Mathematics: Given right triangle MNO below how many units long is NO

Given right triangle MNO below how many units, №13271622, 01.04.2022 02:24

Chemistry

Using the balanced equation below, how many grams of manganese(III) oxide would be produced from the complete reaction of 46.8 g of zinc? Zn + 2MnO2 + H20 — Zn(OH)2 + Mn203

Step-by-step answer

P Answered by Master

The mass of manganese(III) oxide produced is 113.03 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

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

Given mass of zinc = 46.8 g

Molar mass of zinc = 65.38 g/mol

Plugging values in equation 1:

$\text{Moles of zinc}=\frac{46.8g}{65.38g/mol}=0.716 mol$

The given chemical equation follows:

$Zn+2MnO_2+H_2O\rightarrow Zn(OH)_2+Mn_2O_3$

By the stoichiometry of the reaction:

If 1 mole of zinc produces 1 mole of manganese(III) oxide

So, 0.716 moles of zinc will produce = $\frac{1}{1}\times 0.716=0.716mol$ of manganese(III) oxide

Molar mass of manganese(III) oxide = 157.87 g/mol

Plugging values in equation 1:

$\text{Mass of manganese(III) oxide}=(0.716mol\times 157.87g/mol)=113.03g$

Hence, the mass of manganese(III) oxide produced is 113.03 g

Chemistry

Using the balanced equation below, how many grams of zinc would be required to produce 9.65 g of zinc hydroxide? Zn + 2MnO2 + H20 → Zn(OH)2 + Mn203 Please help I need it for a test

Step-by-step answer

P Answered by PhD

5

Mass = 6.538 g

Explanation:

Given data:

Mass of zinc hydroxide produced = 9.65 g

Mass of zinc required = ?

Solution:

Chemical equation:

Zn + 2MnO₂ + H₂O → Zn(OH)₂ + Mn₂O₃

Number of moles of zinc hydroxide:

Number of moles = mass/molar mass

Number of moles = 9.65 g/ 99.42 g/mol

Number of moles = 0.1 mol

now we will compare the moles of zinc and zinc hydroxide,

Zn(OH)₂ : Zn

1 : 1

0.1 : 0.1

Mass of zinc required:

Mass = number of moles × molar mass

Mass = 0.1 mol × 65.38 g/mol

Mass = 6.538 g

Chemistry

Using the balanced equation below, how many grams of manganese(III) oxide would be produced from the complete reaction of 46.8 g of zinc? Zn + 2MnO2 + H20-Zn(OH)2 + Mn2O3

Step-by-step answer

P Answered by PhD

Answer: The mass of manganese(III) oxide produced is 113.03 g.

Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
Number of moles = Given mass/Molar mass (1).
Given mass of zinc = 46.8 g;
Molar mass of zinc = 65.38 g/mol.
Plugging values in equation 1:
Moles of zinc = 46.8g/65.38g/mol = 0.716mol.
The given chemical equation follows:
Zn + 2MnO2 + H20-Zn(OH)2 + Mn2O3.
By the stoichiometry of the reaction:
If 1 mole of zinc produces 1 mole of manganese(III) oxide
So, 0.716 moles of zinc will produce = 1/1*0.716=0.716 mol of manganese(III) oxide.
Molar mass of manganese(III) oxide = 157.87 g/mol.
Plugging values in equation 1:
Mass of marganese (III) oxide = (0.716 mol*157.87g/mol)=113.03g.
Hence, the mass of manganese(III) oxide produced is 113.03 g.

