105/22 • (1.251 - 0.620)=

105/22 (1.251 - 0.620)=

Answers

Answer 1

Answer:

105/22*(1.251-0.620)

105/22*0.631

4.772*0.631

3.011132

Hope it helps

Answer 2

Answer:

3.0

Explanation:

First, complete the operations inside the parenthesis according to the normal rules for significant figures. Because there are subsequent calculations, keep at least one extra significant figure when possible: (4.7727) × (0.631).

The final product will be rounded to two significant figures because it can’t be more precise than the least precise number in the problem, 22. The final product is 3.0.


Related Questions

what is the oxidation state of the oxygen atoms in co2,h2o and o2 and what does this information tell you about photosynthesis and respiration

Answers

Answer:

-2, -2 and 0.

- Respiration is a process in which energy is produced and photosynthesis is a process in which energy is used.

Explanation:

Hello,

In this case, oxygen is a substance that is used for animals and us to acquire the energy necessary for several functions by the cellular respiration (we also need glucose), besides, it is a product of the photosynthesis carried out by vegetable cells (plants). Moreover, carbon dioxide and water are used by the plants to produce oxygen we need as well as glucose via the aforementioned photosynthesis, thus, both chemical reactions are shown below:

[tex]C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O\ \ \ respiration\\\\6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2\ \ \ photosynthesis[/tex]

In such a way, since the oxygen in carbon dioxide and water has an oxidation state of -2 (reduced form) we can say that the respiration is a process in which energy is produced and since the oxygen yielded during the photosynthesis has an oxydation state of 0, we can say that photosynthesis is a process in which energy is used.

Best regards.

.formular for Charles' law​

Answers

Explanation:

Hey there!

Charles law states that the volume of gas is directly proportional to the absolute temperature at constant pressure.

As the fact is, at a constant pressure the volume of fix amount of dry gas is directly proportional to absolute temperature from Charles law.

Since V and T are directly varying directly, we can write the equation using constant "k".

V/T = k

In this case the value of k depends on the pressure of gas, the amount of gas and also unit volume.

V/T = k .........(i)

Let us consider V1 and T1 the Volume and temperature of the ideal gases.

Then the equation is;

V1/T1 =k........(ii)

After this let change the temperature and volume be T2 and V2, respectively.

Then the equation is:

V2/T2 = k.......(iii)

Now; Equating equation (ii) and (iii)…

V1/T1 = V2/T2

So, this the formula. (i.e V1/T1 = V2/T2).

Hope it helps!

2NO + 2H2 ⟶N2 + 2H2O What would the rate law be if the mechanism for this reaction were: 2NO + H2 ⟶N2 + H2O2 (slow) H2O2 + H2 ⟶2H2O (fast)

Answers

Answer:

rate = [NO]²[H₂]

Explanation:

2NO + H2 ⟶N2 + H2O2 (slow)

H2O2 + H2 ⟶2H2O (fast)

From the question, we are given two equations.

In chemical kinetics; that is the study of rate reactions and changes in concentration. The rate law is obtained from the slowest reaction.

This means that our focus would be on the slow reaction. Generally the rate law is obtained from the concentrations of reactants in a reaction.

This means our rate law is;

rate = [NO]²[H₂]

(a) Identify the name of the method used below for the separation.
(b) Give one more application of this method of separation.
(c) What is the name for the line at position B ?
(d) what conclusions can you draw about the colours present in sweets C and D ?

Answers

Answer:

(a) Chromatography

(b) DNA fingerprinting

(c) Origin

(d) Sweet C consists of more colours than sweet D.

  ii. The speed of colours in sweet C are proportional to one another, while that of colours in D is not.

Explanation:

Chromatography is one of the physical method of separating mixtures. This process composed of the ability of the constituents in a mixture to separate by virtue of rate of movement through a medium, thus separates into constituents.

It can be used to determine the soluble constituents of a given mixture. And for purification purpose.

PLEASE HELP WILL GIVE BRAINLIEST!!!
Refining aluminum takes place in steel vats lined with graphite. What function does the graphite lining serve?
A. It is the cathode.
B. It is the anode.
C. It soaks up the waste oxygen released from bauxite (Al2O3)
D. It serves as a grounding layer to prevent electrical surges.

Answers

Answer: It's the anode broski (B)

Explanation: I'm taking the Chem summer course too broski, this was the correct answer. Cheers broski

A baseball has a mass of 0.145 kilograms. If acceration due to gravity is 9.8m/s,what is the weight of the baseball in newtons?

