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AnnZ [28]
3 years ago
7

How many kJ of heat are released by the reaction of 25.0 g of Na2O2(s) in the following reaction? (M = 78.0 g/mol for Na2O2)

Chemistry
1 answer:
solong [7]3 years ago
6 0

-20.16 KJ of heat are released by the reaction of 25.0 g of Na2O2.

Explanation:

Given:

mass of Na2O2 = 25 grams

atomic mass of Na2O2 = 78 gram/mole

number of mole = \frac{mass}{atomic mass of 1 mole}

                          = \frac{25}{78}

                          =0. 32 moles

The balanced equation for the reaction:

2 Na2O2(s) + 2 H2O(l) → 4 NaOH(aq) + O2(g) ∆Hο = −126 kJ

It can be seen that 126 KJ of energy is released when 2 moles of Na2O2 undergoes reaction.

similarly 0.3 moles of Na2O2 on reaction would give:

\frac{126}{2} = \frac{x}{0.32}

x = \frac{126 x 0.32}{2}

 = -20.16 KJ

Thus, - 20.16 KJ of energy will be released.

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For the reaction PC15 (8) PC13 (g) + Cl2 (g) K = 0.0454 at 261 °C. If a vessel is filled with these gases such that the initial
Fiesta28 [93]

Answer:

The correct answer is B.

The K_{eq} is samller than Q of the reaction . So,the reaction will shift towards the left i.e. towards the reactant side.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{eq}

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

The expression for Q is written as:

Q=\frac{[PCl_3][Cl_2]}{[[PCl_5]^1}

Q=\frac{0.20 M\times 2.5 M}{0.20 M}

Q=2.5

Given : K_{eq} = 0.0454

Thus as K_{eq}, the reaction will shift towards the left i.e. towards the reactant side.

4 0
3 years ago
Read 2 more answers
In a healthy pond, the temperature is 16°C (61°F). What is the most likely pH of this pond?
masha68 [24]

Answer:

A . 6.3 In a healthy pond, the temperature is 16°C (61°F). What is the most likely pH of this pond

7 0
2 years ago
You are given a solid that is a mixture of na2so4 and k2so4.
Murljashka [212]

Here we have to calculate the amount of SO_{4}^{2-} ion present in the sample.

In the sample solution 0.122g of SO_{4}^{2-} ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ +  SO_{4}^{2-}

(Na)₂SO₄=2Na⁺ +  SO_{4}^{2-}

Thus, BaCl₂+  SO_{4}^{2-} = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and  K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of  SO_{4}^{2-} ion is precipitated in this reaction.  

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion (SO_{4}^{2-}) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of SO_{4}^{2-} ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of SO_{4}^{2-} ion is present. So in 1 g of BaSO₄ \frac{96.06}{233.3}=0.411 g of SO_{4}^{2-} ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of SO_{4}^{2-} ion is present.        

5 0
3 years ago
How many moles of mg3n2 would be produced when 180.0 g of mg reacts with 85.0 g of n2?
aalyn [17]
2.47 moles are produce if 85g of N2 react with 180g of Mg
7 0
3 years ago
Number of
zzz [600]

Answer:

symbol= F

proton= 9

neutron= 19 I understand this is it ok for you

6 0
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