Chemistry

Using the balanced equation below, how many grams of manganese(III) oxide would be produced from the complete reaction of 46.8 g of zinc? Zn + 2MnO2 + H20 — Zn(OH)2 + Mn203

Step-by-step answer

P Answered by Specialist

The mass of manganese(III) oxide produced is 113.03 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

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

Given mass of zinc = 46.8 g

Molar mass of zinc = 65.38 g/mol

Plugging values in equation 1:

$\text{Moles of zinc}=\frac{46.8g}{65.38g/mol}=0.716 mol$

The given chemical equation follows:

$Zn+2MnO_2+H_2O\rightarrow Zn(OH)_2+Mn_2O_3$

By the stoichiometry of the reaction:

If 1 mole of zinc produces 1 mole of manganese(III) oxide

So, 0.716 moles of zinc will produce = $\frac{1}{1}\times 0.716=0.716mol$ of manganese(III) oxide

Molar mass of manganese(III) oxide = 157.87 g/mol

Plugging values in equation 1:

$\text{Mass of manganese(III) oxide}=(0.716mol\times 157.87g/mol)=113.03g$

Hence, the mass of manganese(III) oxide produced is 113.03 g

Chemistry

Using the balanced equation below, how many grams of zinc would be required to produce 9.65 g of zinc hydroxide? Zn + 2MnO2 + H20 → Zn(OH)2 + Mn203 Please help I need it for a test

Step-by-step answer

P Answered by PhD

5

Mass = 6.538 g

Explanation:

Given data:

Mass of zinc hydroxide produced = 9.65 g

Mass of zinc required = ?

Solution:

Chemical equation:

Zn + 2MnO₂ + H₂O → Zn(OH)₂ + Mn₂O₃

Number of moles of zinc hydroxide:

Number of moles = mass/molar mass

Number of moles = 9.65 g/ 99.42 g/mol

Number of moles = 0.1 mol

now we will compare the moles of zinc and zinc hydroxide,

Zn(OH)₂ : Zn

1 : 1

0.1 : 0.1

Mass of zinc required:

Mass = number of moles × molar mass

Mass = 0.1 mol × 65.38 g/mol

Mass = 6.538 g

Chemistry

How many milliliters of manganese metal, with a density of 7.43 g/ml, would be needed to produce 21.7 grams of hydrogen gas in the single-replacement reaction below? show all steps of your calculation as well as the final answer. mn + h2o → mno + h2 (5 points)

Step-by-step answer

P Answered by PhD

6

First compute the number of grams of manganese metal required to make 21.7 grams of H2.
21.7 g H2 x (1 mole H2/ 2 g H2) x (1 mole Mn/1 mol H2) x (55 grams Mn/1 mol Mn) = 596.75 grams
Now density = mass/volume
7.43 = 596.75/volume
volume = 596.75/7.43 = 80.31 mL
80.31 mL is the amount of manganese needed.

Chemistry

How many milliliters of manganese metal, with a density of 7.43 g/ml, would be needed to produce 21.7 grams of hydrogen gas in the single-replacement reaction below? show all steps of your calculation as well as the final answer. mn + h2o → mno + h2 (5 points)

Step-by-step answer

P Answered by PhD

6

First compute the number of grams of manganese metal required to make 21.7 grams of H2.
21.7 g H2 x (1 mole H2/ 2 g H2) x (1 mole Mn/1 mol H2) x (55 grams Mn/1 mol Mn) = 596.75 grams
Now density = mass/volume
7.43 = 596.75/volume
volume = 596.75/7.43 = 80.31 mL
80.31 mL is the amount of manganese needed.

Chemistry

A sample of oxalic acid is titrated with a standardized solution of KMNO4. A 25 mL sample of oxalic acid required 12.7 mL of 0.0206 M KMnO4 to achieve a pink colored solution. The balanced equation for this reaction is shown below: 6 H+ (aq) + 2 MnO4 - (aq) + 5 H2C2O4(aq) → 10 CO2(g) +8 H2O(l) + 2Mn2+(aq) Required: a. What does the pink color signify in this reaction? b. What is the ratio of MnO4 - ions to H2C2O4 molecules in this reaction? c. How many moles of MnO4 - ions reacted with the given amount of oxalic acid solution? d. How many moles

Step-by-step answer

P Answered by PhD

11

Following are the solution to the given points:

Explanation:

Oxalic acid volume $= 25.00 \ mL = 0.0250 \ litres$

KMnO4 volume $= 12.70 \ ml = 0.0127 \ litres$

KMnO4 molarity $= 0.0206\ M = 0.0206 \ \frac{mol}{l}$

In point a:

Its pink presence after full intake of oxalic acid with attachment to KMnO4 is suggested by the end-point of the process due to the small abundance of KMnO4, As just a self predictor, KMnO4 is used.

