Give the expressions for the disappearance of the reactants first, in the order written in the chemical equation. Then write the expressions for the appearance of the products in the order written in the chemical equation.

Write the expressions in order of appearance in the equation in the form.

± 1/X . Δ[A]/ Δt

where ± is either a plus OR a minus sign, not both,
X is an integer, and
A is a chemical species.

4NH3(g)+ 5O2(g) → 4NO2+ 6H2O(g)

Answers

Answer 1

Answer: Rate of disappearance of [tex]NH_3[/tex] = [tex]\frac{-\Delta [NH_3]}{4dt}][/tex]

Rate of disappearance of [tex]O_2[/tex] = [tex]\frac{-\Delta [O_2]}{5dt}[/tex]

Rate of appearance of [tex]NO_2[/tex] = [tex]\frac{+\Delta [NO_2]}{4dt}[/tex]

Rate of appearance of [tex]H_2O[/tex] = [tex]\frac{+\Delta [H_2O]}{6dt}[/tex]

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

For the given chemical equation :

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO_2(g)+6H_2O(g)[/tex]

Rate of disappearance of [tex]NH_3[/tex] = [tex]\frac{-\Delta [NH_3]}{4dt}][/tex]

Rate of disappearance of [tex]O_2[/tex] = [tex]\frac{-\Delta [O_2]}{5dt}[/tex]

Rate of appearance of [tex]NO_2[/tex] = [tex]\frac{+\Delta [NO_2]}{4dt}[/tex]

Rate of appearance of [tex]H_2O[/tex] = [tex]\frac{+\Delta [H_2O]}{6dt}[/tex]


Related Questions

A solid is dissolved in a liquid, and over time a solid forms again. How can
you confirm the type of change that took place?
A. Testing the new solid to show that its properties are the same as
the starting solid would confirm that a physical change took
place.
B. The solid dissolving in a liquid is confirmation that a chemical
change took place.
C. The solid forming from the liquid is confirmation that a physical
change took place.
D. Showing that the total mass of the solid and liquid changed would
confirm that a chemical change took place.

Answers

I think B is write but even I’m not sure

A certain mass of carbon reacts with 9.53 g of oxygen to form carbon monoxide. ________ grams of oxygen would react with that same mass of carbon to form carbon dioxide, according to the law of multiple proportions.

Answers

Answer: 9.53 *2= 19.06

Explanation:

The law of multiple proportions states that if two elements combines to form more than one compound the ratio of masses of the second element which combines to the fixed mass of the first element will always be the ratios of the small whole numbers.

in case of carbon monoxide, mass of carbon will be the same of mass of oxygen.

But in case of carbon dioxide, if carbon is 9.53 units then oxygen will be twice as that of carbon.

CO2, so 9.53*2= 19.06 grams of oxygen will combine with 9.53 grams of carbon to form carbon dioxide.

Combustion analysis of a 13.42-g sample of estriol (which contains only carbon, hydrogen, and oxygen) produced 36.86 g CO2 and 10.06 g H2O. The molar mass of estriol is 288.38 g/mol . Find the molecular formula for estriol. Express your answer as a chemical formula.

Answers

Answer:

C18H24O3

Explanation:

Step 1:

Data obtained from the question. This include the following:

Mass of estriol = 13.42g

Mass of CO2 = 36.86g

Mass of H2O = 10.06g

Molar mass of estriol = 288.38g/mol

Step 2:

Determination of the mass of Carbon (C), Hydrogen (H) and Oxygen (O) present in the compound. This is illustrated below:

For Carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 36.86 = 10.05g

For Hydrogen, H:

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 10.06 = 1.12g

For Oxygen, O:

Mass of O = 13.42 – (10.05 + 1.12) = 2.25g

Step 3:

Determination of the empirical formula for estriol. This is illustrated below:

