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o-na [289]
2 years ago
8

How many grams of Na2SO4 should be weighed out to prepare 0.5L of a 0.100M solution?​

Chemistry
1 answer:
Nezavi [6.7K]2 years ago
3 0

Answer:

7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.

Explanation:

First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.

    According to the given question we have to prepare 0.100 M solution

1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4

1    ml of solution contain     14.208÷1000= 0.014 gram

0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.

 So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.

     

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Which statements are correct regarding the Law of Conservation of Matter and Energy?
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Answer:

Matter or energy can change from one form to the other

Explanation:

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Likewise, the law of conservation of mass/matter states that matter can not be destroyed or created but can change via physical or chemical means to conserve it. For example, matter can change from liquid state to gaseous state.

From the above two laws, it can be said that "matter or energy can change from one form to the other".

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3 years ago
in order to decrease the freezing point of 500. g of water to 1.00° c how many grams of ethylene glycol (C2H602)must be added (K
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6 0
2 years ago
The free energy change for the following reaction at 25 °C, when [Cr3+] = 1.32×10-3 M and [Fe3+] = 1.14 M, is 131 kJ: Cr3+(1.32×
larisa [96]

Answer:

E°cell = - 1.3575 V

This reaction is spontaneous in the reverse direction

Explanation:

The given cell reaction:

Cr³⁺(1.32 × 10⁻³ M) + Fe²⁺(aq) → Cr²⁺(aq) + Fe³⁺(1.14 M)

The given Gibbs free energy: ΔG = 131 kJ = 131 × 10³ J     (∵ 1 kJ = 10³ J)

As we know,

ΔG = - n F E°cell

Here, n - the number of moles of electrons transferred = 1

F - Faraday constant = 96500

E°cell - cell potential = ?

\therefore E^{\circ }_{cell} = -\frac{\Delta G}{n \: F} = -\frac{131\times 10^{3}\, J}{1\, mol\times96500 \, C.mol^{-1}}

\Rightarrow E^{\circ }_{cell} = -1.3575\, V

<u>For a given chemical reaction if-</u>

1. ΔG = negative and E°cell = positive

⇒ <em>The reaction is spontaneous and proceeds spontaneously in the forward direction.</em>

2.  ΔG = positive and E°cell = negative

⇒ <em>The reaction is non-spontaneous and proceeds spontaneously in the reverse direction.</em>

<u>Since, for this chemical reaction: </u>

Cr³⁺(1.32 × 10⁻³ M) + Fe²⁺(aq) → Cr²⁺(aq) + Fe³⁺(1.14 M)

ΔG = + 131 × 10³ J ⇒ positive

and, E°cell = - 1.3575 V ⇒ negative

<u>Therefore, this reaction is spontaneous in the reverse direction.</u>

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