What is the percent of iron and oxygen in iron(III) oxide?

Answer:

Approximately [tex]0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}[/tex].

Explanation:

The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)

Calculate the increase in the temperature of this sample:

[tex]\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C[/tex].

The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)

It took [tex]2911\; \rm J[/tex] of energy to raise the temperature of this sample by [tex]\Delta T = 92.6\; \rm ^\circ\! C[/tex]. Therefore, raising the temperature of this sample by [tex]1\; \rm ^\circ\! C[/tex] (unit temperature) would take only [tex]\displaystyle \frac{1}{92.6}[/tex] as much energy. That corresponds to approximately [tex]31.436\; \rm J[/tex] of energy.

On the other hand, the energy required to raise the temperature of this material by [tex]1\; \rm ^\circ\! C[/tex] is proportional to the mass of the sample (also assuming no phase change.)

It took approximately [tex]31.436\; \rm J[/tex] of energy to raise the temperature of [tex]57.07\; \rm g[/tex] of this material by [tex]1\; \rm ^\circ C[/tex]. Therefore, it would take only [tex]\displaystyle \frac{1}{57.07}[/tex] as much energy to raise the temperature of [tex]1\; \rm g[/tex] (unit mass) of this material by [tex]1\; \rm ^\circ \! C\![/tex]. That corresponds to approximately [tex]0.551\; \rm J[/tex] of energy.

In other words, it takes approximately [tex]0.551\; \rm J[/tex] to raise [tex]1\; \rm g[/tex] (unit mass) of this material by [tex]1\; \rm ^\circ \! C[/tex]. Therefore, by definition, the specific heat of this material would be approximately [tex]0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}[/tex].

Answer:

B

Explanation:

I did the question before and got it right.

Answer:

Chemical symbols refer to chemical elements only. They do not necessarily refer to atoms of that element, but also to ions.

Explanation:

Answer:

Explanation:

Caffeine is a weak base with pKb = 10.4

Kb = 10⁻¹⁰°⁴ = 3.98 x 10⁻¹¹

molecular weight of caffeine = 194.2

455 x 10⁻³ g / L = 455 x 10⁻³ / 194.2 moles / L

concentration of given solution a = 2.343 x 10⁻³ M

Let the caffeine be represented by B .

B    +   H₂O =  BH + OH⁻

a - x                   x        x  

x² / ( a - x ) = Kb

x² / ( a - x ) = 3.98 x 10⁻¹¹

x is far less than a so a -x is almost equal to a

x² = 3.98 x 10⁻¹¹ x 2.343 x 10⁻³ = 9.32  x 10⁻¹⁴

x = 3.05 x 10⁻⁷

[ OH⁻ ] = 3.05 x 10⁻⁷

pOH = - log ( 3.05 x 10⁻⁷ )

= 7 - log 3.05

= 7 - 0.484 = 6.5

pH = 14 - 6.5 = 7.5  

The pH of 455 mg/L of caffeine is 7.5

Using the formula;

Mass concentration = molar concentration × molar mass

Molar mass of caffeine = 194 g/mol

Mass concentration of caffeine = 455 mg/L

Molar concentration = Mass concentration/molar mass

Molar concentration = 455 × 10^-3g/L/194 g/mol

= 0.00235 M

Let Caffeine by depicted by the general formula BH

We can now set up the ICE table as follows;

           :B + H2O   ⇄ BH       +       OH^-

I      0.00235              0                      0

C           - x                  +x                    +x

E    0.00235  - x           x                      x

Note that water is present in large excess

Again; pKb of caffeine =10.4

Kb = Antilog[-pKb]

Kb = Antilog [-10.4]

Kb = 3.98 × 10^-11

Kb = [BH] [OH^-]/[:B]

3.98 × 10^-11 = [x] [x]/[ 0.00235  - x  ]

3.98 × 10^-11 [ 0.00235  - x  ] = [x] [x]

9.4 × 10^-14 - 3.98 × 10^-11x = x^2

x^2 + 3.98 × 10^-11x  - 9.4 × 10^-14 = 0

x = 3.1 × 10^-7 M

Recall  [BH] = [OH^-] = 3.1 × 10^-7 M

Now;

pOH = - log [OH^-]

pOH = log [3.1 × 10^-7 M]

pOH = 6.5

But;

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 6.5

pH = 7.5

The pH of 455 mg/L of caffeine is 7.5

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Missing parts

Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4.  Calculate the pH of a solution containing a caffeine concentration  of 455 mg/L.

