How many mL of 0.2860 M sulfuric acid are required to react with 58.42 mL of 0.09756 M iron(III) hydroxide

Answers

Answer 1

Answer:

0.05978 mL of 0.2860 M sulfuric acid is required to react with 58.42 mL of 0.09756 M iron (III) hydroxide.

Explanation:

The balanced reaction between tha surfuric acid and iron (III) hydroxide is:

Fe(OH)₃ + 3 H₂SO₄ → Fe(HSO₄)₃ + 3 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:

Fe(OH)₃: 1 mole H₂SO₄: 3 molesFe(HSO₄)₃: 1 mole H₂O: 3 moles

Being molarity (M) the number of moles of solute that are dissolved in a given volume:

[tex]Molarity=\frac{number of moles}{volume}[/tex]

then the number of moles can be calculated as:

number of moles= molarity* volume

So, if 58.42 mL ( 0.05842 L) of 0.09756 M iron (III) hydroxide react, then it means that the number of moles that react are:

number of moles= 0.09756 M* 0.05842 L

number of moles= 0.005699 moles

Now it is possible to apply the following rule of three: if by stoichiometry 1 mole of Fe(OH) ₃ reacts with 3 moles of H₂SO₄, then 0.005699 moles of Fe(OH) ₃ with how many moles of H₂SO₄ does it react?

[tex]moles of H_{2} SO_{4} =\frac{0.005699 moles ofFe(OH)_{3}* 3moles of H_{2} SO_{4} }{1mole ofFe(OH)_{3}}[/tex]

moles of H₂SO₄= 0.017097

Being sulfuric acid 0.2860 M, it is replaced in the definition of molarity:

[tex]0.2860M=\frac{0.017097 moles}{volume}[/tex]

Solving for the volume:

[tex]volume=\frac{0.017097 moles}{0.2860 M}[/tex]

volume= 0.05978 mL

0.05978 mL of 0.2860 M sulfuric acid is required to react with 58.42 mL of 0.09756 M iron (III) hydroxide.


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Answer:

the answer is A

Explanation:

Answer:

a

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the roller coaster is at the highes pont witch is a .

Which the following statement ia correct for the equation shown here​

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B.

As you can see both NO and NH3 have 4 moles therefore it is 4:4 between the molecules or in other words a 1:1 ratio in simplest forms

List 2 differences between an ionic and covalent chemical bond

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1. Ionic bonds result from transfer of electrons, whereas covalent bonds are formed by sharing. 2. Ionic bonds are electrostatic in nature, resulting from that attraction of positive and negative ions that result from the electron transfer process; charge separation between covalently bonded atoms is less extreme.

A sample of an unknown metal has a mass of 22.4g. A graduated cylinder contains 3.2 ml of water. After the metal sample is added to the graduated cylinder, the water level reads 4.7 ml. Calculate the density of the metal

Answers

Answer:

d = 14.9 g/mL

Explanation:

Given data:

Mass of metal = 22.4 g

Volume of eater = 3.2 mL

Volume of water + metal = 4.7 mL

Density of metal = ?

Solution:

Volume of metal:

Volume of metal = volume of water+ metal - volume of water

Volume of metal = 4.7 mL - 3.2 mL

Volume of metal = 1.5 mL

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4. A 40.0L tank of ammonia has a pressure of 12.7 kPa. Calculate the volume of the
ammonia if its pressure is changed to 8.4 kPa while its temperature remains constant.

Answers

Answer:

V₂ = 60.47L

Explanation:

Given data:

Initial volume = 40.0 L

Initial pressure = 12.7 KPa

Final volume = ?

Final pressure = 8.4 KPa

Solution:

The given problem will be solved through the Boyls law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

12.7 KPa× 40.0 L = 8.4 KPa × V₂

V₂ = 12.7 KPa× 40.0 L/ 8.4 KPa

V₂ = 508 KPa. L / 8.4 KPa

V₂ = 60.47L

The volume of the ammonia gas if it's pressure is changed to 8.4 KPa is 60.5 L

From the question given above, The following data were obtained:

Initial pressure (P₁) = 12.7 KPa

Initial volume (V₁) = 40 L

Final pressure (P₂) = 8.4 KPa

Temperature = constant

Final volume (V₂) = ?

