Which is a nonpolar molecule?

Answer:

The hydrogen can be gotten from the added Acid or water during "workup".

Explanation:

Basically we can say that the reaction describe in this question is a Reduction reaction because of the chemical compound called sodium borohydride. In the reaction described above we can see that there is a Reduction of ketone to alcohol by the compound; sodium borohydride.

For the reduction Reaction to occur the C-O bond must break so as to enable the formation of O-H bond and C-H bond.

So, "the reaction mixture is treated with water and H2SO4 to produce the desired alcohol", thus, the oxygen will definitely pick up the hydrogen from H2SO4 or H2O.

Answer:

The answer to your question is given below.

Explanation:

Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:

Al (13) —› 1s² 2s²2p⁶ 3s²3p¹

The orbital diagram is shown on the attached photo.

Answer: screen shot

Explanation:

Answer:

11445.8years

Explanation:

Half-life of carbon-14 = 5720 years

First we have to calculate the rate constant, we use the formula :

Answer:

ΔH = -338.8kJ

Explanation:

it is possible to sum the enthalpy changes of some reactions to obtain the enthalpy change of the whole reaction (Hess's law).

Using the reactions:

R₁ O₂(g) → 2O(g) ΔH₁°= 449.2 kJ

R₂ O₃(g) + Cl(g) → O₂(g) + ClO(g) ΔH₂° = -126 kJ

R₃ ClO (g) + O (g) → O₂ (g) + Cl (g) ΔH₃°= -268 kJ

By the sum 2R₂ + 2R₃:

(2R₂ + 2R₃) = 2O(g) + 2O₃(g) → 4O₂(g)

ΔH = 2ₓ(-126kJ) + (2ₓ-268kJ) = -788kJ

Now, this reaction + R₁

2O₃(g) → 3O₂(g)

ΔH = -768kJ + 449.2kJ

Answer:

0.36 g of N2.

Explanation:

The following data were obtained from the question:

Temperature (T) = 25 °C

Volume (V) = 340 mL

Measured pressure = 730 torr

Vapour pressure = 23.76 torr

Mass of N2 =..?

First, we shall determine the true pressure of N2. This can be obtained as follow:

Measured pressure = 730 torr

Vapor pressure = 23.76 torr

True pressure =..?

True pressure = measured pressure – vapor pressure

True pressure = 730 – 23.76

True pressure = 706.24 torr.

Converting 706.24 torr to atm, we have:

760 torr = 1 atm

Therefore,

706.24 torr = 706.24 / 760 = 0.929 atm

Next, we shall convert 340 mL to L. This is illustrated below:

1000 mL = 1 L

Therefore,

340 mL = 340/1000 = 0.34 L

Next, we shall convert 25 °C to Kelvin temperature. This is illustrated below:

Temperature (K) = Temperature (°C) + 273

T(K) = T (°C) + 273

T (°C) = 25 °C

T(K) = 25 °C + 273

T (K) = 298 K

Next, we shall determine the number of mole of N2. This can be obtained as follow:

Pressure (P) = 0.929 atm

Volume (V) = 0.34 L

Temperature (T) = 298 K

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =...?

PV = nRT

0.929 x 0.34 = n x 0.0821 x 298

Divide both side by 0.0821 x 298

n = (0.929 x 0.34 ) /(0.0821 x 298)

n = 0.0129 mole

Finally, we shall determine the mass of N2 as shown below:

Mole of N2 = 0.0129 mole

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 =.?

Mole = mass /Molar mass

0.0129 = mass of N2/ 28

Cross multiply

Mass of N2 = 0.0129 x 28

Mass of N2 = 0.36 g

Therefore, 0.36 g of N2 was collected.

The given question is incomplete, the complete question is:

A chemistry student weighs out 0.0349g of formic acid HCHO2 into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1500M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.

Answer:

The correct answer is 5.06 ml.

Explanation:

Based on the given information, the weight of formic acid given is 0.0349 grams. The volume of formic acid of V1 given is 250 ml. The molecular mass of formic acid is 46 grams per mole. Now the molarity of formic acid will be,  

[HCOOH] = weight * 1000 / molecular mass * volume (ml)

= 0.0349 * 1000 / 46 * 250

= 0.003035 M or M1

The molarity of NaOH given is 0.1500 M or M2

Let us assume that the volume needed to attain equivalence point is V2 ml.  The volume V2 can be determined by using the dilution equation,  

M1V1 = M2V2

V2 = M1V1/M2  

V2 = 0.003035 * 250 / 0.1500

V2 = 5.06 ml.  

