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3 years ago

Is " Energy created " an Endothermic reaction or an Exothermic reaction?

Chemistry

1 answer:

podryga [215]3 years ago

6 0

A endothermic reaction

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what is the balanced equation when copper metal is placed in a solution when platnium ii chloride is placed. what is the equatio

koban [17]

Answer:

$Cu~+~PtCl_2->Pt~+~CuCl_2$

Explanation:

In this case, we can start with the formula of Platinum (II) Chloride. The cation is the atom at the left of the name (in this case $Pt^+^2$ ) and the anion is the atom at the right of the name (in this case $Cl^-$ ). With this in mind, the formula would be $PtCl_2$ .

Now, if we used metallic copper we have to put in the reaction only the copper atom symbol $Cu$ . So, we have as reagents:

$Cu~+~PtCl_2->$

The question now is: What would be the products? To answer this, we have to remember "single displacement reactions". With a general reaction:

$A~+~BC->AB~+~C$

With this in mind, the reaction would be:

$Cu~+~PtCl_2->Pt~+~CuCl_2$

I hope it helps!

5 0

2 years ago

A chemist needs to make 250 mL of a 2.50 M aqueous solution of ammonium hydroxide from a 6.00 M ammonium hydroxide solution. How

Furkat [3]

Answer:

250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)

Explanation:

Data Given

M1 = 6.00 M

M2 = 2.5 M

V1 = 250 mL

V2 = ?

Solution:

As the chemist needs to prepare 250 mL of solution from 6.00 M ammonium hydroxide solution to prepare a 2.50 M aqueous solution of ammonium hydroxide.

Now

first he have to determine the amount of ammonium hydroxide solution that will be taken from6.00 M ammonium hydroxide solution

For this Purpose we use the following formula

M1V1=M2V2

Put values from given data in the formula

6 x V1 = 2.5 x 250

Rearrange the equation

V1 = 2.5 x 250 /6

V1 = 104 mL

So 104 mL is the volume of the solution which we have to take from the 6.00 M ammonium hydroxide solution to prepare 2.5 M aqueous solution of ammonium hydroxide

But we have to prepare 250 mL of the solution.

so the chemist will take 104 mL from 6.00 M ammonium hydroxide solution and have to add 146 mL water to make 250 mL of new solution.

in this question you have to tell about the amount of water that is 146 mL

250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)

7 0

2 years ago

All of the following are diatomic molecules except? a)H b)Br c) CI d)S

xxTIMURxx [149]

d)S is not one of them.

Here's an easy way to remember them, attached:

8 0

2 years ago

Read 2 more answers

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

2 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$