C = 10.05g

H = 1.12g

O = 2.25g

Divide by their molar mass

C = 10.05/12 = 0.8375

H = 1.12/1 = 1.12

O = 2.25/16 = 0.1406

Divide by the smallest i.e 0.1406

C = 0.8375/0.1406 = 6

H = 1.12/0.1406 = 8

O = 0.1406/0.1406 = 1

Therefore, the empirical formula for estriol is C6H8O

Step 4:

Determination of the molecular formula for estriol. This is illustrated below:

Molecular formula is simply a multiple of the empirical formula i.e

Molecular formula => [C6H8O]n

[C6H8O]n = 288.38g/mol

[(12x6) + (8x1) + 16]n = 288.38

[72 + 8 + 16]n = 288.38

96n = 288.38

Divide both side by 96

n = 288.38/96 = 3

Molecular formula => [C6H8O]n

=> [C6H8O]n

=> [C6H8O]3

=> C18H24O3

Therefore, the molecular formula for estriol is C18H24O3

The compound is C18H24O3.

From the information in the question;

Mass of C = 36.86 g/44 g/mol × 12 g/mol = 10.1 g

Number of moles of carbon = 10.1 g/12 g/mol = 0.84 moles

Mass of hydrogen = 10.06 g/18 g/mol × 2 g/mol = 1.11 g

Number of moles of hydrogen = 1.11 g/1g/mol = 1.11 moles

Mass of oxygen = 13.42 - (10.1 g + 1.11 g) = 2.21 g

Number of moles of oxygen = 2.21g/16 g/mol = 0.14 moles

Dividing through by the lowest number of moles;

C - 0.84 moles/0.14 moles   H -  1.11 moles/0.14 moles   O - 0.14 moles/0.14 moles

C - 6   H - 8    O -1

The empirical formula is C6H8O

The molecular formula of the compound is;

[6(12) + 8(1) + 16]n = 288.38

n =  288.38/86 =3

The compound is C18H24O3

Learn more about molecular formula:https://brainly.com/question/8073802

A certain lightbulb containing argon at 1.20 atm and 18°C is
heated to 85°C at constant volume. Calculate its final pressure
(in atm).

Answers

Answer:

certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? P 1 T 1 P 2 T 2 ... Ideal Gas Equation 5.4 Charles' law: V T (at constant n and P ) ... Consider a case in which two gases, A and B , are in a container of volume V.

Explanation:

A weather balloon is inflated to a volume of 27.6 L at a pressure of 755 mmHg and a temperature of 29.9 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 385 mmHg and the temperature is -14.1 ∘C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

Answers

Answer: The volume of the balloon at this altitude is 46.3 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 755 mm Hg

[tex]P_2[/tex] = final pressure of gas (at STP) = 385 mm Hg

[tex]V_1[/tex] = initial volume of gas = 27.6 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]29.9^0C=(29.9+273)K=302.9K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]-14.1^0C=((-14.1)+273)K=258.9K[/tex]

Putting all the values we get:

[tex]\frac{755\times 27.6}{302.9}=\frac{385\times V_2}{258.9}[/tex]

[tex]V_2=46.3L[/tex]

Thus the volume of the balloon at this altitude is 46.3 L

Convert 150 K to degrees C.

Answers

Answer:

K = 150, C = - 123.15°

Explanation:

Kelvin = Celcius + 273.15 / 0 Kelvin = - 273.14 C

_____________________________________

Thus,

150 K = Celcius + 273.15,

150 - 273.15 = C,

C = -123.15 degrees

Solution, C = - 123.15°

Answer:

C=-123.15

Explanation:

This is easy

need helpp asapp please

Answers

Answer:

B. None of these

Explanation:

Sulfur has less ionization energy than phosphorus because sulfur has a pair of electron in its 3p subshell that increases electron repulsion in sulfur and sulfur electrons can easily remove from its sub-level.

While, there are no electron pairs in 3p subshell of phosphorus, therefore it requires more energy to remove an electron from 3p subshell.