Hence, the volume of NaOH needed is 5.06 ml.  

Answer:

760 mmHg

Explanation:

Step 1: Given data

Step 2: Calculate the atmospheric pressure

Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.

P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg

Explanation:

The mass to charge ratio =136

Answer:

A tetraquark in physics is an exotic meson composed of four valence quarks.

Explanation:

It has been suspected to be allowed by quantum chromodynamics, the modern story of strong interactions.

Hope it helps.

Answer:

Explanation:

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

ionisation constant = 1.36 x 10⁻⁴ .

molecular weight of lactic acid = 90 g

moles of acid used = 20 / 90

= .2222

it is dissolved in one litre so molar concentration of lactic acid formed

C = .2222M

Let n be the fraction of moles ionised  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionisation constant Ka

Ka = nC x nC / C - nC

= n²C ( neglecting n in the denominator )

n² x .2222 = 1.36 x 10⁻⁴

n = 2.47  x 10⁻²

nC = 2.47  x 10⁻² x .2222

= 5.5 x 10⁻³

So concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per litre .

The concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter .

Ionization of lactic acid can be represented as:

CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻  + H⁺

Given:

ionization constant = 1.36 x 10⁻⁴

mass= 20.0 g

Now, Molecular weight of lactic acid = 90 g

[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]

It is dissolved in 1.00L so molar concentration of lactic acid formed will be

C = 0.22M

Consider "n" to be the fraction of moles ionized  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionization constant Ka

[tex]K_a =\frac{nC*nC}{C-nC}[/tex]

[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )

On substituting the values we will get:

[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]

To find the concentration of hydronium ion in the solution,

[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]

So, concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter.  

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Answer:

hi here goes your answer

Explanation:

iv. The lower the PH, the weaker the base

When a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities. The statement is True.

Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.

There are various methods of expressing the concentration of a solution.

Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.

Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.

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Answer:

2.81mL

Explanation:

Based on the reaction:

C₆H₃COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl

Benzoyl chloride + ammonia → Benzamide

1 mole of benzoyl chloride in excess of ammonia produce 1 mole of Benzamide.

Thus, assuming a theoretical yield, to produce 24.1mmoles of benzamide you require 24.1mmoles of benzoyl chloride.

As molar mass of benzoyl chloride is 141g/mol, mg you require are:

mg Benzoyl chloride: 24.1mmol × (141mg / 1mmol) = 3398.1mg = 3.3981g of benzoyl chloride.

to convert this mass to mL, you require density of Benzoyl chloride (1.21g/mL). Thus, mL you need are:

3.3981g × (1mL / 1.21g) =

Answer:

28.13 g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2O4(l) + 2N2H4 (l) → 3N2(g) + 4H2O(g)

Next, we shall determine the masses of N2O4 and N2H4 that reacted and the mass of H2O produced from the balanced equation.

This is illustrated below:

Molar mass of N2O4 = (14x2) + (16x4) = 92 g/mol

Mass of N2O4 from the balanced equation = 1 x 92 = 92g

Molar mass of N2H4 = (14x2) + (4x1) = 32 g/mol

Mass of N2H4 from the balanced equation = 2 x 32 = 64 g

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

Mass of H2O from the balanced equation = 4 x 18 = 72 g

Summary:

From the balanced equation above,

92 g of N2O4 reacted with 64 g of N2H4 to produce 72 g of H2O.

Next, we shall determine the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

92 g of N2O4 reacted with 64 g of N2H4.

Therefore, 40 g of N2O4 will react with = (40 x 64)/92 = 27.83 g of N2H4.

From the calculations made above, we can see that it will take a higher mass i.e 27.83 g than what was given i.e 25 g of N2H4 to react completely with 40 g of N2O4.

Therefore, N2H4 is the limiting reactant and N2O4 is the excess reactant.

Finally, we shall determine the mass of H2O produced from the reaction of 40.0 g of N2O4 and 25.0 g of N2H4.

In this case the limiting reactant will be used because it will produce the maximum amount of H2O as all of it is consumed in the reaction.

The limiting reactant is N2H4 and the mass of H2O produced can be obtained as follow:

From the balanced equation above,

64 g of N2H4 reacted to produce 72 g of H2O.