Answers

Answer:

I hope it works

Explanation:

As we know that

w=m*g

given m=0.145 , g=9.8

hence we get

w= (9.8)*(0.145)

w=1.421 m/sec 2

if its help-full thank hit the stars and brain-list it thank you

what is the zeff of carbon​

Answers

Answer:

Explanation:

The steps to calculate the Zeff is :

1)  Write  the electronic configuration.

Carbon: 1s2 2s2 2p2

Oxygen: 1s2 2s2 2p4

2) there are two core electrons in each atom and four in carbon and six in oxygen.

1s) (2s2p)

3) as mentioned the shielding of electrons within the same shell is negligible.

4) for electron of s or p orbital  the shielding contribution by the electrons having a principal quantum number less by one would be 0.85 each. And all electrons further left would contribute an amount of 1.0 each.

5) For oxygen:

Zeff = Z - S

S = 2X0.85 = 1.7

Zeff = 8- 1.7 = 6.3

For carbon

Zeff = Z - S

S = 2X0.85 = 1.7

Zeff = 6- 1.7 = 4.3

ΔH = +572 kJ for the decomposition of water by the reaction 2 H 2O( ) → 2 H 2(g) + O 2(g). How many grams of water can be decomposed by 5.00 × 10 3 kJ of energy? ΔH = +572 kJ for the decomposition of water by the reaction 2 H 2O( ) → 2 H 2(g) + O 2(g). How many grams of water can be decomposed by 5.00 × 10 3 kJ of energy?

Answers

Answer:

315 g

Explanation:

Step 1: Write the thermochemical equation

2 H₂O(l) → 2 H₂(g) + O₂(g)      ΔH = +572 kJ

Step 2: Calculate the molar of water decomposed by 5.00 × 10³ kJ of energy

According to the thermochemical equation, 572 kJ are required to decompose 2 moles of water.

5.00 × 10³ kJ × (2 mol/572 kJ) = 17.5 mol

Step 3: Calculate the mass corresponding to 17.5 moles of water

The molar mass of water is 18.02 g/mol.

17.5 mol × 18.02 g/mol = 315 g

plsssss help!!! Deep Space 1 was a spacecraft powered by an engine that gave off xenon particles to change velocity. It had a mass of 500 kg. Which expression can be used to find the spacecraft’s acceleration if its engine created a net force of 0.10 N? A) 0.10 N 500 kg B) 500 kg · 0.10 N C) 500 kg 0.10 N D) 500 kg + 0.10 N

Answers

Answer:

Spacecraft’s acceleration (a) = 0.10 N / 500 kg

Explanation:

Given:

Mass of Spacecraft (M) = 500 Kg

Force generate by engine (F) = 0.10 N

Find:

Spacecraft’s acceleration (a)

Computation:

F = Ma

0.10 = 500 (a)

a = 0.10 / 500

Spacecraft’s acceleration (a) = 0.10 N / 500 kg

The expression which can be used to find the spacecraft’s acceleration if its engine created a net force of 0.10 N is  0.10 N/500Kg.

We know that force is the product of the mass a body and its acceleration. The result of motion is the action of an unbalanced force. We have the following information;

Mass of the spacecraft = 500 kg

Force on the engine =  0.10 N

From Newton's law;

F = ma

F = force

m = mass

a = acceleration

a = F/m

acceleration = 0.10 N/500Kg

Learn more: https://brainly.com/question/8592296

For the following reaction, 3.76 grams of iron are mixed with excess oxygen gas . The reaction yields 4.29 grams of iron(II) oxide . iron ( s ) oxygen ( g ) iron(II) oxide ( s ) What is the theoretical yield of iron(II) oxide

Answers

Answer:

4.84g of FeO is the theoretical yield

Explanation:

The Iron, Fe(s), reacts with oxygen, O₂(g), producing Iron (II) oxide, as follows:

2Fe(s) + O₂(g) → 2FeO

Theoretical yield is the yield of a reaction in which you assume the 100% of reactants is converted in products.

To find theoretical yield we need to find moles of Iron, and, knowing 2 moles of Fe produce 2 moles of FeO (Ratio 1:1), we can find theoretical yield of FeO as follows:

Moles Fe (Molar mass: 55.845g/mol)

Using the molar mass of the compound we can convert grams to moles, thus:

3.76g Fe × (1mol / 55.845g) = 0.0673 moles of Fe

Moles and mass of FeO

As there are in reaction 0.0673 moles Fe, assuming a theoretical yield (And as ratio of the reaction is 1:1), you will obtain 0.0673 moles of FeO.

