Two samples of the same rainwater are tested using two indicators at an environrnental lab. The first indicator, Methyl Orange, reveals a distinct yellow color when added to the sample. The second indicator, Litmus, turns red when placed in contact with the water sample.

Required:
a. Identify a possible pH value for the rainwater.
b. Explain, in terms of hydronium ions and hydroxide ion concentrations, the pH value of the rainwater.

Answer:

[tex]2.11\ * 10^{-2}[/tex]  is the correct answer to the given question.

Explanation:

Given k=6.40 x 10-3 min-1.

According to the first order reaction .

The concentration of time can be written as

[tex][\ A\ ]\ = \ [\ A_{0}\ ] * e \ ^\ {-kt}[/tex]

Here [tex][\ A\ ]_{0}[/tex] = Initial concentration.

So  [tex][\ A\ ]_{0}= 0.0314 M[/tex]

Putting this value into the above equation.

[tex]0.0314 \ *\ e^{6.40 x 10^{-3} \ * \ 62.0 }[/tex]

=0.211 M

This can be written as

[tex]=\ 2.11 *\ 10^{-2}[/tex]

Answer:

30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.

Explanation:

M1 = 3.00 M

M2 = 1.80 M

V2 = 50 .0 mL = 50 /1000 L = 0.05 L

V1 = unknown

In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.

So we use the equation:

                                  M1 V1 = M2 V2

V1 = M2 V2 / M1

V2 = 1.80 * 0.05 / 3.0

V2 = 0.09 /3.0

V2 = 0.03 L or 30 mL

To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.

Answer:

Explanation:

i dont like reading

Answer: The solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.

Explanation:

The given data is as follows.

Volume of sample water = 46 ml

Temperature = [tex]21^oC[/tex]

After vaporization, washes and then drying the weight of mineral X = 0.87 g

This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.

          1 ml of water = [tex]\frac{0.87 g}{46.0 ml}[/tex]

                                = [tex]1.891 \times 10^{-2}[/tex] g/ml

Therefore, we can conclude that solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.

Answer:

I think that the fire will continue burning, because the oil and water don't mix and the water is heavier (denser) than oil, so the oil will go up and the fire with it. That's why because the gas station have sand instead of water

Water is heavier than oil. Because oil is lighter and immiscible with water, it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire. As a result, the fire won't be put out.

A small amount of water will instantly sink to the bottom of a pan or deep fryer filled with hot, burning oil and explode there. The Scientific American claims that the characteristics of oils explain why they do not mix with water.

Oil or petroleum-related fires cannot be put out with water. Water sinks below the oil because it is heavier than oil and does not float, allowing the fire to continue to burn. Oil and petroleum fires can be put out with fire extinguishers or sand.

The temperature of the burning substance is lowered by water. The fire goes out when the temperature drops below the burning substance's ignition temperature. Here, the water serves as an acclimatizer.

Thus,  it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire.

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Answer:

The answer is Hg.

Explanation:

Symbol for Mercury is Hg.

Answer:

0.031 m

HgO(s) ⇒ Hg(l) + 1/2 O₂(g)

Chemical change

Element

Explanation:

A 75 gram solid cube of mercury (II) oxide has a density of 2.4 × 10³ kg/m³.  What is the length of one side of the cube in cm?

Step 1: Convert the mass to kilograms

We will use the relationship 1 kg = 1,000 g.

[tex]75g \times \frac{1kg}{1,000g} = 0.075kg[/tex]

Step 2: Calculate the volume (V) of the cube

[tex]0.075kg \times \frac{1m^{3} }{2.4 \times 10^{3} kg} = 3.1 \times 10^{-5} m^{3}[/tex]

Step 3: Calculate the length (l) of one side of the cube

We will use the following expression.

[tex]V = l^{3} \\l = \sqrt[3]{V} = \sqrt[3]{3.1 \times 10^{-5} m^{3} }=0.031m[/tex]

The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?

