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dalvyx [7]
3 years ago
5

Which equilibrium reaction will experience a shift towards the products in equilibrium position when the concentration of ni2+ i

s increased? view available hint(s) which equilibrium reaction will experience a shift towards the products in equilibrium position when the concentration of is increased? ni2+(aq)+6nh3(aq)⇌[ni(nh3)6]2+ [ni(h2o)6]2+(aq)+3en(aq)⇌[ni(en)3]2+(aq)+6nh3(aq) ni(oh)2(s)⇌ni2+(aq)+2oh−(aq) nis(s)⇌ni2+(aq)+s2−(aq)?
Chemistry
1 answer:
juin [17]3 years ago
8 0

Answer is <span>
      Ni²</span>⁺(aq) + 6NH₃(aq) ⇌ [Ni(NH₃)₆]²⁺<span>

</span>

<span>When the concentration of Ni²⁺</span><span>(aq) increases, according to the Le Chatelier’s principle system tries to become equilibrium by reducing the increased factor. To do that, the concentration of Ni²⁺</span><span>(aq) should be reduced. Hence, the forward reacted should be promoted to reduce the Ni²⁺</span><span>(aq) concentration</span>.
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Determine the number of moles of C5H12 that are contained in 357.4 g of the compound.
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Answer: 4.96 moles

Explanation:

C5H12 is the chemical formula for pentane, the fifth member of the alkane family.

Given that,

number of moles of C5H12 = ?

Mass in grams = 357.4 g

Molar mass of C5H12 = ?

To get the molar mass of C5H12, use the atomic mass of carbon = 12g; and Hydrogen = 1g

i.e C5H12 = (12 x 5) + (1 x 12)

= 60g + 12g

= 72g/mol

Now, apply the formula

Number of moles = Mass / molar mass

Number of moles = 357.4g / 72g/mol

= 4.96 moles

Thus, 4.96 moles of C5H12 that are contained in 357.4 g of the compound.

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3 years ago
Complete the isotope symbol with mass numbers of 35
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The answer would be
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If you have 155 mL solution of a 0.762 M FeCl3 solution, how many
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Moles are the product of the molar concentration and the volume of the solution. The moles of Ferric chloride is 0.118 moles.

<h3>What are moles?</h3>

Moles are the smallest unit of the measurement of atoms and molecules. It is calculated by the molar concentration and the volume of the solution.

Given,

Molar concentration of ferric chloride = 0.762 M

Volume of solution = 0.155 L

Moles are calculated as:

\begin{aligned} \rm Moles &= \rm Molarity \times Volume (L) \\\\&= 0.762 \times 0.155\\\\&= 0.118\;\rm mol\end{aligned}

Therefore, 0.118 moles of ferric chloride are present in the sample.

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