In an oven, 7000 kg of limestone are heated to over 900ºC to give lime and carbon dioxide. Assuming all calcium carbonate reacts. How much lime do you get? What volume of CO2 is emitted into the atmosphere? Atomic weights: Ca: 40; C: 12; O: 16.

Answer:

B

Explanation:

Answer:

Scientific notation is a system in which quantities are too big or too tiny to compose in decimal form.

Key words:

1: Scientific

2: Quantities

3: Decimal

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Hope this helps.

Answer:

The formula is as follows

2Cu + O2 ---> 2CuO

When two copper atoms react with a diatonic oxygen, they form copper oxide (rust)

Answer:

60.288 grams of O₂ gas are contained in 890.0 mL at 21.0 degrees Celsius and 750.0 psi.

Explanation:

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

Replacing:

51.0345 atm*0.890 L= n*0.082 [tex]\frac{atm*L}{mol*K}[/tex] *294 °K

Solving:

[tex]n=\frac{51.0345 atm*0.890 L}{0.082\frac{atm*L}{mol*K} *294 K}[/tex]

n=1.884 moles

Being 32 g / mole the molar mass of oxygen O₂, the following rule of three applies: if 1 mole contains 32 grams, 1,884 moles, how much mass does it have?

[tex]mass=\frac{1.884 moles*32 grams}{1mole}[/tex]

mass= 60.288 grams

60.288 grams of O₂ gas are contained in 890.0 mL at 21.0 degrees Celsius and 750.0 psi.

Answer:

The mass of the O2 gas is 60.16 grams

Explanation:

Step 1: Data given

Volume of O2 gas = 890.0 mL = 0.890 L

Temperature of the O2 gas = 21.0 °C = 294 K

Pressure = 750.0 psi = 51.03 atm

Step 2: Calculate moles of O2 gas

p*V = n*R*T

⇒with p = the pressure of the O2 gas = 51.03 atm

⇒with V = the volume = 0.890 L

⇒with R = the gas constant == 0.08206 L*atm/mol*K

⇒with T = the temperature = 294 K

⇒with n = the number of moles of O2 gas

n = (p*V) / (R*T)

n = (51.03 atm * 0.890 L) / (0.08206L*atm/mol*K * 294K)

n = 1.88 moles of O2

Step 3: Calculate the mass of O2 gas

Mass = moles * molar mass O2

Mass O2 gas = 1.88 moles * 32.0 g/mol

Mass O2 gas = 60.16 grams

The mass of the O2 gas is 60.16 grams

Answer:

there are 5 ways this could happen

Autosomal dominant inheritance:  a child recieves a normal gene from one parent and a defective gene from the other parent.

can occur on any of the 22 non-sex chromosomes and have a 50% inheritence rate, gender is not a factor, and disorder differs with inheritance.

examples:  Huntington's disease, neurofibromatosis, achondroplasia, familial hypercholesterolemia

Autosomal recessive inheritance:  both parents carry the defective gene but they are not affected by the disorder.

there is a 25% chance of defective gene from both parents, a 50% chance of inheriting one gene to become a carrier, gender is not a factor in the pattern of the defective gene.

examples:  Tay-Sachs disease, sickle cell anemia, cystic fibrosis, phenylketonuria (PKU)

X-linked (sex-linked) recessive inheritance:  mother carries the affective gene on one of the two X chromosomes.

males inherite X chromosomes from their mothers and Y from their father; which gives the son a 50% chance of inheriting the disorder.

daughters have a 50% chance, but they are not affected by the disorder.

examples:  Hemophilia A, Duchenne muscular dystrophy

X-linked Dominant:  females are affected more so than males; more common for males if they are in the same generation if the mom is affected (because females have two X-chromosomes)

example:  Hypophatemic rickets (Vitiamin Dresistant rickets, ornithine transcarbamylase deficiency.