The balanced chemical equation is:

HgO(s) ⇒ Hg(l) + 1/2 O₂(g)

This is a chemical change because new substances are formed.

The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?

The liquid gray substance is Hg(l), which is an element because it is formed by just one kind of atoms.

Answer: 125 g

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles[/tex]

The balanced reaction is:

[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]

According to stoichiometry :

1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]

Thus 1.30 moles of [tex]B_2H_6[/tex] will require=[tex]\frac{3}{1}\times 1.30=3.90moles[/tex]  of [tex]O_2[/tex]

Mass of [tex]O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g[/tex]

Thus 125 g of [tex]O_2[/tex] will be needed to burn 36.1 g of [tex]B_2H_6[/tex]

Answer:

Monosodium Glutamate (MSG) is a chemical which is used in agricultural of food chemistry.

Explanation:

Monosodium Glutamate (MSG) is a chemical that is used in types of different  food as food additives. The functional group that is present in Glutamate are carboxylic acid and amine. This chemical is used in different types of foods which is responsible for enhancing the taste of the food. Monosodium Glutamate is safe if it is used in moderate dose but adversely affected when it is used in large amount.

Answer : The correct answer is 1.46

Explanation :

The following rules are used to round off a number to the required number of significant figures:

(1) If the rightmost digit to be removed is more than 5, the preceding number is increased by one.

(2) If the rightmost digit to be removed is less than 5, the preceding number is not changed.

(3) If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number but it is increased by one if it is an odd number.

(4) The same procedure is follow for decimal values.

As we are given, 1.4593

In the given answer, there are 5 significant figures. Now we have to convert it into 3 significant figures.

According to the rules, round off the given measurement in three significant figures as 1.46

Therefore, the correct answer is 1.46

Answer:

a) ortho-para isomers predominates

b) 3-nitrobenzoic acid ( meta isomer predominates)

c) 3-bromo nitrobenzene ( meta isomer predominates)

d) the ortho- para isomers predominates

Explanation:

a) Toluene contains -CH3 which is an ortho- para- director hence the major product of the sulphonation of toluene should be the ortho- para isomers.

b) The major product of the nitration of benzoic acid is 3-nitrobenzoic acid. This is an electrophilic substitution in which the meta isomer predominates.

c) The meta isomer predominates giving 3-bromo nitrobenzene as the major product.

d) The isopropyl group is an ortho- para director hence the ortho- para isomers predominates .

Answer:

[tex]M=0.0637M[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)[/tex]

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

[tex]n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2} *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}[/tex]

Finally, the resulting molarity in 30.8 mL (0.0308 L):

[tex]M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M[/tex]

Regards.

Answer:

Mass of Ga = 0.73694 gram

Explanation:

Given:

Current = 0.850 A

Time = 60 minutes

Find:

Amount of gas deposit.

Computation:

Total charge = Current × Time in second

Total charge = 0.850 × 60 × 60

Total charge = 3,060 C

Mole of electron = Total charge / Faraday constant         [Faraday constant = 96,485.3329]

Mole of electron = 3,060 / 96,485.3329

Mole of electron = 0.0317146

Moles of Ga = 1/3 [Mole of electron]

Moles of Ga = 1/3 [0.0317146]

Moles of Ga = 0.01057

Mass of Ga = molar mass × Moles of Ga

Mass of Ga = 69.72 × 0.01057

Mass of Ga = 0.73694 gram

Answer:

b. the heating of 10 g of liquid water from 0°C to 100°C.

Explanation:

Hello,

In this case, we must notice a., c. and d. processes are not actually absorbing heat but releasing it since cooling, freezing and condensation are processes with negative heat sign since matter changes from a state of more energy to a state of less energy. We can prove this by realizing that freezing enthalpy of water is -6.00 kJ/mol, condensation enthalpy of eater is -40.8 kJ/mol and a change of temperature from 100 °C to 0 °C is negative.

