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3 years ago

1) Which electrons are involved in bonding between atoms?|

Chemistry

1 answer:

EleoNora [17]3 years ago

8 0

Answer:

<u>VALENCE ELECTRONS </u>are involved in the bonding of atoms

Explanation:

Hope that helps

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300 moles of sodium nitrite are needed for a reaction. the solution is 0.450 m. how many ml are needed?

miv72 [106K]

Answer:

$\boxed{\text{667 000 mL}}$

Explanation:

$\text{Molar concentration} = \dfrac{\text{moles}}{\text{litres}}\\\\c = \dfrac{n}{v}$

Data:

c = 0.450 mol·L⁻¹

n = 300 mol

Calculation:

$\begin{array}{rcl}0.450 & = & \dfrac{300}{V}\\\\0.450V& = & 300\\\\V & = & \dfrac{300}{0.450}\\\\V & = & \text{667 L} =\textbf{667 000 mL}\end{array}\\\text{The volume of solution needed is }\boxed{\textbf{667 000 mL}}$

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Suppose you have a mixture of solid mg(oh)2 and baso4, how do you separate the mixture

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This separation technique is a 4-step procedure. First, add H₂SO₄ to the solution. Because of common ion effect, BaSO₄ will not react, only Mg(OH)₂.

Mg(OH)₂ + H₂SO₄ → MgSO₄ + 2 H₂O

The aqueous solution will now contain MgSO₄ and BaSO₄. Unlike BaSO₄, MgSO₄ is soluble in water. So, you filter out the solution. You can set aside the BaSO₄ on the filter paper. To retrieve Mg(OH)₂, add NaOH.

MgSO₄ + 2 NaOH = Mg(OH)₂ + Na₂SO₄

Na₂SO₄ is soluble in water, while Mg(OH)₂ is not. Filter this solution again. The Mg(OH)₂ is retrieved in solid form on the filter paper.

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Given that kb for c6h5nh2 is 1.7 × 10-9 at 25 °c, what is the value of ka for c6h5nh3 at 25 °c?

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Answer: $k_a$ for $C_6H_5NH_3^+$ at 25°C is $0.588\times 10^{-4}$

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$k_w=k_a\times k_b$

$k_w\text{ at }25^o=1\times 10^{-14}$

Putting values in above equation, we get:

$1\times 10^{-14}=k_a\times (1.7\times 10^{-9})\\k_a=0.588\times 10^{-5}$

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