Can you give me the answers to questions 28 and 29 please?

Answer:

4.7K

Explanation:

Given parameters:

Initial volume  = 75L

Initial pressure  = 125psi  

        to atm gives 8.5atm

Initial temperature  = 288K

New volume  = 6.1L

New pressure  = 25psi

        to atm gives 1.7atm

Unknown:

New temperature = ?

Solution:

To solve this problem, we use the combined gas law which is given below:

        [tex]\frac{P1 V1}{T1}[/tex]    = [tex]\frac{P2 V2}{T2}[/tex]  

P, V and T are pressure, volume and temperature

1 and 2 are initial and new states

                  [tex]\frac{8.5 x 75}{288}[/tex]   = [tex]\frac{1.7 x 6.1}{T2}[/tex]

        T2  = 4.7K

2265 g Fe₃O₄

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

[RxN - Balanced] Fe₃O₄ + 4H₂ → 3Fe + 4H₂O

[Given] 705.0 g H₂O

Step 2: Identify Conversions

[RxN] 4 mol H₂O → 1 mol Fe₃O₄

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Fe - 55.85 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Molar Mass of Fe₃O₄ - 3(55.85) + 4(16.00) = 231.55 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 4 sig figs.

2264.74 g Fe₃O₄ ≈ 2265 g Fe₃O₄

Answer:

6.5256×10¹⁴

Explanation:

n= v/24

v= n×25

v= 2.719×10¹³ × 24= 6.5256×10¹⁴ moles/dm³

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According to ideal gas equation, the volume occupied by 2.719 x 10¹³ moles of methane gas is 6174.76×10¹³cm³.

The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.Ideal gases exist in hypothetical state.The law was proposed by Benoite Paul Emile Clapeyron in 1834.

Substituting values of pressure=1 atmosphere,number of moles=2.719 x 10¹³ ,R=8.314 ,T=273.15 K.

Therefore volume=nRT/P=2.719x 10¹³×8.314×273.15/1=6174.76×10¹³ cm³.

Thus, the volume occupied by 2.719 x 10¹³ moles of methane gas is 6174.76×10¹³cm³.

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Answer:

Glucose molecules

Explanation:

Answer:

I would say, what helps me is really paying attention in class and asking questions, also making sure you study for upcoming test's and quizzes and completely assingments on time

Explanation:

Answer:

3

Explanation:

The reaction expression is given as:

      Al(OH)₃  +   HNO₃   →    H₂O   +  Al(NO₃)₃  

To solve this problem, let us assign coefficient a,b,c and d to each specie;

           aAl(OH)₃  +   bHNO₃   →    cH₂O   +  dAl(NO₃)₃  

Conserving Al : a  = d

                   O:  3a + 3b  = c + 9d

                   H: 3a + b  = 2c

                   N: b  = 3d

let a = 1 , d  = 1, b  = 3 , c = 3

  Multiply through by 3;

 a  = 1, b  = 3, c = 3 and d  = 1

   Al(OH)₃  +   3HNO₃   →    3H₂O   +  Al(NO₃)₃  

Answer:

Explanation:

H₃PO₄ + 3NaOH = Na₃PO₄ + 3 H₂O.

1 mole     3 mole

22.85 mL of the standardized 0.0995M NaOH solution will contain

.0995 x .02285 moles of NaOH

= 22.7357 x 10⁻⁴ moles of NaOH

3 moles of NaOH reacts with 1 mole of H₃PO₄

22.7357 x 10⁻⁴ moles of NaOH reacts with 1 x 22.7357 x 10⁻⁴  / 3 mole of H₃PO₄

= 7.5785 x 10⁻⁴ moles of H₃PO₄

7.5785 x 10⁻⁴ moles of H₃PO₄  is contained in 10 mL or 10⁻² L

molarity of H₃PO₄ solution = 7.5785 x 10⁻⁴ / 10⁻²

= 7.58 x 10⁻² M

= .0758 M .

Answer:

Introduction

As a diligent student of chemistry, you will likely encounter tons of reactions that occur in aqueous solution (perhaps you are already drowning in them!). When ions are involved in a reaction, the equation for the reaction can be written with various levels of detail. Depending on which part of the reaction you are interested in, you might write a molecular, complete ionic, or net ionic equation.

