

Answered by Benjemin360 :)

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2 years ago

Write a hypothesis that answers the lesson

steposvetlana [31]

Answer:

the reaction will come to a halt and the other reactant will still be present.

7 0

3 years ago