Using the Boyle's law equation, the final volume of the ammonia can be obtained as illustrated below:

P₁V₁ = P₂V₂

12.7 × 40 = 8.4 × V₂

508 = 8.4 × V₂

Divide both side by 8.4

V₂ = 508 / 8.4

V₂ = 60.5 L

Therefore, the final volume of the ammonia gas is 60.5 L

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Fe2O3(s) 2Al(s)Al2O3(s) 2Fe(s) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.73 moles of Fe2O3(s) react at standard conditions.

Answers

Answer:

[tex]\Delta S=54.3\frac{J}{K}[/tex]

Explanation:

Hello!

In this case, for the given reaction, we can write the equation to compute the entropy change as shown below:

[tex]\Delta s=2\Delta S_{Fe}+\Delta S_{Al_2O_3}-2\Delta S_{Al}-\Delta S_{Fe_2O_3}[/tex]

Letting:

[tex]\Delta s_{Fe}=27.3\frac{J}{mol*K}\\\\ \Delta s_{Fe_2O_3}=84.4\frac{J}{mol*K}\\\\\Delta s_{Al}=28.3\frac{J}{mol*K}\\\\\Delta s_{Al_2O_3}=51.00\frac{J}{mol*K}[/tex]

We obtain the entropy change per mole of Fe2O3(s):[tex]\Delta s=2*27.3\frac{J}{mol*K}+84.4\frac{J}{mol*K}-2*28.3\frac{J}{mol*K}-51.0\frac{J}{mol*K} \\\\\Delta s=31.4\frac{J}{mol*K}[/tex]

Finally, the total entropy change when 1.73 moles of Fe2O3(s) react turns out:

[tex]\Delta s=31.4\frac{J}{mol*K}*1.73mol\\\\\Delta S=54.3\frac{J}{K}[/tex]

Best regards!

After burning your candle for a few hours you and your friend notice that it's mass has decreased. Your friend claims that the mass "disappeared". Why do you know he is wrong (Hint: it's a Scientific Law)? What actually happened to the mass? What form is that mass in now? What is this process called?

Answers

Answer:

The law of conversion of mass is a scientific law that says that mass or matter can not be created or destroyed like energy but changes its form in a different type of state. So, by the law of conversion of mass, you can say that mass is not disappeared but change in different forms of states of matter like gas, soot, or melted residues of candle and ash of thread.

Due to this, the mass of the solid or waxy state due to burning changes to gas released by burning and residue of wax that will be equal to the mass of the original mass.

H E L P I NEED HELPPP!!!! The primary purpose of the experimental method is to *

solve problems.
design experiments.
make observations.
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Answers

Answer:

The answer is make observations

Explanation:

Observing is the first step of the experimental method.

Match the functions with the corresponding cell structure.

Column A
1.
Cell Membrane:
Cell Membrane
2.
Nucleus:
Nucleus
3.
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Cytoplasm
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1c 2a 3b

Explanation:

Answer:

1.C

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3. B

Explanation:

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Answers

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Answers

Answer:

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   o o

o  O  o o

     o

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A) 1
B) 2
C) 8
D) 11

Answers

Answer:

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Answer: A)1

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Valence electrons are always in Thebes last shell and last number of the configuration.

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Answers

There is no comparable easy way to experimentally measure the change in entropy for a reaction

Answer:

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Explanation:

Hope it helps

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Answer: the last one

Explanation:

What class of organic product results when 1-heptyne is reacted with disiamylborane followed by treatment with basic hydrogen peroxide?

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Answer:

aldehyde

Explanation:

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So, when terminal alkynes, for example, 1-heptyne react on Hydroboration oxidation(i.e. disiamylborane followed by treatment with basic hydrogen peroxide), the formation of aldehyde occurs.

According to the ,cell theory where do cells come from?