In such a way, the only process absorbing heat is b. the heating of 10 g of liquid water from 0°C to 100°C since energy must be added to the system, or absorbed by it in order to attain the heating.

Regards.

The process having the greatest amount of heat is:

b. the heating of 10 g of liquid water from 0°C to 100°C.

The options a., c. and d. processes are not actually absorbing heat but releasing it since cooling, freezing and condensation are processes with negative heat sign since matter changes from a state of more energy to a state of less energy.

The freezing enthalpy of water is -6.00 kJ/mol, condensation enthalpy of eater is -40.8 kJ/mol and a change of temperature from 100 °C to 0 °C is negative.

So out of all the options, only process at b is a heating process thus it will absorb greatest amount of heat.

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Answer:

[CH₂Cl₂] = 7.07x10⁻² M

[CH₄] = 0.319 M

[CCl₄] = 0.164 M  

Explanation:

The equilibrium reaction is the following:

2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)  

The equilibrium constant of the above reaction is:

[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]

When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:

[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]

[tex]C_{CH_{4}} = 0.328 M[/tex]      

Now, the concentrations at the equilibrium are:

2CH₂Cl₂(g)   ⇄   CH₄(g)  +  CCl₄(g)

5.35x10⁻² - 2x   0.328 + x   0.173 + x    

[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]

[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]

Solving the above equation for x:  

x₁ = 0.076 and x₂ = -0.0086

Hence, the concentration of the three gases once equilibrium has been reestablished is:

[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M

[CH₄] = 0.328 + (-0.0086) = 0.319 M

[CCl₄] = 0.173 + (-0.0086) = 0.164 M  

We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.

I hope it helps you!

Answer:

if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.!

Explanation:

Oxidation refers to the loss of electrons. Any specie that looses electrons in a redox reaction is said to be the reducing agent. Hence the reducing agent participates in the oxidation half equation. In this case, magnesium is the reducing agent.

Reduction has to do with the gain of electrons. The oxidizing agent participates in the reduction half equation. Hence the oxidizing agent is reduced in the redid reaction. The reducing agent in this case is the oxygen molecule.

Oxidation half equation;

Mg(s)-----> Mg^2+(aq) + 2e

Reduction half equation;

O2(g) + 2e ------> 2O^2-(aq)

From the balanced reaction equation, two moles of electrons is transferred.

Hence if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.

Answer:

The correct answer is 32.2 grams.

Explanation:

Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,  

ΔHrxn = 1676/2 = 838 kJ/mol

Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,  

(1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum

The grams of aluminum produced can be obtained by using the formula,  

mass = moles * molecular mass

= 1.19 * 26.98

= 32.2 grams.  

In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.

A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change.

Al₂O₃(s) ⇒ 2 Al(s) + 3/2 O₂(g)     ΔH rxn = 1676 kJ

According to the thermochemical equation, 2 moles of Al are formed when 1676 kJ of heat is transferred.

1.000 × 10³ kJ × (2 mol Al/1676 kJ) = 1.193 mol Al

The molar mass of Al is 26.98 g/mol.

1.193 mol × 26.98 g/mol = 32.19 g

In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.

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Answer:

Here's what I get  

Explanation:

A Lewis structure shows the valence electrons surrounding the atoms.

Your structure has two problems:

Here's one way to draw a Lewis structure.

1. Draw a trial structure

Make F and O terminal atoms and give each one an octet (Fig. 1).

2. Count the valence electrons in the trial structure

5 BP + 15 LP  = 10 + 30 = 40 electrons

3. Check the number of valence electrons available  

1 S =   1 × 6 =  6 electrons

1 O =   1 × 6 =   6

4 F  = 4 × 7 = 28                    

     TOTAL = 40 electrons

The trial structure has the correct number of electrons.

4. Determine the formal charge on each atom.

To get the formal charges, we cut the covalent bonds in half.

Each atom gets the electrons on its side of the cut.