Definitions of molecular, complete ionic, and net ionic equations

A molecular equation is sometimes simply called a balanced equation. In a molecular equation, any ionic compounds or acids are represented as neutral compounds using their chemical formulas. The state of each substance is indicated in parentheses after the formula. [Huh?]

Let's consider the reaction that occurs between \text{AgNO}_3AgNO

3

​

start text, A, g, N, O, end text, start subscript, 3, end subscript and \text{NaCl}NaClstart text, N, a, C, l, end text. When aqueous solutions of \text{AgNO}_3AgNO

3

​

start text, A, g, N, O, end text, start subscript, 3, end subscript and \text{NaCl}NaClstart text, N, a, C, l, end text are mixed, solid \text{AgCl}AgClstart text, A, g, C, l, end text and aqueous \text{NaNO}_3NaNO

3

​

start text, N, a, N, O, end text, start subscript, 3, end subscript are formed. Using this information, we can write a balanced molecular equation for the reaction:

\text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)AgNO

3

​

(aq)+NaCl(aq)→AgCl(s)+NaNO

3

​

(aq)start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis, plus, start text, N, a, C, l, end text, left parenthesis, a, q, right parenthesis, right arrow, start text, A, g, C, l, end text, left parenthesis, s, right parenthesis, plus, start text, N, a, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis

[What kind of reaction is this?]

If we could zoom in on the contents of the reaction beaker, though, we wouldn't find actual molecules of \text{AgNO}_3AgNO

3

​

start text, A, g, N, O, end text, start subscript, 3, end subscript, \text{NaCl}NaClstart text, N, a, C, l, end text, or \text{NaNO}_3NaNO

3

​

start text, N, a, N, O, end text, start subscript, 3, end subscript. Since \text{AgNO}_3AgNO

3

​

start text, A, g, N, O, end text, start subscript, 3, end subscript, \text{NaCl}NaClstart text, N, a, C, l, end text, and \text{NaNO}_3NaNO

3

​

start text, N, a, N, O, end text, start subscript, 3, end subscript are soluble ionic compounds, they dissociate into their constituent ions in water. For example, \text{NaCl}NaClstart text, N, a, C, l, end text dissociates into one ion of \text{Na}^+Na

+

start text, N, a, end text, start superscript, plus, end superscript for every ion of \text{Cl}^-Cl

−

start text, C, l, end text, start superscript, minus, end superscript; these ions are stabilized by ion-dipole interactions with the surrounding water molecules. [I don't get it!]

Image of crystalline sodium chloride next to image of chloride and sodium ions dissociated in water. Each chloride ion is interacting with multiple water molecules through the positive dipole of the water, and each sodium ion is interacting with water molecules through the negative dipole of the water.

Image of crystalline sodium chloride next to image of chloride and sodium ions dissociated in water. Each chloride ion is interacting with multiple water molecules through the positive dipole of the water, and each sodium ion is interacting with water molecules through the negative dipole of the water.

Sodium chloride dissociates into sodium and chloride ions in water, and these ions become solvated by the highly polar water molecules. Image credit: "Salts: Figure 1" by OpenStax Anatomy and Physiology, CC-BY-NC-SA 4.0.

From the molecular formula, we can rewrite the soluble ionic compounds as dissociated ions to get the complete ionic equation:

\text{Ag}^+(aq) + \blueD{{\text{NO}_3}^-(aq)} + \maroonD{\text{Na}^+(aq)} + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s) + \maroonD{\text{Na}^+(aq)} + \blueD{{\text{NO}_3}^- (aq)}Ag

+

(aq)+NO

3

​

−

(aq)+Na

+

(aq)+Cl

−

(aq)→AgCl(s)+Na

+

(aq)+NO

3

​

−

​

−

(aq)

​

+

Na

+

(aq)

​

+Cl

−

(aq)→AgCl(s)+

Na

+

(aq)

​

+

NO

3

​

−

(aq)

​

Answer:

B) separation by boiling points

Explanation:

According to the diagram depicted, crude oil is separated into its different units using boiling points.