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Explanation:

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Genchemium, a hypothetical metal with a atomic mass of 150 g/mol, has a body-centered (bcc) cubic unit cell and a density of 1.93 g/cm3. What is the edge length (in pm) of genchemium

Answers

Answer:

edge length a = 636.7 pm

Explanation:

If you check the image below, you would see that there is a technical error that makes it unable to submit the answer, but to curb that effect, I have attached the screenshot of my answer showing how explained it by using the text-editor.

Thanks!

The edge length (in pm) of the hypothetical metal genchemium is;

a = 0.6367 pm

We are given;

Atomic mass; M = 150 g/mol

Density; ρ = 1.96 g/cm³

This is a BCC structure and as such formula to find the edge length is;

ρ = (z × M)/(N × a³)

where;

z = number of atoms available in each unit cell = 2 atoms

M is atomic mass

N is avogadro's number = 6.023 × 10²³ mols

a is edge length

Thus, making a the subject gives;

a = ∛(z × M)/(N × ρ)

Plugging in the relevant values;

a = ∛(2 × 150)/(6.023 × 10²³ × 1.96)

a = 63.67 × 10⁻⁹ cm

converting to meters gives;

a = 0.6367 × 10⁻⁹ m

converting to pm gives;

a = 0.6367 pm

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a 125 torr mixture of c2h4(g) and c3hg(g) at a given temperature was burned in excess oxygen in a rigid, sealed container. the product c02( g) was collected under the same initial reaction conditions and determined to be 280 torr. what was the mole fraction of c2h4(g) in the original mixture

Answers

Answer:

0.24

Explanation:

C2H4(g) + 3O2(g) ------> 2CO2(g) + 2H2O(g)

C3 H8 (g) + 5O2 (g) → 3CO2(g)  +4H2 O(g)

Initially,

PV=nRT

125×V=(a+b)RT .....(i)

After combustion, pressure is due to the total moles of CO2  

280×V=(3a+2b)RT .....(ii)

Dividing equation ii by i

280/125 = 3a+2b/a+b

2.24 = 3a+2b/a+b

2.24a + 2.24b =3a+2b

2.24b - 2b = 3a - 2.24a

0.24b = 0.76a

b= 3.2a

mole fraction of C3H8 = a/a + b = a/a + 3.2a = 1/4.2 = 0.24

Compare and contrast lithium and potassium and also fluorine !!!??? Pls need help

Answers

lithium and potassium are both in the alkali family and they are metals. Fluorine is a non-metal and lives in the halogens family. But, each of them have 2 valence electrons in the innermost ring

How would I find the number of protons, neutrons and electrons when I only have the atomic number and atom (assignment says assume the net charge is 0)

Answers

Explanation:

Atomic number , protons and electrons have the same value / their value is same .

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Answers

Answer:

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Explanation:

HELPP A GIRLLL OUTT!! PLEASEEE I WOULD APPRECIATE THAT ! <3
Which element has the same number of electrons in the outermost energy level at Lithium?
A. Hydrogen
B. Beryllium
C. Carbon
D. Magnesium

Answers

Answer:

Hydrogen is your answer because lithium and hydrogen both have 1 electron in their outermost energy level


What volume in milliliters of 100 proof vodka (50% v/v ethanol) would be likely to kill a typical 360-g rat?

Answers

The given question is incomplete, the complete question is:

The LD50 value for ethyl alcohol given orally to rats is 10.3 g/kg. What volume in milliliters of 100 proof vodka (50% v/v ethanol) would be likely to kill a typical 360-g rat?

Answer:

The correct answer is 9.4 ml.

Explanation:

Based on the given information, in rats the LD50 of ethyl alcohol is 10.3 g/kg.

