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2 years ago

Which of the following phase changes is exothermic?

1 answer:

8 0

Answer:

gas to solid

Explanation:

Exothermic means the release of heat. Any transformation from a less ordered to a more ordered state releases energy. This is represented by the change of a liquid to solid, gas to liquid, or gas to solid. Vise versa endothermic is the transformation from a more ordered state to a less ordered state represented by solid to liquid, liquid to gas, and solid to gas.

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How many grams of lithium chloride can be produced from 14.3 g of lithium chlorate when it decomposes into lithium chloride and

vampirchik [111]

6.7 grams of lithium chloride will be produced.

<h3><u>Explanation:</u></h3>

Lithium chlorate is LiClO₃ and lithium chloride is LiCl. The reaction is,

2LiClO₃ = 2LiCl +3O₂.

So here, 2 moles of lithium chlorate produces 2 moles of lithium chloride.

Or, one molecule of lithium chlorate will produce one mole of lithium chloride.

Molecular weight of lithium chlorate = $6.9 + 35.5 + 16\times3$ = 90.4.

So, 14.3 grams of lithium chlorate has 0.16 moles of lithium chlorate.

Thereby, moles of lithium chloride produced is 0.16 moles.

Molecular weight of lithium chloride = $6.9+35.5$ = 42.4 grams.

So weight of lithium chloride produced = $42.4 \times 0.16$ = 6.7 grams.

Thus, weight of lithium chloride produced will be 6.7 grams.

3 0

3 years ago

Please Please Please Please Please Help Me (please answer this question completely and 1 will friend you and help you with your

pav-90 [236]

Answer:

Static electricity is a type of energy that is produced by friction.

Explanation:

Static electricity may be the result when there is an imbalance between the negative charges and the positive charges in the surroundings. They are non contact forces. They pull of push without actually touching the body. The charged particles interact by pulling or pushing the uncharged particles.

These charges are released when they come closer to some uncharged or other charged particles like the electrostatic charges. It is the charge build up in an area.

6 0

3 years ago

How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2

Leya [2.2K]

Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

i.e. ,

moles = ( mass / molecular mass )

since,

mass of KNO₃ = 58.6 g ( given )

Molecular mass of KNO₃ = 101 g / mol

Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1 mol of KNO₃ gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4 * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

Therefore,

58.6g of KNO₃ gives 0.725 mol of O₂.

3 0

2 years ago

Determine the mass of CuSO4 • 5H20 that must be used to prepare 250mL of 2.01 M CuSO4(aq).

mario62 [17]

Given parameters:

Volume of CuSO₄ = 250mL

Concentration of CuSO₄ = 2.01M

Unknown:

Mass of CuSO₄.5H₂O = ?

To solve this problem, we must write the chemical relationship between both species.;

CuSO₄.5H₂O → CuSO₄ + 5H₂O

Now that we know the expression, it is possible to solve for the unknown mass.

First find the number of moles of CuSO₄;

Number of moles = Concentration x Volume

Take 250mL to L so as to ensure uniformity of units;

Volume = 250 x 10⁻³L

Input the parameters and solve for number of moles;

Number of moles = 250 x 10⁻³ x 2.01 = 0.5mol

From the equation;

1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O

So 0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O

Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16) = 249.6g/mole

Mass of CuSO₄.5H₂O = number of moles x molar mass

= 0.5 x 249.6

= 124.8g

The mass of CuSO₄.5H₂O is 124.8g

5 0

3 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 $(\Delta S^o)$ .

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

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