Answer:

A. Theoretical yield of CaO is 11.59 g

B. Percentage yield of CaO = 58.76%

Explanation:

The following data were obtained from the question:

Mass of CaCO₃ = 20.7 g

Actual yield of CaO = 6.81 g

Theoretical yield of CaO =?

Percentage yield of CaO =?

The equation for the reaction is given below:

CaCO₃ —> CaO + CO₂

Next, we shall determine the mass of CaCO₃ that decomposed and the mass of CaO produced from the balanced equation. This can be obtained as follow:

Molar mass of CaCO₃ = 40 + 12 + (3×16)

= 40 + 12 + 48

= 100 g/mol

Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g

Molar mass of CaO = 40 + 16 = 56 g/mol

Mass of CaO from the balanced equation = 1 × 56 = 56 g

SUMMARY:

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

A. Determination of the theoretical yield of CaO.

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

Therefore, 20.7 g of CaCO₃ will decompose to produce =

(20.7 × 56)/100 = 11.59 g of CaO.

Thus, the theoretical yield of CaO is 11.59 g

B. Determination of the percentage yield.

Actual yield of CaO = 6.81 g

Theoretical yield of CaO = 11.59 g

Percentage yield of CaO =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 6.81/11.59 × 100

Percentage yield of CaO = 58.76%

Answer:

ion-dipole

Explanation:

Let us remember that the potassium cation is an ion, a positive ion to be precise.

There is a dipole existing in the hydrogen cyanide molecule. The positive end of the dipole is on hydrogen while the negative end of the dipole is on the cyanide moiety.

The only possible interaction between the potassium cation and the hydrogen cyanide is an ion dipole interaction. The cation interacts with the cyanide moiety having a partial negative charge in the molecule.

The number of atoms in 1 mole of carbon equals the number of atoms in 1 mole of boron is false because the molar mass of carbon is different from that of boron. Option A is correct.

From the listed option the number of atoms in 1 mole of carbon equals the number of atoms in 1 mole of boron is false because the molar mass of carbon is different from that of boron

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Answer: 720 g

Explanation:

To solve this we must be knowing each and every concept related to mole. Therefore, shivani measures out 6.0 moles of epsom salt (MgSO[tex]_4[/tex]) to put her bath. 1,566g MgSO[tex]_4[/tex] went in the bath.

A mole is merely a measuring unit. In reality, it is one of its International System of Units' seven foundation units. When current units are insufficient, new units are created.

Chemical reactions frequently occur at levels where employing grams would be inappropriate, but using absolute quantities of atoms/molecules/ions would also be misleading.

For all practical reasons, one mole of a substance in grams is nearly equal to 1 molecule for the compound per daltons.

Mathematically,

number of mole =given mass ÷ molar mass

number of mole=6.0 moles

molar mass of epsom= 261.47g/mol

substituting all the given values in the above equation, we get

6.0 moles =mass of epsom ÷ 261.47g/mol

mass of epsom =1,566g

Therefore, 1,566g  MgSO[tex]_4[/tex] went in the bath.

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The mass of the boat is equal to the mass of the liquid which is displaced by placing the boat in the liquid.

Archimedes' Principle states that the buoyant force of the object is equal to the weight of the water which is displaced by the boat. If the weight of the water that is displaced by the boat is less than that of the weight of the boat, then in this case the boat sinks. Archimedes' principle or the buoyant force is very useful for the calculation of the volume of an object that does not have a regular shape.

The mass of the displaced liquid is always equal to the volume of the liquid multiplied by the density of the fluid. When a boat is placed in the water, the volume of displaced water is equal to the mass of the boat which is placed in it.

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Answer:

Photosynthesis transforms light energy into chemical energy by making sugars. Cellular respiration releases the chemical energy from food by breaking down sugars.

Explanation:

I'm actually learning this in school right now

Hope this helps!

~PurpleMist

Explanation:

The half-life is the time taken for a radioactive substance to decay to half of its original composition.

