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inn [45]
3 years ago
7

Which equation shows how to calculate how many grams (g) of Mg(OH)2 would be produced from 4mol KOH? The balanced reaction is:

Chemistry
1 answer:
Simora [160]3 years ago
6 0

Answer:

4 mol of KOH would produce 116.6 g of Mg(OH)₂

Explanation:

According to the following balanced equation:

  • MgCl₂+ 2 KOH -----> Mg(OH)₂ + 2 KCL

One can note that 2 mol of KOH react with MgCl₂ to produce 1 mol of Mg(OH)₂.

using cross multiplication  

2 mol of KOH → 1 mol of Mg(OH)₂.

4 mol of KOH → ?? mol of Mg(OH)₂.

no of moles of  Mg(OH)₂ = (1 mol* 4 mol) / 2 mol =2 mol

Now we can convert moles of  Mg(OH)₂ to grams using the formula

mass of Mg(OH)₂= (no. of moles * molar mass) = (2 mol * 58.3g/mol) = 116.6 g

  • So, 4 mol of KOH would produce 116.6 g of Mg(OH)₂.
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Gold has a density of 19.32 is this a chemical change or physical change
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How many moles of copper are there in 6.93 g of copper sample<br>pls be quick​
Alex_Xolod [135]
  • Molar mass of copper=63.5g/mol
  • Given mass=6.93g

\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:mass}{Molar\:mass}

\\ \sf\longmapsto No\:of\:moles=\dfrac{6.93}{63.5}

\\ \sf\longmapsto No\:of\:moles=0.109\approx 1.11moles

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2 years ago
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gregori [183]
Zn⁰ ----> Zn⁺²   +   2e⁻ -  oxidation
Hg⁺² + 2e⁻ ----> Hg⁰ - reduction

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7 0
3 years ago
What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe
Deffense [45]

Answer:

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]

<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒  [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

7 0
3 years ago
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