Answer:

380

Explanation:

Answer:

electron shell is the answer

Answer:

The Group 2 metals become more reactive towards the water as you go down the Group.

Explanation:

These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.

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Answer:

0.75 moles of CO2

Explanation:

molar mass of CO2⇒ 44.01 g/mol

molar mass Oxygen ⇒ 31.998 g/mol

divide the mass given by molar mass of oxygen  so

48÷31.998= 1.50009376 moles of O

then you use the ration of oxygen to carbon to find the moles of CO2 which is one mole of CO2 for 2 moles of oxygen

1.50009376 moles of O ×[tex]\frac{1}{2}[/tex] = 0.75004688

with sig figs

0.75 moles of CO2

Answer:

1 ➡️ Cells

2 ➡️ Arteries

3 ➡️ Veins

4 ➡️ Heart

Explanation:

The parts of the electric circuit that best correspond to the heart, arteries, veins, and cells have been properly labeled.

The circulatory system involves the transportation of nutrients, oxygen and water by blood to other the parts of the body.

From the electric circuit, we see that arteries transport blood away from the heart to the other cells in the body. The veins actually return the blood back to the heart from the cells. The heart pumps the blood

The electric circuity diagram has the label 1 bulb analogous to cell, label 2 analogous to arteries, label 3 analogous to veins, and label 4 cell analogous to heart.

The electric circuit has been given as the power source and the conducting wires that allows the flow of the current in the circuit.

In the human body, the heart has been transported the oxygenated blood through the arteries to the cell and carried the deoxygenated blood from the cells back to the heart via veins.

In the circuit, the battery has been the source of the power/blood. The current has been carried from the heart to the cell/bulb through the arteries labeled, 2, and transported back to the battery via veins labeled 3.

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Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

Answer:

Explanation:

In my opinion the answer should be SO2

Answer:

a should be answer i think.

Answer:

K loses one electron to CI

Explanation:

The lewis electron dot notation shows only the chemical symbol of the element surrounded by dots to represent the valence electrons.

  We have atom of K with one valence electrons

   Cl with 7 valence electrons

For an electrostatic attraction to occur, both particles must be charged. To do this, one of the species must lose an electron, and the other gains it.

This will make both species attain a stable octet;

   Hence, K will lose 1 electron and Cl will gain the electrons.

Answer:

C: K loses one electron to CI

Explanation:

I took the test and got it correct!!

The heat of vaporization of benzene is required.

The heat of vaporization of benzene is 33009 J/kg.

[tex]T_0[/tex] = Normal boiling point = 80.1+273.15 K

[tex]T_B[/tex] = Boiling point at given pressure = 26.1+273.15 K

[tex]R[/tex] = Gas constant = 8.314 J/mol K

[tex]P[/tex] = Pressure at given [tex]T_B[/tex] = 100 torr

[tex]\Delta H[/tex] = Heat of vaporization

From the Clausius–Clapeyron equation

[tex]\dfrac{1}{T_B}=\dfrac{1}{T_0}-\dfrac{R\ln(\dfrac{P}{P_0})}{\Delta H}\\\Rightarrow \Delta H=\dfrac{R\ln\dfrac{P}{P_0}}{\dfrac{1}{T_0}-\dfrac{1}{T_B}}\\\Rightarrow \Delta H=\dfrac{8.314\times \ln\left(\frac{100}{760}\right)}{\frac{1}{80.1+273.15}-\frac{1}{26.1+273.15}}\\\Rightarrow \Delta H=33008.99\ \text{J/kg}[/tex]

The heat of vaporization of benzene is 33009 J/kg.

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Answer:

K and ba

Explanation:

Answer: K and Ba

Explanation:

Answer:

7

Explanation:

Let x represent the number of moles of water in the hydrated salt i.e MgSO₄.xH₂O

The following data were obtained from the question:

Mass of MgSO₄.xH₂O = 1.547 g

Mass of anhydrous MgSO₄ = 0.7554 g

Mole of H₂O = x =?

Next, we shall determine the mass of water, H₂O in the hydrated salt, MgSO₄.xH₂O. This can be obtained as follow:

Mass of MgSO₄.xH₂O = 1.547 g

Mass of anhydrous MgSO₄ = 0.7554 g

Mass of H₂O =?