Theoretical yield is given in grams, As molar mass of FeO is 71.844g/mol, theoretical yield of the reaction is:

0.0673 moles FeO × (71.844g / mol) =

4.84g of FeO is the theoretical yield

What is the mole fraction of urea, CO(NH2)2, in a solution prepared by dissolving 4.0 g of urea in 32.0 g of methanol, CH3OH

Answers

Answer:

0.0630

Explanation:

The molar mass of urea = 60 g/mol

we all know that:

[tex]\mathtt{number \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]

Then; the number of moles of urea

= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]

= 0.0667 mol

Similarly; the number of moles of methanol

= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]

= 0.9988 mol

The total number of moles = (0.0667 + 0.9988) mol

= 1.0655 mol

Finally,the mole fraction of urea  [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]

[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]

= 0.0630

For each bond, show the direction of polarity by selecting the correct partial charges. _________ Si-P _________ _________ Si-Cl _________ _________ Cl-P _________ The most polar bond is _______

Answers

Answer:

Siδ⁺ -- Pδ⁻⁻

Clδ⁻⁻ -- Pδ⁺

Siδ⁺ -- Clδ⁻⁻

Of the mentioned bonds the most polar bond is Si -- Cl

The polarity of the bond primarily relies upon the electronegativity difference between the two atoms that forms the bond. Therefore, if the electronegativity difference between the two atoms that forms the bond is more the bond will be more polar, and if it is less then the bond will be less polar. The electronegativity of the atoms mentioned is Si = 1.8 , P = 2.1 and Cl = 3.00.  

Therefore, the Si - Cl atoms exhibit more electronegativity difference, thus, the Si - Cl bond will be the most polar bond.  

sodium bisulfite is an ionic compound that is used as a mild bleaching agent and a food preservative. It contains the bisulfate ion, an amphiprotic ion with the chemical formula HSO₃-.

a. Write the chemical equation for the reaction of HSO₃- with OH-. Is the bisulfite ion functioning as an acid or a base in this reaction?
b. Write the chemical equation for the reaction of HSO₃- with H₃O+. Is the bisulfite ion functioning as an acid or a base in this reaction?

Answers

a)

HSO₃⁻ + OH⁻ → H₂SO₄⁽²⁻⁾

In this reaction, we know that OH will act as a base. which means that the bisulfite ion will act as an acid

also because SO₃⁽²⁻⁾ will be relative more stable than O⁽²⁻⁾ because of larger valence shell. so HSO₃ will protonate. making it the acid

b)

HSO₃⁻ + H₃O⁺ → H₂SO₃

Here, the hydronium ion will be the one to protonate because that will help it form water, which is a much stabler compound

so the hydronium ion will protonate, making it the acid and the bisulfite will take that proton, making it the base

How many mL of a 0.130 M aqueous solution of chromium(II) nitrate, Cr(NO3)2, must be taken to obtain 5.08 grams of the salt

Answers

Answer:

222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.

Explanation:

Being:

Cr: 52 g/moleN: 14 g/moleO: 16 g/mole

the molar mass of chromium (II) nitrate, Cr(NO₃)₂ is:

Cr(NO₃)₂ = 52 g/mole + 2* (14 g/mole + 3* 16 g/mole)= 176 g/mole

So: if 176 grams are present in 1 mole of the compound, 5.08 grams in how many moles of the compound will be present?

[tex]amount of moles=\frac{5.08 grams* 1 mole}{176 grams}[/tex]

amount of moles=0.0289 moles

Molarity (M) is the number of moles of solute that are dissolved in a given volume. It is then calculated by dividing the moles of the solute by the volume of the solution:

[tex]molarity (M)=\frac{number of moles of solute}{volume}[/tex]

Molarity is expressed in [tex]\frac{moles}{liter}[/tex]

So in this case:

molarity= 0.130 Mnumber of moles of solute= 0.0289 molesvolume= ?

Replacing:

[tex]0.130 M=0.130 \frac{moles}{liter} =\frac{0.0289 moles}{volume}[/tex]

Solving:

[tex]volume=\frac{0.0289 moles}{0.130 \frac{moles}{liter} }[/tex]

volume=0.2223 liters

Being 1 L= 1,000 mL:

volume=0.222 liters= 222.3 mL

222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.

