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o-na [289]
3 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]3 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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Answer:

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Explanation:

Given that :

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Thus; the reduced mass μ = \dfrac{m_1 \times m_2}{m_1 + m_2}

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∵ 1 μ = 1.66 × 10⁻²⁷ kg

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The rotational level Energy can be expressed by the equation:

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I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

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Explanation:

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<u>Balancing the chemical reaction;</u>

2H_{2} +O_{2}<u>                        </u>2H_{2}O<u />

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You typed 3. Is it accidental? If so, then the answer is above. If not, then you could be trying to type 56Fe +3, which means an ionic iron with charge +3. Charges are formed when you have too many or too few electrons to counter-balance the prositive charge of the nucleus. Charge +3 means you're 3 electrons short to negate the nucleus positive charge.

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7 0
3 years ago
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