Hence, the reason is electron repulsion and the correct answer is B.

what type of matter is toluene

Answers

Answer:

is an organic chemical conpond

The amount of calcium in a 15.0-g sample was determined by converting the calcium to calcium oxalate, CaC2O4. The CaC2O4 weighed 40.3 g. What is the percent of calcium in the original sample

Answers

Answer:

128 gram of CaC2O4 contains 40 gram of Calcium

40.3 gram of CaC2O4 cotnains = 40*40.3/128 = 12.59 gram of Calcium

out of 15 gram 12.59 gram is Calcaim that means around 50% of orginal sample has Calcium

Do you think there is a limit to the size of Atoms scientist can make? prove with evidence.

Answers

Answer:

Yes, there is a limit to the size of atoms that scientists can make.

Explanation:

In the nucleus, atoms contain protons and neutrons. It is known that as the number of protons in the nucleus increases, the atom becomes unstable due to the repulsion of positively charged protons clumped together in the small space of the nucleus.

However, an attractive force exists between neutrons and protons which binds the nucleus together and minimizes repulsion between protons. Even neutrons have recently been found to slightly repel each other.

Several attempts made at synthesizing many very heavy elements lately have proved abortive because the elements only exist for a few fractions of a second owing to large repulsion between the particles in the nucleus.

This goes a long way to show that there is a limit to the number of protons and neutrons that can be assembled together to form a new nucleus. We cannot bring an unlimited number of nucleons together to form new atoms due to inter particle repulsive forces.

In the diagram below, particles of the substance are moving from the liquid phase to the gas phase at the same rate as they move from the gas phase to the liquid phase. A number of balls are loosely packed in the bottom of a container, beneath a line across the middle of the container, and a few balls above the line. 2 balls below the line have arrows pointing upward through the centerline; a few of the balls above the line have arrows pointing down through the centerline. The gas and liquid are at equilibrium. a high vapor pressure. a low vapor pressure. zero vapor pressure.

Answers

Answer:

The gas and liquid is in equilibrium.

Explanation:

liquids within a container undergoes state change, changing into gas. If this container is left open, these gases will escape into the external environment. In a situation in which the container is closed, the molecules that leave the liquid surface as gas will eventually condense on contact with the cover wall and change back into the liquid state. Some of these gases will reenter the liquid surface. At first, more of the liquid is transformed into gas and escape into the space above the liquid surface. Eventually, the available space becomes saturated with vapor, and then some of the gases start entering the liquid phase at the same rate as the liquid enters the gas phase. At this stage, the gas and liquid phase now exists in equilibrium.

The lock-and-key model and the induced-fit model are two models of enzyme action explaining both the specificity and the catalytic activity of enzymes. Following are several statements concerning enzyme and substrate interaction. Indicate whether each statement is part of the lock-and-key model, the induced-fit model, or is common to both models.

a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.
c. Enzyme active site has a rigid structure complementary
d. Substrate binds to the enzyme through noncovalent interactions

Answers

Answer:

The lock-and-key model:

c. Enzyme active site has a rigid structure complementary

The induced-fit model:

a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.

Common to both The lock-and-key model and The induced-fit model:

b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.

d. Substrate binds to the enzyme through non-covalent interactions

Explanation:

Generally, the catalytic power of enzymes are due to transient covalent bonds formed between an enzyme's catalytic functional group and a substrate as well as non-covalent interactions between substrate and enzyme which lowers the activation energy of the reaction. This applies to both the lock-and-key model as well as induced-fit mode of enzyme catalysis.

The lock and key model of enzyme catalysis and specificity proposes that enzymes are structurally complementary to their substrates such that they fit like a lock and key. This complementary nature of the enzyme and its substrates ensures that only a substrate that is complementary to the enzyme's active site can bind to it for catalysis to proceed. this is known as the specificity of an enzyme to a particular substrate.

The induced-fit mode proposes that binding of substrate to the active site of an enzyme induces conformational changes in the enzyme which better positions various functional groups on the enzyme into the proper position to catalyse the reaction.