Mass of H₂O = (Mass of MgSO₄.xH₂O) – (Mass of anhydrous MgSO₄)

Mass of H₂O = 1.547 – 0.7554

Mass of H₂O = 0.7916 g

Finally, we shall determine the value of the x as illustrated below:

Mass of MgSO₄.xH₂O = 1.547 g

Molar mass of MgSO₄.xH₂O = 24 + 32 + (16×4) + x[(2×1) + 16]

= 24 + 32 + 64 + x(2 + 16)

= 120 + 18x

Mass of H₂O = 0.7916 g

Molar mass of xH₂O = 18x

Molar Mass of xH₂O/ Molar mass of MgSO₄.xH₂O = mass of xH₂O /Mass of MgSO₄.xH₂O

18x/ 120 + 18x = 0.7916/1.547

Cross multiply

0.7916 (120 + 18x) = 18x × 1.547

94.992 + 14.2488x = 27.846x

Collect like terms

94.992 = 27.846x – 14.2488x

94.992 = 13.5972x

Divide both side by 13.5972

x = 94.992 / 13.5972

x = 7

Thus, the formula for the hydrated salt, MgSO₄.xH₂O is MgSO₄.7H₂O

Number of moles of water, H₂O in the hydrated salt MgSO₄.7H₂O is 7.

The number of moles of attached water molecules is 7.

Mass of hydrated MgSO4 = 1.547 grams

Mass of anhydrous MgSO4 = 0.7554 grams

Number of moles of hydrated MgSO4 = 1.547 grams/120 + 18x

Number of moles of anhydrous MgSO4 = 0.7554 grams /120

Number of moles of anhydrous salt = Number of moles of hydrated salt

0.7554 grams /120 = 1.547 grams/120 + 18x

0.7554(120 + 18x) = 1.547 × 120

90.6 + 13.6x = 185.6

185.6 - 90.6 /13.6 = x

x = 7

The number of moles of attached water molecules is 7.

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Answer:

A chemical formula tells us the number of atoms of each element that is in a compound. It contains the symbols of the atoms for the elements present in the compound as well as how many there are for each element in the form of subscripts.

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Answer:

3.65 mol O₂

Explanation:

Step 1: RxN

2H₂O₂ → 2H₂O + O₂

Step 2: Define

Given - 7.30 mol H₂O₂

Solve - x mol O₂

Step 3: Stoichiometry

[tex]7.30 \hspace{3} mol \hspace{3} H_2O_2(\frac{1 \hspace{3} mol \hspace{3} O_2}{2 \hspace{3} mol \hspace{3} H_2O_2} )[/tex] = 3.65 mol O₂

Answer:

Rubidium

Answer: Rubidium is the most reactive metal. Explanation: Metals are the elements that looses electrons and thus, their chemical reactivity will be the tendency to loose electrons.

Explanation:

Hello. You forgot to say that this question is about the possibility of using some type of product that prevents cut apples from becoming darker.

Answer and Explanation:

1. The problem with this activity is "What can you put on an apple slice to keep it from turning brown?" The problem in a scientific experiment is the element that provides a question about what is being observed.

2. Hypotheses are assumptions made about what was observed, in addition, hypotheses can answer the question shown in the problem. In the case of apples, the hypotheses can be: "Lemon juice is the most efficient product in delaying the blackening of cut apples." "Water is inefficient in preventing blackened cut apples." "Milk accelerates the blackness of cut apples."

3. The dependent viable is one that needs to be influenced by another element to bring about a result. In the case of the experiment with apples, the dependent variable is the blackening rate of the apple slices.

4. The independent variable, on the other hand, refers to the element that does not need any influence to cause a result, on the contrary, this variable acts on the dependent variable, influencing it. In the case of this experiment, the independent variable is the types of liquid tested to delay the blackening of cut apples.

5. Lemon was the most efficient liquid. Probably this efficiency is related to the lemon's ability to decrease the strength of antioxidants, due to its very low pH.

Answer:

costal erosion

Explanation:

can cause erosion

Answer:

The water piles up with nowhere to go but onto land when it gets to the coast. The rising water, called storm surge, can submerge low-lying areas and towns along the coast. Combined with the crashing waves of the storm, the storm surge can cause demolishing docks, houses, roads, and erode beaches.

Explanation:

Answer:

A. It occupies 22.4 L

Explanation:

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm

Avogadro's Law states that 1 mol at 1 atm occupies 22.4 L.