Formal charge = valence electrons in isolated atom - electrons on bonded atom

FC = VE - BE  

(a) On S

VE = 6

BE = 5 bonding electrons = 5

FC = 6 - 5 = +1

(b) On O:

VE = 6

BE = 3 LP(six electrons) + 1 bonding electron  = 7

FC = 6 - 7 = -1

(c) On F:

VE = 6

BE = 3 lone pairs(6 electrons) + 1 bonding electron = 6 + 1 =7

FC = 7 - 7 = 0

5. Minimize the formal charges

We must rearrange the valence electrons so that S gets one more and O gets one fewer.

Move a lone pair from the O to make an S=O double bond (Fig. 2).

6. Recalculate the formal charges

(a) On S

VE = 6

BE =  (3 bonding electrons) = 6

FC = 6 - 6 = 0

(b) On O:

VE = 6

BE = 2 LP(four electrons) + 2 bonding electrons = 6  

FC = 6 - 6 = 0

Fig. 2 shows the Lewis structure in which all atoms have a formal charge of  zero.

The formal charge of the atoms can be concluded zero with the bond formation between the sulfur and oxygen atom.

The lewis structure can be defined as the dot structure of the valence bond with the bonded atoms. The formal charge can be calculated with the difference in the valence electrons and the bonding electrons.

The formal charge of an atom can be zero when the valence electrons and the bonding electrons are equal. In the structure of [tex]\rm OSF_4[/tex], the formal charge has been assigned zero with the bond formation resulting in the valence electrons and bonding electrons being equal.

The lewis structure with the central S atom has been attached.

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Answer:

The correct answer is 5.30 * 10^-4 mol per L.

Explanation:

Based on Henry's law, in a solution solubility of the gas is directly proportional to the pressure, that is, C is directly proportional to P. Here P is the pressure and C is the concentration of the dissolved gases.  

Therefore, it can be written as,  

C2/C1 = P2/P1

Here, C1 is 6.8 * 10^-4 mol/L, P1 is 1 atm and P2 is 0.78 atm, then the value of C2 obtained by putting the values in the equation,  

C2/(6.8*10^-4) = 0.78/1

C2 = 0.78 * 6.8*10^-4

C2 = 5.30 * 10^-4 mol per L.  

Hence, the concentration of dissolved nitrogen at 0.78 atm is 5.30*10^-4 mol/L.  

Answer:

  1703 mmHg

Explanation:

Volume and pressure are presumed to be inversely proportional. Hence a change in volume by a factor of 125/325 = 5/13 is expected to change the pressure by a factor of 13/5:

  (13/5)(655 mmHg) = 1703 mmHg

Answer: HA has lowest pH and it is the most acidic as compared to the rest of given acids.

Explanation:

We know that relation between [tex]pK_a[/tex] and [tex]K_a[/tex] is as follows.

       [tex]pK_a = -log K_a[/tex]

This means that more is the value of [tex]K_a[/tex], smaller will be the [tex]pK_a[/tex]. Also, more is the value of [tex]K_a[/tex] smaller will be the pH of a solution.

As, larger is the value of [tex]K_a[/tex] more negative will be the [tex]pK_a[/tex] value. Hence, stronger will be the acid.

In the given options, HA has the smallest [tex]pK_a[/tex] value.

Therefore, we can conclude that HA has lowest pH and it is the most acidic as compared to the rest of given acids.

Answer:

D

Explanation:

trust me its correct i think

Answer:

10 g of CO2

Explanation:

Equation of the reaction:

CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2

Fom the above balanced equation,

1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2

Molar mass of Octane = 114 g/mol

Molar mass of oxygen gas = 32 g/mol

Molar mass of CO2 = 44 g/mol

Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.

From the given mass of reactants;

3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.

Therefore oxygen is the limiting reactant.

15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.