Mitochondrial:  can affect both males and femlaes, can only be passed by females due to all mitochondria of all children is from the mother, and can appear in every generation.

examples:  Lebrer's hereditary optic neuropathy and Kearns-Sayre syndrome

Explanation:

Answer:

vacuum

Explanation:

vacuum is the form by which matter cant exist

Answer:

The reaction mentioned in the question is:  

CO (g) + Cl₂ (g) ⇔ COCl₂ (g), the equilibrium constant or Kc will be 77.5 at 600 K. In the given mixture, the moles of CO given is 4.24 × 10⁻² moles, the moles of Cl₂ given is 4.48 × 10⁻² moles, and the moles of COCl₂ given is 0.114 moles. The volume of the container is 1.00 L.  

The concentration or Molarity can be determined by using the formula, concentration (M) = mole/volume (L),  

The molarity of CO = 4.24 × 10⁻² moles / 1.00 L = 4.24 × 10⁻² M

The molarity of Cl₂ = 4.48 × 10⁻² moles / 1.00 L = 4.48 10⁻² M

The molarity of COCl₂ = 0.114 moles / 1.00 L = 0.114 M

The reaction quotient or Qc,  

Qc = [COCl2]/[CO][Cl2]  

Qc = 0.114 / 4.24 × 10⁻² × 4.48 × 10⁻²

Qc = 60.03

Thus, Qc is less than Kc. Hence, the reaction will proceed in forward direction so that equilibrium can be attained, that is, until Qc becomes equal to Kc.  

Therefore, the product (COCl₂) will be produced, and the reactants CO and Cl₂ will get consumed.  

1. The given statement is true.  

2. The given statement is true.  

3. The given statement is false. In order to reach equilibrium, the reactants must be consumed.  

4. The given statement is true.  

5. The given statement is false, that is Qc is not equal to Kc.  

Answer:

0.00316

Explanation:

You have to use the following equation:

[tex]pH=-log[H^+][/tex]

You are given the pH and need to find the concentration of H+.  Plug in the given components and solve.

[tex]2.5=-log[H^+]\\H^+ = 10^{-2.5}\\H^+=0.00316[/tex]

The concentration of H is 0.00316.

Answer:

The reaction of FeBr3/Br2 with benzaldehyde will yield a major product in which the -Br is attached to the benzaldehyde at the meta position.

Explanation:

In chemistry, resonance is a way of describing bonding in certain molecules or ions by the combination of several contributing structures into a resonance hybrid in valence bond theory. Resonance structures often explain the formation of certain major and minor products in organic chemistry reactions.

Aromatic aldehydes and ketones undergo electrophilic substitution reactions such as nitration, sulphonation and halogenation. Since the aldehydic group (-CHO) and ketonic group (-COR or -COAr) are electron-withdrawing, they are deactivating and m-directing.

In benzaldehyde, the ring becomes deactivated at ortho & para positions due to an electron withdrawing aldehyde group. Hence electrophilic substitution is favored at Meta position.

This implies that the reaction of FeBr3/Br2 with benzaldehyde will yield a major product in which the -Br is attached to the benzaldehyde at the meta position.

Explanation:

methyl orange turns pink in an acidic solution

yellow in basic and orange in neutral

Answer: The substance would be a strong acid.

Explanation:

Methyl orange turns orange when it is added to a medium acid and turns red when it is added to a strong acid. The pH of the acid would be above 4 because it is a strong acid. Methyl orange turns yellow when it is added to a medium base.

Answer:

This is true because oxygen belongs to the non-metal part of the periodic table.

Answer:hello

Oxygen is a non-metal it's exactly true.

Explanation:

There's a good rule that if the number of electrons in the last layer is 1,2,3,and sometimes 4 it's a metal but if this number increase it's a non-metal like Oxygen or Nitrogen.

Good luck.