The partial Lewis structure that follows is for a hydrocarbon molecule. In the full Lewis structure, each carbon atom satisfies the octet rule, and there are no unshared electron pairs in the molecule. The carbon-carbon bonds are labeled 1, 2, and 3.
A) How many hydrogen atoms are in the molecule?
B) Rank the carbon-carbon bonds in order of increasing bond length.
C) Which carbon-carbon bond is the strongest one?

Answers

Answer:

A) How many hydrogen atoms are in the molecule?

Cl2 + F2 → ClF3, 5. How many moles of Cl2 are needed to react with 3.44 moles of F2? 6. How many grams of ClF3 form when 0.204 moles of F2 react with excess Cl2? 7. How many grams of ClF3 form from 130.0 grams of Cl2 when F2 is in excess? 8. How many grams of F2 are needed to react with 3.50 grams of Cl2?

Answers

Answer:

5) 1.147 moles Cl2

6) 12.57 grams ClF3

7)  339.10 grams ClF3

8) 5.63 grams F2

Explanation:

Step 1: Data given

Number of moles F2 = 3.44 moles

Molar mass F2 = 38.00 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles F2

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 3.44 moles F2 we'll need 3.44/3 = 1.147 moles Cl2

Step 1: Data given

Number of moles F2 = 0.204 moles

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles ClF3

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 0.204 moles F2 we'll have 2/3 * 0.204 = 0.136 moles

Step 4: Calculate mass ClF3

Mass ClF3 = Moles ClF3 * molar mass ClF3

Mass ClF3 = 0.136 moles * 92.448 g/mol

Mass ClF3 = 12.57 grams ClF3

Step 1: Data given

Mass of Cl2 = 130.0 grams

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles Cl2

Moles Cl2 = mass Cl2 / molar mass Cl2

Moles Cl2 = 130.0 grams / 70.9 g/mol

Moles Cl2 = 1.834 moles

Step 4: Calculate moles

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 1.834 moles Cl2 e'll have 2*1.834 = 3.668 moles ClF3

Step 5: Calculate mass ClF3

Mass ClF3 = Moles ClF3 * molar mass ClF3

Mass ClF3 = 3.668 moles * 92.448 g/mol

Mass ClF3 = 339.10 grams ClF3

Step 1: Data given

Mass of Cl2 = 3.50 grams

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles Cl2

Moles Cl2 = Mass Cl2 / molar mass Cl2

Moles Cl2 = 3.50 grams / 70.9 g/mol

Moles Cl2 = 0.0494  moles

Step 4: Calculate moles F2

For 1 mol Cl2 we need 3 moles F2

For 0.0494 moles we need 3*0.0494 = 0.1482 moles

Step 5: Calculate mass F2

Mass F2 = moles F2 * molar mass F2

Mass F2 = 0.1482 moles * 38.00 g/mol

Mass F2 = 5.63 grams F2

What is the net ionic equation of the reaction of MgSO4 with Ba(NO3)2 ?

Answers

Answer:

Ba(+2)(aq) + SO4(-2)(aq) -----> BaSO4(s)

Explanation:

Take a look at the attachment below;

can a kind human being help me with this table at least only with the first burning fire wood someone please ​

Answers

Answer:

See below

Explanation:

* Burning fire wood is given to be our first option. Now burning tends to be a property of wood, and it does effect the chemical compositions of it. Wood, in the presence of fire / oxygen, turns into ash and carbon dioxide.

* Decomposition is recognized as a chemical change, and heating copper carbonate is a perfect example of decomposition. When energy is added to this chemical process, the copper carbonate decomposes into copper oxide.

* Mixing sodium chloride solution and silver nitrate solution. When this reaction takes place, a white precipitate of AgCl is formed. And of course, this is a chemical reaction.

* When acids or bases come in contact with citric acid, the pH degree changes much. Due to this, carbon dioxide bubbles are formed.

* When eggs are fried they absorb the heat in the pan. Doing so the egg starts to curl a bit, resulting in the formation of new particles.

