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3 years ago

Would you expect manganese(II) oxide, MnO, to react more readily with HF(aq) or NaOH(aq)?

Chemistry

1 answer:

ella [17]3 years ago

3 0

Answer:

NaOH(aq)

Explanation:

NaOH(aq) is known to precipitate Mn^2+ ions according to the following reaction; Mn^2+(aq)+2OH^−(aq)↽−−⇀Mn(OH)2(s)

Hence, manganese(II) oxide reacts more readily with NaOH(aq) under ordinary conditions precipitating the metal hydroxide solid. This is one of the characteristic reactions of Mn^2+.

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Manoa falls drops 153 feet, measured the bottom to be 663 liters per second

oksian1 [2.3K]

Answer:

Manoa

waimea

likeke

Explanation:

4 0

3 years ago

1. When frequency increases energy:

liraira [26]

B

When frequency increases, as does the energy, but wavelength decreases. It also works vise versa; if wavelength were to increase, its frequency and energy will decrease.

8 0

3 years ago

A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

Anit [1.1K]

Answer: The percent yield of the reaction is 32.34 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For benzoic acid:

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol$

The chemical equation for the reaction of benzoic acid and methanol is:

$\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}$

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0295=0.108$ moles of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g$

To calculate the percentage yield of methyl benzoate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%$

Hence, the percent yield of the reaction is 32.34 %

6 0

3 years ago

Calculate the hydroxide ion consentration of a solution with pH = 3.25. show all calculations leading to your answer.

lidiya [134]

First you calculate the pOH of the solution:

pH+ pOH = 14

3.25 + pOH = 14

pOH = 14 - 3.25

pOH = 10.75

Concentration of [OH]⁻ in solution:

[ OH⁻ ] = $10 ^{-pOH}$

[ OH⁻ ] = 10^ - 10.75

[OH⁻] = 1.778 x 10⁻¹¹ M

hope this helps !

6 0

3 years ago

Write the chemical formula for

nataly862011 [7]

Ba2Cl
NaS2
The numbers are in subscript

6 0

3 years ago

Read 2 more answers