Explanation:

Ag+(aq) + 2NH3 (aq) ⇄ Ag(NH3)2 (aq)

When it comes to question of this sort, the LeChatelier principle should come to mind. The LeChatelier principle states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

increasing the concentration of Ag+

This would lead to an increase in concentration of reactants, the system would annul or oppose this change by moving towrds the right, favouring the forward reaction.

decreasing the concentration of NH3

This would lead to a decrease in concentration of reactants, the system would annul this change by moving to the left, favourin the backward reaction.

increasing the concentration of Ag(NH3)2

This would lead to an increase in concnetration of products, the system would annul or oppose this change by moving towards the left, favouring the backward reaction.

Answer:

ΔH°c = -2219.9 kJ

Explanation:

Let's consider the combustion of propane.

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)

We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.

ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]

ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]

ΔH°c = -2219.9 kJ

Answer:

Collision theory states that the rate constant for a chemical reaction is composed

of three factors, (1) the absolute number of collisions, Z, between molecules; (2) The

fraction of collisions, f, with an energy greater than the activation energy; and (3) the

fraction of molecules, p, in which the molecules are in the correct orientation to react.

k = Zfp Equation 1

The absolute number of collisions, Z, increases with temperature. However, it has

been shown that at 25o

C, the increase in the number of collisions accompanying a 10o

C

increase in temperature accounts for only about 2% of the increase in the reaction rate.

Similarly, while it is important that molecules be in the proper orientation to react when

they collide, molecular orientation is independent of temperature. Thus it follows that the

major factor controlling reaction rates is the fraction, f, of molecules in the reaction

mixture with an energy greater than the activation energy. This factor, f, depends on the

absolute temperature. It has been shown that f is related to Ea by the following equation

Explanation:

Answer:

The correct answer is 596.5 kJ.

Explanation:

The mass of ethanol or C2H5OH mentioned in the question is 20 gm.  

The molar mass of ethanol is 46 g/mol.  

The moles of the compound can be determined by using the formula,  

n = weight of the compound/molar mass

= 20/46 = 0.435 moles

It is mentioned in the question that standard heat of combustion of ethanol is 1372 kJ/mole, that is, one mole of ethanol is producing 1372 kilojoules of energy at the time of combustion.  

Therefore, the energy liberated by completely burning the 20 grams of ethanol is 0.435*1372 = 596.5 kJ.  

Since the standard heat of combustion of ethanol is -1372 kJ/mol, the heat released by the combustion of a 20.0 g sample is -593 kJ.

Let's consider the thermochemical equation for the combustion of ethanol.

C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)   ΔH°c = -1372 kJ/mol

We can calculate the heat released by burning 20.0 g of ethanol considering the following relationships.

[tex]20.0 g EtOH \times \frac{1molEtOH}{46.07gEtOH} \times \frac{(-1367kJ)}{1molEtOH} = -593 kJ[/tex]

Since the standard heat of combustion of ethanol is -1372 kJ/mol, the heat released by the combustion of a 20.0 g sample is -593 kJ.

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The standard heat of combustion of ethanol, C₂H₅OH, is -1372 kJ/mol ethanol. How much heat (in kJ) would be liberated by completely burning a 20.0 g sample.

Answer:

pOH = -log[OH-]

Explanation:

Hello,

To calculate the pH or pOH of a substance, we relate it the negative logarithm of either the H+ or OH- concentration.

pOH is used to find the alkalinity of a solution my measuring the concentration of the hydroxide ion [OH-]. Solutions having a pOH value that is greater than 7 are said to be acidic while solutions with value less than 7 are alkaline.

pH = -log[H+]

While

pOH = -log[OH-]

Note: the pOH value of a solution depends on the concentration of the hydroxide ion in the solution.

Answer:

An expression for the combined total dollar amount he earned this month is $(7x + 14y) where x + y = 103.