In point b:

H_2C_2O_4 molecules mole ratio to MnO_4^- ions:

The equilibrium for both the oxalic acid and KMnO4 reaction is suggested:

$6H+ (aq) + 2MnO_4- (aq) + 5H_2C_2O_4 (aq) \rightarrow 10CO_2 (g) + 8H_2O (l) + 2Mn_2+ (aq)$

The reaction of 5 mol of oxalic acid is 2 mol MnO_4^- ions

H_2C_2O_4 : molecules mole proportion to MnO_4^- ions:

5 H_2C_2O_4 : : 2MnO_4^-

In point c:

The Moles of MnO_4^- ions reacted with the H_2C_2O_4 :

The molar mass of the solution is the number of solute moles in each volume of water

$Molarity =\frac{moles}{Volume}\\\\Moles \ of\ KMnO_4 = Molarity \times volume$

Moles with ions reacted to mol with both the amount of : supplied.

In point d:

H_2C_2O_4 moles in the sample present:

H_2C_2O_4 moles = moles MnO_4^- ions $\times$ mole ratio

H_2C_2O_4 moles in the sample = $2.6162 \times 10^{-4}\ mol \times (\frac{5}{2})$

H_2C_2O_4 molecules = $6,5405\times 10^{-4}$ mol are present in the sample

In point e:

Oxalic acid molarity = $\frac{mole}{volume}$

$=\frac{ 6.54 \times 10^{-4} mol}{0.025\ L} \\\\ = 0.0260 \ M$

In point f:

Oxalic acid level by mass in the solution:

Oxalic acid mass calculation:

Oxalic acid molar weight = 90.0349 $\frac{g}{mol}$ .

Oxalic acid mass per liter = oxalic acid moles per liter $\times$ molar mass

$= 0.0260 \frac{mol}{L} \times 90.0349 \frac{g}{mol}\\\\= 2.3409 \frac{g}{L}\\\\ = 2.3409 \frac{g}{1000 \ mL}\\\\= 0.2409 \frac{g}{100 \ mL}$

When Oxalic acid solution density $= 1.00 \ \frac{g}{mL}$

Mass oxalic acid percentage = $0.2409 \%$

Oxalic acid mass proportion $= 0.24\% \ \frac{W}{v} \ \ Mass$

Chemistry

A sample of oxalic acid is titrated with a standardized solution of KMNO4. A 25 mL sample of oxalic acid required 12.7 mL of 0.0206 M KMnO4 to achieve a pink colored solution. The balanced equation for this reaction is shown below: 6 H+ (aq) + 2 MnO4 - (aq) + 5 H2C2O4(aq) → 10 CO2(g) +8 H2O(l) + 2Mn2+(aq) Required: a. What does the pink color signify in this reaction? b. What is the ratio of MnO4 - ions to H2C2O4 molecules in this reaction? c. How many moles of MnO4 - ions reacted with the given amount of oxalic acid solution? d. How many moles

Step-by-step answer

P Answered by PhD

11

Following are the solution to the given points:

Explanation:

Oxalic acid volume $= 25.00 \ mL = 0.0250 \ litres$

KMnO4 volume $= 12.70 \ ml = 0.0127 \ litres$

KMnO4 molarity $= 0.0206\ M = 0.0206 \ \frac{mol}{l}$

In point a:

Its pink presence after full intake of oxalic acid with attachment to KMnO4 is suggested by the end-point of the process due to the small abundance of KMnO4, As just a self predictor, KMnO4 is used.

