Solve the following mole-mass problem. Refer to the periodic table. How much mass does 0.50 mole of copper contain?

PLEASE HELP!! What is the concentration [H3O+] of a solution with a pH of 13.2? Is the solution acidic, basic, or neutral?

Ilya [14]

The concentration of [H3O+] will be 6.3 x $10^{-14}$ M

Mathematically, pH = -log [H+] or -log [H3O+]

With a pH of 13.2:

-log [H3O+] = 13.2

log [H3O+] = -13.2

[H3O+] = 6.3 x $10^{-14}$ M

More on pH can be found here: brainly.com/question/491373

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3 0

2 years ago

Read 2 more answers

Select the correct electron configuration for Boron. (Atomic Number 5) 1s 22s 12p 2 1s 22s 3 1s 32s 12p 1 1s 22s 22p 1

kodGreya [7K]

ans is: 1s2, 2s2 ,2p1,...

7 0

3 years ago

3. When the hydronium ion concentration of a solution is 1x10-4 M, What

rewona [7]

Answer:

pH = 4

Solution is acidic

Explanation:

Given data;

Hydronium ion concentration = 1 × 10⁻⁴ M

pH of solution = ?

Solution:

pH = -log [H₃O⁺]

pH = -log [ 1 × 10⁻⁴ ]

pH = 4

According to pH scale the pH 7 is neutral while the pH less than 7 is acidic and greater than 7 is basic. The given solution has pH 4 it means solution is acidic.

7 0

3 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.217 mole of $HCl$ react to give 0.217 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

Molar mass of $NaCl$ = 58.5 g/mole

$\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g$

Now we have to calculate the percent yield of this reaction.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = $\frac{5.24g}{12.7g}\times 100$

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0

4 years ago

Write the balanced NET IONIC equation for the reaction that occurs when ammonium nitrate and potassium hydroxide are combined. N

vova2212 [387]

Answer:

Net Ionic equation

NH₄⁺ + OH⁻ → NH₃ + H₂O

Option B is correct.

Weak Acid Strong Base

Check Explanation for the extent of the reaction.

Explanation:

Ammonium nitrate = NH₄NO₃

Potassium Hydroxide = KOH

Ammonium salts combine with alkalis to liberate NH₃ and form water.

The two reactants combine to give

NH₄NO₃ + KOH → KNO₃ + NH₃ + H₂O

In ionic form,

- NH₄NO₃ exists as NH₄⁺ and NO₃⁻

- KOH exists as K⁺ and OH⁻

- KNO₃ as K⁺ and NO₃⁻

And NH₃ and H₂O stay as they are, as per covalent compounds.

So, we have

NH₄⁺ + NO₃⁻ + K⁺ + OH⁻ → K⁺ + NO₃⁻ + NH₃ + H₂O

Eliminating the ions that exist on both sides, we have the net ionic equation to be

NH₄⁺ + OH⁻ → NH₃ + H₂O

which shows that this reaction is essentially a neutralization reaction in which the Bronsted Lowry acid, NH₄⁺, loses its proton to the base, OH⁻ and gives conjugate base, NH₃ and conjugate acid, H₂O.

This reaction is classified as a Weak acid versus Strong Base reaction as NH₄⁺ is from a Weak acid and OH⁻ is from a strong base.

Since this reaction is between a Weak base and a strong acid, the ionization isn't expected to be 100%, Hence, the extent of this reaction will be any option that is not 100%, a couple pieces of information might be required for the correct estimate, but above 50% seems correct.

Hope this Helps!!!

8 0

3 years ago