Explanation:

Let the no. of hours worked as tutor be x

Earning as tutor in 1 hour = $7

Earning as tutor in x hour = $7*x = $7x

Let the no. of hours worked as waiter be y

Earning as waiter in 1 hour = $14

Earning as waiter in y hour = $14*y = $14y

Given that total hours worked in the month = 103 hours

therefore, x + y = 103 ----> (1)

Total amount earned in the month = Earning as tutor in x hour + Earning as waiter in y hour = $(7x + 14y)

An expression for the combined total dollar amount he earned this month is $(7x + 14y) where x + y = 103.

Answer:

4>3>1>2.

That is, the C-C bond length in ethane > benzene > ethene > ethyne.

Explanation:

The C-C bond in ethane is single, the C-C bond in ethene is double and the C-C bond in ethyne is triple. As the number of bonds between a C-C increases, the length of the bond decreases with an increase in strength. This explains why the C-C bond length in ethane > ethene > ethyne. For benzene, all the C-C bonds in the aromatic compound has been found to have an identical length of 1.40 Å, compared to ethane (1.54Å), ethene (1.34Å) and ethyne (1.20Å). Hence the trend in bond lengths: ethane > benzene > ethene > ethyne.

Answer:

= 4 * 10-8    = 400 Angstrom

Explanation:

EF = 5 eV, K = 4.29 x 102 J/(s m K), and Cvel = 1% of the lattice heat capacity

K= 1/3 (Cv)*v*l

v is fermi velocity which is equal to [tex]v = (2E_f/m)^{0.5}[/tex]

after putting mass of electron as [tex]9.1 * 10^{-31}kg[/tex] and [tex]E_f = 5 eV[/tex] we get [tex]v= 1.33 * 10^6 m/s[/tex]

[tex]C_v[/tex] is 1% of lattice heat capacity

Heat Capacity of Aluminium is [tex]0.897 J g^{-1}K^-1[/tex]

Density = [tex]2.6989 g \ cm^{-3}[/tex]

For  lattice heat capacity you need to use the heat capacity for alimunium given  and then multiply with density to get per unit volume term

Heat Capacity per unit volume =   [tex]0.897 J g^{-1}K^-1[/tex] * [tex]2.6989 g \ cm^{-3}[/tex]

[tex]= 2.42 J K^{-1} cm^{-3} \\\\= 2.42* 10^6 J K^{-1} m^{-3}[/tex]

Cv = 1% of heat capacity per unit volume

[tex]=0.01 * 2.42* 10^8 J K^{-1} m^{-3} \\\\= 2.42* 10^4 J K^{-1} m^{-3}[/tex]

Putting values in this equation K= 1/3 (Cv)*v*l

[tex]l = 3K/(C_v * v )\\\\ = 3 * 4.29 * 10^2 / (2.42* 10^4 * 1.33 * 10^6 )[/tex]

[tex]= 4 * 10^{-8 }[/tex]

  = 400 Angstrom

Answer:

The volume of base required is = 25.66 mL

Explanation:

Volume of titrant required is given by:

Final titre value - Initial titre value

=26.78 - 1.12

= 25.66 mL

Note: repeat and average volumes for accuracy of result.

Answer:

C = 4.60x10⁻⁵ M

Explanation:

The concentration of the compound can be calculated using Beer-Lambert Law:

[tex]A = \epsilon*C*l[/tex]

Where:

A: is the absorbance of the ethanol = 0.58  

ε: is the molar absorptivity of the ethanol = 12600 M⁻¹cm⁻¹

C: is the concentration of the compound =?

l: is the optical path length = 1 cm

Hence, the concentration of the compound is:

[tex]C = \frac{A}{\epsilon*l} = \frac{0.58}{12600 M^{-1}cm^{-1}*1 cm} = 4.60 \cdot 10^{-5} M[/tex]

Therefore, the concentration of the compound is 4.60x10⁻⁵ M.

I hope it helps you!

Answer: The right conjugate of [tex]OH^-[/tex] is [tex]H_2O[/tex]

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

For the given chemical equation:

[tex]OH^-(aq)+HCO_3^-(aq)\rightleftharpoons H_2O(l)+CO_3^{2-}(aq)[/tex]

Here, [tex]OH^-[/tex] is gaining a proton, thus it is considered as a brønsted-lowry base and after gaining a proton, it forms [tex]H_2O[/tex] which is a conjugate acid.