    Let us assume that the original amount is x atoms;

                     Half - life   = 60days

Days                          Composition           Half-life number

0                                         x                                   0

60                                   x/2                                    1

120                                  x/4                                    2

180                                   x/8                                   3

240                                  x/16                                   4

300                                  x/32                                  5

360                                   x/64                                6

The fraction that would be left after 360 days will be [tex]\frac{1}{64}[/tex]

The mole fraction of oxygen gas : 0.381

Given

2.31 atm Oxygen

3.75 atm Hydrogen

Required

The mole fraction of Oxygen

Solution

Dalton Law's of partial pressure

P tot = P₁ + P₂ + ..Pₙ

Input the value :

P tot = P O₂ + P H₂

P tot = 2.31 atm + 3.75 atm

P tot = 6.06 atm

The mole fraction of O₂ (X O₂):

P O₂ = X O₂ x P tot

X O₂ = P O₂ / P tot

X O₂ = 2.31 /6.06

X O₂ = 0.381

The mole fraction of oxygen gas in the given mixture of oxygen and hydrogen gas is 0.381.

If the details is present in terms of partial pressure then by using the below equation, we can calculate the mole fraction as:

p = (x)(P), where

p = partial pressure

P = total pressure

x = mole fraction

Total pressure of the mixture will be calculated by adding the partial pressure of oxygen and hydrogen gases as:

P = 2.31 + 3.75 = 6.06 atm

Now mole fraction of oxygen will be calculated as:

x = 2.31 / 6.06 = 0.381

Hence mole fraction of oxygen is 0.381.

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Answer:

UMMMMMMMMMMM

Explanation:

Answer:

3729

Explanation:

Answer:

wave length=speed*frequency

=36*6

=216 m

Answer:

The magnification is 10x

Explanation:

Answer: The magnification of the eyepiece on the microscope is the lens at the top of the microscope that the user looks through. This lens most commonly magnifies a sample by 10x

Explanation: The eyepiece magnifies the primary image produced by the objective; the eye can then use the full resolution capability of the objective. The microscope produces a virtual image of the specimen at the point of the distinct vision.

Answer:

0.768g of Fe are in the tablet = 38.4% by mass of Fe in the tablet

Explanation:

The balanced redox equation of Fe²⁺ with KMnO₄ in acidic medium is:

5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Where 5 moles of Fe²⁺ react per mole of KMnO₄

The moles required of KMnO₄ are:

27.50cm³ * (1dm³ / 1000cm³) = 0.0275dm³ * (0.100mol / dm³) = 0.00275mol KMnO₄

Moles of Fe²⁺ are:

0.00275mol KMnO₄ * (5 moles Fe²⁺ / 1mol KMnO₄) = 0.01375moles of Fe²⁺

In grams (Molar mass Fe: 55.845g/mol):

0.01375moles of Fe²⁺ * (55.845g/mol) =

And percentage is:

0.768g Fe / 2.000g * 100 =

Answer:

Louis braille was a french educator and inventor of a sytem of reading and writing for use use by the blind or visually impaired . his system remains virtually unchanged to this day and is know worldwide simply as braille

Answer:

vgffffgggggvghvxgyfdgubvhumigrrdghg

Answer:

Molar mass of the substance

Answer:

3

Explanation:

The three things that can affect a condition to enforce a slide is friction, weight, and slope of an area. Solutions that can make flow easily are those that can create less friction depending on the condition of an environment where it is present.

Answer:

22.6%

Explanation:

I just answered and got it right

Answer:

See detailed explanation.

Explanation:

Hello!

In this case, for the described chemical reaction, we can proceed as follows:

A) For the complete chemical reaction we note down every reacted and produced species as well as the proper balancing process:

[tex]2NaBr(aq)+Cl_2(aq)\rightarrow 2NaCl(aq)+Br_2(g)[/tex]

In which gaseous bromine may give off.

B) The dissociated ionic equation requires the ionization of the aqueous species in ions, expect for chlorine which is not ionized:

[tex]2Na^+(aq)+2Br^-(aq)+Cl_2(aq)\rightarrow 2Na^+(aq)+2Cl^-(aq)+Br_2(g)[/tex]

C) For the net ionic equation we cancel out the sodium ions as they are at both reactants and products:

[tex]2Br^-(aq)+Cl_2(aq)\rightarrow +2Cl^-(aq)+Br_2(g)[/tex]

D) Based on C) we infer that the spectator ions here are the sodium ions.

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