_______________________________________________________

I hope this gave you a start!

artaric acid, H2C4H4O6, has two acidic hydrogens. The acid is often present in wines and precipitates from solution as the wine ages. A solution containing an unknown concentration of the acid is titrated with NaOH. It requires 21.65 mL of 0.3500 M NaOH solution to titrate both acidic protons in 50.00 mL of the tartaric acid solution. You may want to reference (Pages 149 - 153) Section 4.6 while completing this problem. Part A Write a balanced net ionic equation for the neutralization reaction. Express your answer as a chemical equation including phases.

Answers

Answer:

2H⁺(aq) + + 2OH⁻(aq) → 2H₂O(l)

Explanation:

A acid that has two acidic hydrogens requires to moles of NaOH per mole of acid producing the salt of the acid and water.

If the acid is tartaric acid, H₂C₄H₄O₆, the reaction with NaOH is:

H₂C₄H₄O₆(aq) + 2NaOH(aq) → 2H₂O(l) + 2NaC₄H₄O₆(aq)

Now, ions in solution in the reaction are:

2H⁺(aq) + C₄H₄O₆²⁻(aq) + 2Na⁺(aq) + 2OH⁻(aq) → 2H₂O(l) + 2Na⁺(aq) +C₄H₄O₆²⁻(aq)

Thus, net ionic equation (Lisiting only the ions involved in the reaction) is:

2H⁺(aq) + + 2OH⁻(aq) → 2H₂O(l)

In the presence of a strong base, the following reaction between (CH3)3CCl and OH- occurs: (CH3)3CCl + OH- → (CH3)3COH + Cl- Studies have suggested that the mechanism for the reaction takes place in 2 steps: Step 1) (CH3)3CCl → (CH3)3C+ + Cl- (slow) Step 2) (CH3)3C+ + OH- → (CH3)3COH (fast) What is the rate law expression for the overall reaction? Group of answer choices

Answers

Answer:

D. rate = k [(CH3)3CCl]

Explanation:

(CH3)3CCl + OH- → (CH3)3COH + Cl-

The mechanisms are;

Step 1)

(CH3)3CCl → (CH3)3C+ + Cl- (slow)

Step 2)

(CH3)3C+ + OH- → (CH3)3COH (fast)

In kinetics, the slowest step is the ratee determining step.

For a given reaction;

A → B + C, the rate law expression is given as;

rate = k [A]

In this problem, from step 1. The rate expression is;

rate = k [(CH3)3CCl]

What is an ion?
A. An atom that has lost or gained 1 or more electrons
O B. An atom that has lost or gained 1 or more neutrons
O C. An atom that has lost or gained 1 or more protons
D. An atom that differs in mass from another atom of the same
element

Answers

Answer:

An ion is an atom that has lost or gained one or more electrons.

Explanation:

Ions are positively or negatively charged atoms of elements. This is because they can give, take, or share electrons with other elements to encourage the formation of chemical bonds.

Protons are what decide the chemical identity of the element. So, for example, if an atom has 11 protons, we know that will be a Sodium (Na) atom. A loss or gain of protons completely changes the chemical identity of the element and it will then become another element.

Electrons are what give an atom a neutral electrical charge (if that atom has the number of protons and neutrons normally described for the element - otherwise, a discrepancy or gain in neutrons is referred to as an isotope and declares that ions have nothing to do with the mass of an element).

With this information, you can realize that neutrons and protons have nothing to do with ions and you can confirm that ions are atoms that have lost or gained one or more electrons.

When an automobile engine starts, the metal parts immediately begin to absorb heat released during the combustion of gasoline. How much heat will be absorbed by a 165 kg iron engine block as the temperature rises from 15.7°C to 95.7°C? (The specific heat of iron is 0.489 J/g·°C.)

Answers

Answer:

H = 4,034,250 J

Explanation:

Mass, m = 165kg = 165,000g (Converting to grams)

Initial temperature = 15.7°C

Final temperature = 95.7°C

Temperature change, ΔT = 95.7 - 15.7 = 50°C

Specific heat capacity, c = 0.489 J/g·°C

Heat = ?