A 5-column table with 2 rows. Column 1 is labeled number of protons, with entries 20 and 9; column 2 is number of neutrons, with entries 20 and D; Column 3 is atomic number, with entries A and E; Column 4 is Mass Number, with entries B and 19, and Column 5 is Element (symbol) with entries C and F. Using the periodic table, complete the table to describe each atom. Type in your answers

Answers

Answer:

A    ⇒ 20

B    ⇒ 40

C    ⇒ Ca

D    ⇒ 10

E     ⇒ 9

F     ⇒ F

Explanation:

edge 2021

Answer:

the person above is correct

Explanation:

a jogger runs a mile in 8.92 minutes. 1 mi=1609m; calculate her speed in km/hr

Answers

Answer:

[tex]Speed = 3.30 \frac{km}{hr}[/tex]

Explanation:

Given

Distance = 1 mile

Time = 8.92 minutes

Required

Calculate Speed in km/hr

Speed is calculated as thus;

[tex]Speed = \frac{Distance}{Time}[/tex]

Substitute 1 mile for distance and 8.92 minutes for time

[tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex]

Convert Miles to Kilometres

If

[tex]1\ mile = 1609\ m[/tex];

Then

[tex]1\ mile = \frac{1609\ km}{1000}[/tex]

[tex]1\ mile = 1.609\ km[/tex] --- (1)

Convert minutes to hour

[tex]1\ minutes = 0.0167\ hour[/tex]

Multiply both sides by 8.92

[tex]8.92 * 1\ minutes = 0.0167\ hour * 8.92[/tex]

[tex]8.92 \ minutes = 0.148964\ hour[/tex] ---- (2)

By substituting (1) and (2) in [tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex], we have

[tex]Speed = \frac{1.609\ km}{0.48694\ hour}[/tex]

[tex]Speed = 3.304309 \frac{km}{hr}[/tex]

[tex]Speed = 3.30 \frac{km}{hr}[/tex] -- Approximated

Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answerβ-1,4- and α-1,6-glycosidicβ-1,4-glycosidicgalactosean unbranchedglucosea branchedfructoseα-1,6-glycosidicAmylose is ......... polymer of ....... units joined by ........ bonds. Amylopectin is ....... polymer of .......units joined by ........ bonds.

Answers

The words given are not clear, so the clear question is as follows:

Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answer:

A. β-1,4- and α-1,6-glycosidic

B. α-1,4-glycosidic

C. α-1,4-galactose

D. an unbranched glucose

E. a branched fructose

F. α-1,6-glycosidic

Amylose is ......... polymer of ....... units joined by ........ bonds.

Amylopectin is ....... polymer of .......units joined by ........ bonds.

Answer:

D. an unbranched glucose

C. α-1,4-galactose

B. α-1,4-glycosidic

E. a branched fructose

A. β-1,4- and α-1,6-glycosidic

F. α-1,6-glycosidic

Explanation:

Amylose and amylopectin are two types of polysaccharides that can be found in starch granules.

Amylose is linear or unbranched glucose polymer of α-1,4-galactose units that are joined by α-1,4-glycosidic.

Amylopectin is a branched fructose polymer of β-1,4- and α-1,6-glycosidic units joined by α-1,6-glycosidic bonds.

Hence, the correct answers in the sequential order are:

Amylose:

D. an unbranched glucose

C. α-1,4-galactose

B. α-1,4-glycosidic

Amylopectin:

E. a branched fructose

A. β-1,4- and α-1,6-glycosidic

F. α-1,6-glycosidic

How many valence electrons must two atoms share to form a single covalent bond? answers A.2 B.4 C.3 D.1

Answers

Answer:

2

Explanation:

A single covalent bond is formed when two electrons are shared between the same two atoms, one electron from each atom.

Answer:

the answer is 2

Explanation:

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution before the addition of any HNO3. The Kb of NH3 is 1.8 × 10-5.

Answers

Answer:

[tex]pH=11.12[/tex]

Explanation:

Hello,

In this case, ammonia dissociation is:

[tex]NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)[/tex]

So the equilibrium expression:

[tex]Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]

That in terms of the reaction extent and the initial concentration of ammonia is written as:

[tex]1.8x10^{-5}=\frac{x*x}{0.10M-x}[/tex]

Thus, solving by using solver or quadratic equation we find:

[tex]x=0.00133M[/tex]

Which actually equals the concentration of hydroxyl ion, therefore the pOH is computed:

[tex]pOH=-log([OH^-])=-log(0.00133)=2.88[/tex]

And the pH from the pOH is:

[tex]pH=14-pOH=14-2.88\\\\pH=11.12[/tex]

Best regards.