Mass of CO2 produced will be

(352 * 15.6)/544 = 10 g of CO2

Answer:

Explanation:

NH3 + H3O+ --> NH4+ + H2O

equllibrium constant =K = [ H2O] [NH4+] / [NH3] [H3O+ ]

                                      =

by inserting thier respecive values can you calcaulte, by the way coniseder  [ H2O] =1 ,

Answer:

pH = 9.08

Explanation:

Quinine, C₂₀H₂₄O₂N₂, Q, is a weak base that, in water, has as equilibrium:

Q + H₂O ⇄ QH⁺ + OH⁻

Where pKb is 5.10

Using H-H equation for weak bases:

pOH = pKb + log₁₀ [QH⁺] / [Q]

The reaction of quinine with HCl is:

Q + HCl → QH⁺ + Cl⁻

Initial moles of quinine are 0.125 moles and moles added of HCl are:

0.05000L × (1.00mol / L) = 0.05000moles.

That means after the addition of 50.00mL of the HCl solution, moles of Q and QH⁺ are:

Q = 0.125mol - 0.050mol = 0.075 moles

QH⁺ = 0.050 moles

Replacing in H-H equation:

pOH = 5.10 + log₁₀ [0.050] / [0.075]

pOH = 4.92

As pH = 14 - pOJ

Answer:

J. J. Thomson

Explanation:

First proposed by J. J. Thomson  in 1904 soon after the discovery of the electron, but before the discovery of the atomic nucleus, the model tried to explain two properties of atoms then known: that electrons are negatively-charged particles and that atoms have no net electric charge.

Answer: 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]{\text{Moles of} H_2}=\frac{1.35g}{2.01g/mol}=0.672moles[/tex]

[tex]\text{Moles of} S_8=\frac{6.86g}{256.5g/mol}=0.0267moles[/tex]

[tex]8H_2+S_8\rightarrow 8H_2S[/tex]

According to stoichiometry :

1 mole of [tex]S_8[/tex] require = 8 moles of [tex]H_2[/tex]

Thus 0.0267 moles of [tex]S_8[/tex] will require=[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex]  of [tex]H_2[/tex]

Thus [tex]S_8[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.

As 1 mole of [tex]S_8[/tex] give = 8 moles of [tex]H_2S[/tex]

Thus 0.0267 moles of [tex]S_8[/tex] give =[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex]  of [tex]H_2S[/tex]

Mass of [tex]H_2S=moles\times {\text {Molar mass}}=0.214moles\times 34.08g/mol=7.29g[/tex]

Thus 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.

Answer:

Translation energy of 1 mole of H2 molecules = KE x Avogadros number

[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]

Explanation:

Photon wavelength [tex]= 58.4 nm = 58.4 * 10^{-9} m[/tex]

Photon momentum = h/wavelength

[tex]= (6.626 * 10^{-34})/(58.4 * 10^{-9})\\\\ = 1.1346 * 10^{-26} \ kg.m/s[/tex]

Mass of H2 molecule m = molar mass/Avogadros number

[tex]= (2.016)/(6.022 * 10^{23})\\\\= 3.3477 * 10^{-24} \ g = 3.3477 * 10^{-27} \ kg[/tex]

Since momentum is conserved:

Photon momentum = H2 molecule momentum = mass x velocity of H2

[tex]1.1346 * 10^{-26} = 3.3477 * 10^{-27} * v[/tex]

velocity [tex]v = 3.389 m/s = 3.39 m/s[/tex]

Translation energy of 1 H2 molecule = kinectic energy (KE) = (1/2)mv^2

[tex]= 1/2 * 3.3477 * 10^{-27} * 3.389^2\\\\= 1.923 * 10^{-26} J[/tex]

Translation energy of 1 mole of H2 molecules = KE x Avogadros number

[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]

Answer: The oxygen's oxidation number is -2.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

In [tex]Fe_2O_3[/tex], Fe is having an oxidation state of +3 called as  cation and oxygen  is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]Fe_2O_3[/tex]

The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.