All the parameters are related with the equation below;

H = m * c * ΔT

H = 165000 * 0.489  * 50

H = 4,034,250 J

Benny Beaver wants to determine what dyesare present in his favorite sports drink. He analyzesa sample witha UV-visiblespectrophotometer and sees absorbance peaks at 415.2nm and 519.6nm. What colordyesare present in his drink

Answers

Answer:

At 415.2nm and 519.6nm, the dyes observed by the instrument are violet and green respectively.

Explanation:

In the electromagentic spectrum, visible wavelengths cover a range from approximately 400 to 800 nm. The colours of the spectrum range from red to violet (Red, Orange, Yellow, Green, Blue, Indigo and violet: a.k.a ROGBIV), in order of decreasing wavelength.

I hope this explanation would suffice.

Consider the insoluble compound nickel(II) hydroxide , Ni(OH)2 . The nickel ion also forms a complex with cyanide ions . Write a balanced net ionic equation to show why the solubility of Ni(OH)2 (s) increases in the presence of cyanide ions and calculate the equilibrium constant for this reaction. For Ni(CN)42- , Kf = 1.0×1031 . Use the pull-down boxes to specify states such as (aq) or (s).

Answers

Answer: Equilibrium constant for this reaction is [tex]2.8 \times 10^{15}[/tex].

Explanation:

Chemical reaction equation for the formation of nickel cyanide complex is as follows.

[tex]Ni(OH)_{2}(s) + 4CN^{-}(aq) \rightleftharpoons [Ni(CN)_{4}^{2-}](aq) + 2OH^{-}(aq)[/tex]

We know that,

      K = [tex]K_{f} \times K_{sp}[/tex]

We are given that, [tex]K_{f} = 1.0 \times 10^{31}[/tex]

and,    [tex]K_{sp} = 2.8 \times 10^{-16}[/tex]

Hence, we will calculate the value of K as follows.

     K = [tex]K_{f} \times K_{sp}[/tex]

     K = [tex](1.0 \times 10^{31}) \times (2.8 \times 10^{-16})[/tex]

        = [tex]2.8 \times 10^{15}[/tex]

Thus, we can conclude that equilibrium constant for this reaction is [tex]2.8 \times 10^{15}[/tex].

When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:
sulfur dioxide (g) + water (l) __________sulfurous acid (H2SO3) (g) + water (I)

Answers

Answer:

Sulfur dioxide + 2 ( water ) -----> sulfurous acid + water /

SO2 + 2 ( H2O ) -----> H2SO3 + H2O

Explanation:

This formula may not be right. Sulfur dioxide tends to react with water to produce sulfurous acid as per it's formula, but then again that chemical reaction need not be balanced. However, I will solve for either case here -

Sulfur dioxide + water -----> sulfurous acid,

Sulfur dioxide + water -----> sulfurous acid + water

_______________________________________________________

As I mentioned before, Sulfur dioxide + water -----> sulfurous acid is a chemical reaction that need not balancing as the number of each element present on the reactant and product side are the same. To help, let me rewrite this reaction -

SO2 + H2O -----> H2SO3,

Reactant                  |                Product

Sulfur = 1,                                  Sulfur = 1,

Oxygen = 3,                              Oxygen = 3,

Hydrogen = 2                           Hydrogen = 2

And hence the equation is already balanced. Now let us consider the case we supposedly have at hand - Sulfur dioxide + water -----> sulfurous acid + water. Take a look at the attachment below;

The element silver has an atomic weight of 108 and consists of two stable isotopes silver-107 and silver-109. The isotope silver-107 has a mass of 107 amu and a percent natural abundance of 51.8 %. The isotope silver-109 has a percent natural abundance of 48.2 %. What is the mass of silver-109

Answers

Answer:

109

Explanation:

Let silver-107 be isotope A

Let silver-109 be isotope B

Let silver-107 abundance be A%

Let silver-109 abundance be B%

The following data were obtained from the question:

Atomic weight of silver = 108

Mass of isotope A (silver-107) = 107

Abundance of isotope A (silver-107) = 51.8%

Abundance of isotope B (silver-109) = 48.2%

Mass of isotope B (silver-109) =?