A sample of water at 21.5°C required an input of 1.69 x 104 of heat to reach its boiling point, 100.0 °C. What was the mass of the
water?

Answers

Fawn spends hours each week managing employee shifts and schedules, time she should be spending on other restaurant operations. What can she do? O a) Start leaving more time in her week for scheduling-related tasks b) Write everything down in a weekly planner O c) Consider online inventory management software O d) Consider online scheduling software

Give 3 examples of chemical change and physical change in daily life

Answers

Explanation:

Physical change

★ Boiling of water

❤︎ freezing of water

❣︎ washing of clothes

♫︎ mixing black and brown stones

Chemical change

➪ making of dough

㋛︎ burning of wood

❥︎ cooking of the raw food

❁ burning of paper

ꨄ︎ rusting of iron

★★★➪➪➪➪ it was help to you

Ozone (O 3) in the atmosphere can react with nitric oxide (NO): O 3(g) + NO(g) → NO 2(g) + O 2(g). Calculate the ΔG° for this reaction at 25°C. (ΔH° = –199 kJ/mol, ΔS° = –4.1 J/K·mol)

Answers

Answer:

ΔG°  = 1022. 8 kJ

Explanation:

ΔH° = –199 kJ/mol

ΔS° = –4.1 J/K·mol

T = 25°C = 25 + 273 = 298K (Converting to kelvin temperature)

ΔG° = ?

The relationship between these varriables are;

ΔG° = ΔH°  - TΔS°

ΔG° = –199 - 298 (–4.1)

ΔG° = -199 + 1221.8

ΔG°  = 1022. 8 kJ

To calculate ΔG, the following equation should be used:

ΔG° = ΔH°  - TΔS°

Given:

ΔH° = –199 kJ/mol,  

ΔS° = –4.1 J/K·mol

T=25+273K=298K  

Substitute the respective values:

ΔG° = ΔH°  - TΔS°

       =–199 kJ/mol-(298K×(-4.1 J/K·mol*(1KJ/1000J)))

       =-197.78kJ/mol

Thus, we can conclude the value of ΔG°=-197.78kJ/mol  

Learn more about ΔG° here:

https://brainly.com/question/2491198

State what would be observed when the following pairs of reagents are mixed in a test tube.
C6H2COOH and Na2CO3(aq)
(ii) CH3CH2CH2OH and KMnO4 /H
(iii) CH3CH2OH and CH3COOH + conc. H2SO4 (iv) CH3CH = CHCH3 and Br2 /H2O​

Answers

Answer:

(i). C6H2COOH and Na2CO3(aq)

observation: Bubbles of a colourless gas (carbon dioxide gas)

(ii) CH3CH2CH2OH and KMnO4 /H

observation: The orange solution turns green.

[This is because oxidation of propanol to propanoic acid occurs]

(iii) CH3CH2OH and CH3COOH + conc. H2SO4

observation: A sweet fruity smell is formed.

[This is because an ester, diethylether is formed]

(iv) CH3CH = CHCH3 and Br2 /H2O

observation: a brown solution is formed.

When 91.96g of Na reacts with 32.o g of O2 how many grams of NaO2 are produced

Answers

Answer:

123.96 g Na₂O

Explanation:

4 Na  +  O₂  ⇒  2 Na₂O

You first need to find the limiting reagent.  Convert the reactants to moles and see which produces the least amount of product using the mole ratios in the chemical equation.

(91.96 g Na)/(22.99 g/mol Na) = 4 mol Na

(4 mol Na) × (2 mol Na₂O/4 mol Na) = 2 mol Na₂O

(32.0 g O₂)/(32.0 g/mol) = 1 mol O₂

(1 mol O₂) × (2 mol Na₂O/1 mol O₂) = 2 mol Na₂O

Since they both produce the same amount of product, you don't need to pick a limiting reagent.  Now, convert moles of Na₂O to grams.

(2 mol Na₂O) × (61.98 g/mol Na₂O) = 123.96 g Na₂O

How many grams of Br are in 445 g CaBr2 ?

Answers

Answer:

156 g Br. Explanation: In order to figure out how many grams of bromine you get in that many grams of calcium bromide, CaBr2 , you must find ...

Answer:

156 g

Explanation:156 g

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Calculate the pH of a buffer that is 0.13 M in lactic acid and 0.10 M in sodium lactate. Express your answer using two decimal places.