In point b:

H_2C_2O_4 molecules mole ratio to MnO_4^- ions:

The equilibrium for both the oxalic acid and KMnO4 reaction is suggested:

$6H+ (aq) + 2MnO_4- (aq) + 5H_2C_2O_4 (aq) \rightarrow 10CO_2 (g) + 8H_2O (l) + 2Mn_2+ (aq)$

The reaction of 5 mol of oxalic acid is 2 mol MnO_4^- ions

H_2C_2O_4 : molecules mole proportion to MnO_4^- ions:

5 H_2C_2O_4 : : 2MnO_4^-

In point c:

The Moles of MnO_4^- ions reacted with the H_2C_2O_4 :

The molar mass of the solution is the number of solute moles in each volume of water

$Molarity =\frac{moles}{Volume}\\\\Moles \ of\ KMnO_4 = Molarity \times volume$

Moles with ions reacted to mol with both the amount of : supplied.

In point d:

H_2C_2O_4 moles in the sample present:

H_2C_2O_4 moles = moles MnO_4^- ions $\times$ mole ratio

H_2C_2O_4 moles in the sample = $2.6162 \times 10^{-4}\ mol \times (\frac{5}{2})$

H_2C_2O_4 molecules = $6,5405\times 10^{-4}$ mol are present in the sample

In point e:

Oxalic acid molarity = $\frac{mole}{volume}$

$=\frac{ 6.54 \times 10^{-4} mol}{0.025\ L} \\\\ = 0.0260 \ M$

In point f:

Oxalic acid level by mass in the solution:

Oxalic acid mass calculation:

Oxalic acid molar weight = 90.0349 $\frac{g}{mol}$ .

Oxalic acid mass per liter = oxalic acid moles per liter $\times$ molar mass

$= 0.0260 \frac{mol}{L} \times 90.0349 \frac{g}{mol}\\\\= 2.3409 \frac{g}{L}\\\\ = 2.3409 \frac{g}{1000 \ mL}\\\\= 0.2409 \frac{g}{100 \ mL}$

When Oxalic acid solution density $= 1.00 \ \frac{g}{mL}$

Mass oxalic acid percentage = $0.2409 \%$

Oxalic acid mass proportion $= 0.24\% \ \frac{W}{v} \ \ Mass$

English

In writing a text, these two elements must be present: content and answer . question 2 not yet answered marked out of 1.00 not flaggedflag question question text refer to the given paragraph below on the great wall of china: the great wall, or chang cheng in chinese, is massive. it begins in the east at the yellow sea, travels near china’s capital, beijing, and continues west through numerous provinces. for thousands of miles, it winds like a snake through china’s varied terrain. smaller walls extend from the main wall. according to conservative

Step-by-step answer

P Answered by PhD

1

In writing a text, these two elements must be present: content and Answer

.Question 2

Refer to the given paragraph below on The Great Wall of China:

What type of description is used in the paragraph?

It is travel writing and it is a location description.

Question text

Which of the following statements is true?

b. You must come up with all the possible ideas from the pre-writing stage.

Question 4

Question text

Refer to the given paragraph below, entitled “Picturing Don Quixote”:

What type of description is used in the paragraph?

Character description and location description

Question 5

Question text

Learning to write is a process.

Question 6

The term that refers to the consequences or events caused by the climax.

The outcome at this point can be predicted.

Question 7

Which of the following questions is not beneficial in exploring your topic?

Select one:

a. Why is it an issue or problem at all?

b. How does the issue relate to other public issues?

c. At what place is the cause or effect of the problem most visible?

d. When is the issue most apparent?

Question 8

Question text

What type of figure of speech is used in the following sentence?

The lady in the water screamed like a banshee.

The correct answer is d. Simile

The writer presented what point of view?

c. First person

Given right triangle MNO below how many units long is NO

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