Now, we shall determine the mass silver-109 as follow:

Atomic weight = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

108 = [(107 x 51.8)/100] + [(Mass of B x 48.2)/100]

108 = 55.426 + (Mass of B x 0.482)

Collect like terms

Mass of B x 0.482 = 108 – 55.426

Mass of B x 0.482 = 52.574

Divide both side by 0.482

Mass of B = 52.574/0.482

Mass of B = 109

Therefore, the mass of silver-109 is 109.

an auto of an element has 17 protons in its nucleus.a) write the electronic configuration of the atom.b)to what period and group does the element belong​

Answers

Answer:

i hope it will help you

Explanation:

electronic configuration 1s²,2s,²2p^6,3s²3p^6,4s^1

as it has one electron in its valence shell so it is the member of group 1A(ALKALI METALS)  and the number of shells is 4 so it is in period 4

A sample of chlorine gas starting at 686 mm Hg is placed under a pressure of 991 mm Hg and reduced to a volume of 507.6 mL. What was the initial volume of the chlorine gas container if the process was performed at constant temperature

Answers

Answer:

The initial volume of the chlorine gas [tex]V1=733.28mL[/tex]

Explanation:

Given:

P1= 686mmHg

P2= 991mmHg

V2= 5076mL

V1=?

According to Boyle's law which states that at a constant temperature, the pressure on a gas increases as it's volume decreases.

It can be expressed as : P1V1 = P2V2

Where P1 is the initial pressure

P2= final pressure

V1= initial volume

V2 = final volume

[tex]V1= (P2V2)/P1[/tex]

V1= (991mmHg*507.6mL)/686mmHg

V1=503031.6/686

[tex]V1=733.28mL[/tex]

Therefore, The initial volume of the chlorine gas [tex]V1=733.28mL[/tex]

When 1-iodo-1-methylcyclohexane is treated with NaOCH2CH3 as the base, the more highly substituted alkene product predominates. When KOC(CH3)3 is used as the base, the less highly substituted alkene predominates. Give the structures of the two products and offer an explanation.

Answers

Answer:

See explanation

Explanation:

In this case, we have 2 types of reactions. [tex]CH_3CH_2ONa[/tex] is a strong base but only has 2 carbons therefore we will have less steric hindrance in this base. So,  the base can remove hydrogens that are bonded on carbons 1 or 6, therefore, we will have a more substituted alkene (1-methylcyclohex-1-ene).

For the  [tex]KOC(CH_3)_3[/tex] we have more steric hindrance. So, we can remove only the hydrogens from carbon 7 and we will produce a less substituted alkene (methylenecyclohexane).

See figure 1

I hope it helps!

all compounds are neutral true or false​

Answers

Answer:

Even all compounds are neutral.

Explanation:

Some of them exhibit polarity. Because of the difference in electron affinity of the constituent atoms, the shared electrons are pulled towards the atom with high affinity to electrons.

What is the coefficient for oxygen in the balanced equation? C 5H 12 + ? O2 → ? CO2 + ? H2O. 2 4 5 6 8

Answers

Answer:

8

Explanation:

When you balance the entire equation, you should get:

C5H12 + 8O2 ---> 5CO2 + 6H2O

The reason for the dramatic decline in the number of measles cases from the 1960s to 2010 in the United States was because the vaccine

Answers

Answer:

It was because the vaccine generated actively acquired immunity, that is, inoculation of a portion of the measles virus so that the body forms the antibodies for a second contact and thus can destroy it without triggering the pathology.

Explanation:

Vaccines are methods of active acquired immunity since the antibody is not passively inoculated, it is manufactured by the body with a physiological process once part of the virus is inoculated.

The measles virus most of all affected the lives of infants or newborn children with severe rashes and high fevers that led to death.

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