Answers

Answer:

pH of the buffer is 3.75

Explanation:

It is possible to find pH of a buffer using Hendersson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is molar concentration of the conjugate base and [HA] concentration of the weak acid

In the lactic buffer, pKa = 3.86. Lactic acid is the weak acid and its conjugate base is tha lactate salt. H-H equation for this buffer is:

pH = 3.86 + log [Lactate] / [Lactic acid]

Replacing with the concentrations of the problem:

pH = 3.86 + log [0.10M] / [0.13M]

pH = 3.75

pH of the buffer is 3.75

Determine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ... is ... ___ . The oxidation state of sulfur in sulfur dioxide SO2 is ___ . The oxidation state of iron in iron(

Answers

Answer:

The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].

The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].

Explanation:

The oxidation states of atoms in a compound should add up to zero.

Ag₂O

There are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:

[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].

SO₂

Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:

[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.

Therefore:

[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the oxidation state of [tex]\rm S[/tex] here:

[tex]\text{Oxidation state of $\rm S$} = 4[/tex].

Rectangular cube 3.2 m length 1.2 m in height and 5 m in length is split into two parts. The container has a movable airtight divider that divides its length as necessary. Part A has 58 moles of gas and part B has 165 moles of a gas.

Required:
At what length will the divider to equilibrium?

Answers

Answer:

The length the divider is to  equilibrium from Part A = 1.30 m and from Part B = 3.70 m

Explanation:

Given that:

A rectangular cube with 3.2 m breadth, 1.2 m height and 5 m in length is splitted into two parts.

The diagrammatic expression for the above statement can be found in the attached diagram below.

The container has a movable airtight divider that divides its length as necessary.

Part A has 58 moles of gas

Part B has 165 moles of a gas.

Thus, the movable airtight divider will stop at a length where the pressure on it is equal on both sides.

i.e

[tex]\mathtt{P = P_A = P_B}[/tex]

Using the ideal gas equation,

PV = nRT

where, P,R,and  T are constant.

Then :

[tex]\mathsf{\dfrac{V_A}{n_A}= \dfrac{V_B}{n_B}}[/tex]

[tex]\mathsf{\dfrac{L_A \times B \times H}{n_A}= \dfrac{L_B \times B \times H}{n_B}}[/tex] --- (1)

since Volume of a cube = L × B × H

From the question; the L = 5m

i,e

[tex]\mathsf{L_A +L_B}[/tex] = 5

[tex]\mathsf{L_A = 5 - L_B}[/tex]

From equation (1) , we divide both sides by (B × H)

Then :

[tex]\mathsf{\dfrac{L_A }{n_A}= \dfrac{L_B }{n_B}}[/tex]

[tex]\mathsf{\dfrac{5-L_B}{58}= \dfrac{L_B }{165}}[/tex]

By cross multiplying; we have:

165 ( 5 - [tex]\mathsf{L_B}[/tex] )  = 58 (

825 - 165[tex]\mathsf{L_B}[/tex]  = 58

825 = 165[tex]\mathsf{L_B}[/tex] +58

825 = 223[tex]\mathsf{L_B}[/tex]

[tex]\mathsf{L_B}[/tex] = 825/223

[tex]\mathsf{L_B}[/tex]  = 3.70 m

[tex]\mathsf{L_A = 5 - L_B}[/tex]

[tex]\mathsf{L_A = 5 - 3.70}[/tex]

[tex]\mathsf{ L_A}[/tex] = 1.30 m

The length the divider is to  equilibrium from Part A = 1.30 m and from Part B = 3.70 m

Please help me! I am a bit stuck on this.

Answers

Option 2 is the correct answer

Which of the following compounds is more soluble in a 0.10 M NaCN solution than in pure neutral water? Ca3(PO4)2 AgBr CaCO3 Mg(OH)2 NH4ClO4

Answers

Answer:

AgBr

Explanation:

Silver bromide has a very low solubility product constant of about 7.7 ×10^-13 in pure water hence it is not quite soluble in pure water.

However, with NaCN, the AgBr forms the complex [Ag(CN)2]^2- which has a formation constant of about 5.6 ×10^8. This very high formation constant implies that the complex is easily formed leading to the dissolution of AgBr in NaCN.

The equation for the dissolution of AgBr in cyanide is shown below;

AgBr(s) + 2CN^-(aq) ----> [Ag(CN)2]^2-(aq) + Br^-(aq)

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