Thus the right conjugate of [tex]OH^-[/tex] is [tex]H_2O[/tex]

Answer:24 kilometers

Explanation:Density divided by time

Answer:

OH−(aq), and H+(aq)

Explanation:

Redox reactions may occur in acidic or basic environments. Usually, if a reaction occurs in an acidic environment, hydrogen ions are shown as being part of the reaction system. For instance, in the reduction of the permanganate ion;

MnO4^-(aq) + 8H^+(aq) +5e-------> Mn^2+(aq) + 4H2O(l)

The appearance of hydrogen ion in the reaction equation implies that the process takes place under acidic reaction conditions.

For reactions that take place under basic conditions, the hydroxide ion is part of the reaction equation.

Hence hydrogen ion and hydroxide ion are included in redox reaction half equations depending on the conditions of the reaction whether acidic or basic.

Explanation:

a) E = P × t

E = 750 × 1 s = 750 J

b) E = P × t

E = 750 × 3s = 2250 J

c) E = P × t

1500 = 750 × t

t = 1500/750

t = 2 s

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C[tex]_{V}[/tex] You can work out C[tex]_{V}[/tex]

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC[tex]_{V}[/tex] ∆T

Polyatomic gas: C[tex]_{V}[/tex] = 3R

∆U= nC[tex]_{V}[/tex] ∆T

∆U= 28g x C[tex]_{V}[/tex] x (350K - 58.3K)

∆U = 28C[tex]_{V}[/tex] x 291.7

∆U = 10967.6 x C[tex]_{V}[/tex]

Answer: His and Lys are deprotonated but Asp will be protonated.

Explanation:

As the pH is given as 7.4 and pK of His is given as 6.00. There will occur a positive charge on His when it's pH < pK therefore, it is neutral at the given pH.

As the pK value of Lys is 10.53 that is greater than the pH of 7.40. Therefore, charge on Lys is positive.

As the pK value of Asp is 3.65 which is less than the pH value of 7.40. Hence, Asp has a negative charge.

Therefore, we can conclude that His and Lys are deprotonated but Asp will be protonated.

Explanation:

faster and the substance will be a gas.

According to the forces of attraction and principles of phase change the substance molecules will move faster and the substance will be a gas.

Forces of attraction  is a force by which atoms in a molecule  combine. it is basically an attractive force in nature.  It can act between an ion  and an atom as well.It varies for different states  of matter that is solids, liquids and gases.

The forces of attraction are maximum in solids as  the molecules present in solid are tightly held while it is minimum in gases as the molecules are far apart . The forces of attraction in liquids is intermediate of solids and gases.

The physical properties such as melting point, boiling point, density  are all dependent on forces of attraction which exists in the substances.

Learn more about forces of attraction,here:

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The question is incomplete; the complete question is;

Which of the following statements is/are true for a 0.10 M solution of a weak acid HA?

a. [ H+] >> [ A−]

b. [ H+] = [ A−]

c. The pH is 1.00.

d. The pH is less than 1.00.

Answer:

b. [ H+] = [ A−]

Explanation:

Given the acid as HA, we know that being a weak acid, its dissociation in water can never be 100%. If it were a strong acid, then it could have undergone a 100% dissociation in solution. The conjugate base of a weak acid is a always a weak base hence A^- is expected to act as a weak base. At the same concentration, weak acids have a higher pH value than strong acids. Hence if the pH of a strong acid HA is 1, then the pH of a weak acid HA must be greater than 1.

But, we look at the equation for the dissociation of the weak acid HA

HA(aq)⇄H^-(aq) + A^-(aq). This implies that the HA dissociates in a 1:1 ratio therefore; [H+] = [ A−], hence the answer given above.