Answer:

1) first order

2) second order

3) zero order

Explanation:

The curve of a first order reaction shows it to be exponential. In fact for a first order reaction, the concentration at a time t is an exponential function;

[A]t= [Ao] e^-kt

Where

[A]t = concentration at time =t

[Ao]= initial concentration

k= rate constant

t= time

For a second order reaction, the rate of reaction is directly proportional to the concentration of reactants.

For zero order reactions, rate of reaction is independent of concentration hence rate = k(rate constant)

The concentration data has been required for the determination of the rate of reaction. Based on the concentration of reactant and product, the rate has been determined.

1. For the first-order reaction, there has been an exponential increase in the rate of the reaction with the increase in the reactant concentration. The rate has been dependent on the concentration of the reactant.

Thus the correct option is A.

2. Irrespective of the first-order kinetic, in the second-order reaction, the rate of reaction has been directly proportional to the concentration of the reactant in the solution.

Thus option B is correct.

3. The zero-order reaction has been independent of the concentration of the reactant. The rate of reaction has been constant with an increase in the reactant concentration.

Thus option C is correct.

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Answer:

0.05 g/cm^3.

Explanation:

The volume of the block = 9*2*6 = 108 cm^3.

Density = mass/volume

= 5.4 / 108

= 0.05 g/cm^3.

Answer. After cytokinesis is completed at end of meiosis - I two haploid cells are formed.on:

Answer:

C. TWO HAPLOID CELLS ARE FORMED

Explanation:

I TOOK THE EDGUNITY TEST AND I GOT IT CORRECT

Answer:

Magnesiums atomic number is 12

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Answer:

false you can not get a exact location of electrons from just modern science

Answer: Potassium Bromide (KBr) The Ionic bond formed between Potassium and Bromine is created through the transfer of electrons from Potassium (metal) to Bromine (nonmetal).

Explanation: this type of structure departs strongly from that expected for ionic bonding and ... whose roots go back to Max Planck's explanation in 1900 of the properties of ... types of interactions between elementary particles (the strong force, the weak force, ...

Answer:

molar mass of unknown gas = 1.987 g/mol

Explanation:

First, the number of moles of the unknown gas is found

Using the ideal gas equation: PV = nRT

P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹

n = PV/RT

n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)

n = 0.2109 moles

Molar mass = mass/ number of moles

molar mass = 0.419 g/ 0.2109 mols

molar mass of unknown gas = 1.987 g/mol

The molar mass of unknown gas by using ideal gas equation = 1.987 g/mol.

This equation gives the relation between pressure, volume, temperature as given below:

[tex]PV = nRT[/tex]

P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹

Substitute the above values in the above equation as follows:

n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)

n = 0.2109 moles

[tex]Molar mass = mass/ number of moles[/tex]

Calculate molar mass by using the above equation,

molar mass = 0.419 g/ 0.2109 moles

The molar mass of unknown gas = 1.987 g/mol

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Answer:

first one is already balanced

2.   MgN2+6HCl->3MgCl2+2NH3

Explanation:

To balance the equations, we need to ensure that the same number of each type of atom is present on both sides of the equation. Here's how we can balance the given equations:

Equation 1: PCl₃ + Cl₂ -> PCl₅

In this equation, there are 1 phosphorus (P) atom, 3 chlorine (Cl) atoms, and 5 chlorine (Cl) atoms. To balance the equation, we can put a coefficient of 2 in front of PCl₃ to have 2 phosphorus atoms, and a coefficient of 5 in front of Cl₂ to have 10 chlorine atoms:

2PCl₃ + 5Cl₂ -> PCl₅

Now the equation is balanced with 2 phosphorus atoms and 10 chlorine atoms on both sides.

Equation 2: Mg₃N₂ + HCl -> MgCl₂ + NH₃

In this equation, there are 3 magnesium (Mg) atoms, 2 nitrogen (N) atoms, 2 hydrogen (H) atoms, and 1 chlorine (Cl) atom.

To balance the equation, we can put a coefficient of 3 in front of HCl to have 3 hydrogen atoms, and a coefficient of 2 in front of NH3 to have 2 nitrogen atoms:

Mg₃N₂ + 3HCl -> MgCl₂ + 2NH₃

Now the equation is balanced with 3 magnesium atoms, 2 nitrogen atoms, 6 hydrogen atoms, and 2 chlorine atoms on both sides.

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