As 1 Kg is 1000 grams

Now the value of LD50 of ethyl alcohol in rats is 10.3 grams/ 1000 grams

The volume of the lethal dose of vodka for a 360 gram rat would be,  

= 360 g × (10.3 g/1000 g)

= 3708 × 10⁻³ g

= 3.708 g

Now the volume of vodka can be determined by using the formula,  

V = mass/density

The density of ethanol is 0.789 g/ml

The volume would be,  

V = 3.708 g/0.789 g/ml

V = 4.70 ml

As 50 percent of Vodka is ethanol, the volume of vodka needed to kill a rat is,  

= 2 × Volume of ethanol

= 2 × 4.70 g/ml

= 9.4 ml

What is the atom of CO2

Answers

Answer:

Carbon dioxide is composed of one carbon atom covalently bonded to two oxygen atoms. It is a gas (at standard temperature and pressure) that is exhaled by animals and utilized by plants during photosynthesis. Carbon dioxide, CO2, is a chemical compound composed of two oxygen atoms and one carbon atom

Explanation:

Answer:

Carbon dioxide is composed of one carbon atom covalently bonded to two oxygen atoms. It is a gas (at standard temperature and pressure) that is exhaled by animals and utilized by plants during photosynthesis. Carbon dioxide, CO2, is a chemical compound composed of two oxygen atoms and one carbon atom.

Explanation:

hope this help

pick me as the brainliest

Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodium in g/cm3

Answers

Answer:

Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

The atomic mass of rhodium = 102.90 amu

For a face-centered cubic unit cell,

[tex]r = \dfrac{a}{2\sqrt{2}}[/tex]

where;

a = edge length.

Making "a" the subject of the formula:

[tex]a = 2 \sqrt{2} \times r[/tex]

[tex]a = 2 \times 1.414 \times 135 \ pm[/tex]

a = 381.8 pm

to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

[tex]density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}[/tex]

where;

mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

[tex]mass\ of\ atom =\dfrac{ atomic \ mass}{Avogadro \ number}[/tex]

[tex]mass\ of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}[/tex]

Recall also that number of atoms in a unit cell for a  face-centered cubic = 4

So;

[tex]mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4[/tex]

mass of unit cell = 6.83380375 × 10⁻²² g

[tex]Density \ of \ unit \ cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}[/tex]

Density of unit cell ( rhodium) = 12.279 g/cm³

The density of rhodium is equal to 12.4g/cm^3

Data;

radius = 135pmatomic mass = 102.90 amu

The radius of a FCC is calculated as

[tex]r = \frac{a}{2\sqrt{2} }\\a = 2\sqrt{2} * r\\a = 2 * 1.414 * 135 = 381.8pm = 381.8 * 10^-^1^0cm\\[/tex]

The Density of Rhodium

The formula of density is given as

[tex]\rho = \frac{mass}{volume}\\[/tex]

The mass of a unit cell = mass of atom * number of atoms per unit cell.

[tex]mass of atom = \frac{atomic mass}{avogadro's number} = \frac{102.90}{6.023*10^2^3}\\[/tex]

For BCC number of atom in unit cell = 4

mass of unit cell = [tex]\frac{102.90}{6.02*10^2^3} * 4[/tex]

This makes the density of the atom equal to

[tex]\rho = \frac{\frac{102.90}{6.02*10^2^3} }{(381*10^-^1^0)^3}* 4\\ \rho = 12.4 g/cm^3[/tex]

The density of Rhodium is 12.4 g/cm^3

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The following reaction shows calcium chloride reacting with silver nitrate.
CaCl2 + 2AgNO3 → 2AgCl + Ca(NO3)2
How many grams of AgCl are produced from 30.0 grams of CaCl2?
(Molar mass of Ca = 40.078 g/mol, Cl = 35.453 g/mol, O = 15.999 g/mol, Ag = 107.868 g/mol, N = 14.007 g/mol)
A. 19.4 grams
B. 38.8 grams
C. 58.2 grams
D. 77.5 grams

Answers

Explanation:

Molar mass of CaCl2 = 110.98g/mol

Moles of CaCl2 = 30.0 / 110.98 = 0.270mol

Moles of AgCl = 0.270mol * 2 = 0.540mol

Mass of AgCl = 0.540mol * (143.32g/mol) = 77.39g.

Answer:

77.5

Explanation:

You gotta round.

Hope this helps :)

A student applied a force of 200 N to push a box 15 m down the sidewalk. Given the formula. W = F xd, how much work has
been done?
O 30 J

Answers

Answer:

3,000 J

W = f x d

W = 200 x 15

W = 3,000 J

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