Therefore, 25 g of N2H4 will react to produce = (25 x 72)/64 = 28.13 g of H2O.

Therefore, 28.13 g of H2O were obtained from the reaction.

Answer:

The pressure of the gas sample will be  0.954 atm.​

Explanation:

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases; conversely if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

To determine the change in pressure or volume during a transformation at constant temperature, the following is true:

P1 · V1 = P2 · V2

That is, the product between the initial pressure and the initial volume is equal to the product of the final pressure times the final volume.

In this case:

Replacing:

0.609 atm* 19.9 L= P2* 12.7 L

Solving:

[tex]P2=\frac{0.609 atm* 19.9 L}{12.7 L}[/tex]

P2= 0.954 atm

The pressure of the gas sample will be  0.954 atm.​

Answer:

b, H2SO4 with HCl, as they are both liquid acids

Answer:

Based on the Modern Periodic table, there is an increase in the electropositivity of the atom down the group as well as increases across a period. On comparing the electropositivities of the mentioned oxides central atom, it is seen that Ca is most electropositive followed by Al, Si, C, P, and S is the least electropositive.  

With the decrease in the electropositivity, there is an increase in the acidity of the oxides. Thus, the increasing order of the oxides from the least acidic to the most acidic is:  

CaO > Al2O3 > SiO2 > CO2 > P2O5 > SO3. Hence, CaO is the least acidic and SO3 is the most acidic.  

Since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:

[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]

Note the following:

Using the periodic table diagram given in the attachment below, we can rank the given oxides according to their increasing acidity.

Therefore, since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:

[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]

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Answer:

0.1110 mol

Explanation:

Mass = 2g

Molar mass = 18.016 g/mol

moles = ?

These quantities are realted by the following equation;

Moles = Mass / Molar mass

Substituting the values of the quantities and solving for moles, we have;

Moles = 2 / 18.016 = 0.1110 mol

Answer:

1. The coefficients are: 1, 3, 2

2. From the balanced equation, we obtained the following:

3 moles oxygen, O2 reacted.

2 moles of Hydrogen sulfide, H2S reacted.

2 moles of water were produced.

2 moles of sulphur dioxide, SO2 were produced.

Explanation:

1. Determination of the coefficients of the equation.

This is illustrated below:

P2O5(s) + H2O(l) <==> H3PO4(aq)

There are 2 atoms of P on the left side and 1 atom on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:

P2O5(s) + H2O(l) <==> 2H3PO4(aq)

There are 2 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

P2O5(s) + 3H2O(l) <==> 2H3PO4(aq)

Now the equation is balanced.

The coefficients are: 1, 3, 2.

2. We'll begin by writing the balanced equation for the reaction. This is given below:

2H2S(g) + 3O2(g) => 2H2O(l) + 2SO2(g)

From the balanced equation above,

3 moles of oxygen, O2 reacted with 2 moles of Hydrogen sulfide, H2S to produce 2 moles of water, H2O and 2 moles of sulphur dioxide, SO2.

Answer:

Explanation:

RX + AgNO₃ = R⁺ ( carbocation ) + AgX + NO₃⁻

C₂H₅OH ( a nucleophile ) + R⁺ = ROC₂H₅

C₅H₁₁X + AgNO₃ = C₅H₁₁⁺ + AgX + NO₃⁻

In the first case carbocation produced is CH₃CH₂CH₂CH₂CH₂⁺

CH₃CH₂CH₂CH₂CH₂⁺ ⇒  CH₃CH₂CH₂C⁺HCH₃ ( secondary carbocation more stable )

CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃

Hence option D is correct .

b )

In the second case carbocation produced is

CH₃CH₂CH₂CH⁺CH₃

CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃

The product formed is same as in case of first

Option B is correct

Answer:

Ksp = 2.06x10⁻¹⁰

Explanation:

For AB₂. solubility product constant, Ksp, is written as follows:

AB₂(s) ⇄ A²⁺ + 2Br⁻

Ksp = [A²⁺] [Br⁻]²

Molar solubility represents how many moles of AB₂ are soluble per liter of solution. and is obtained from Ksp:

AB₂(s) ⇄ A²⁺ + 2Br⁻

AB₂(s) ⇄ X + 2X

where X are moles that are soluble (Molar solubility)

Ksp = [X] [2X]²

As molar solubility of the salt is 3.72x10⁻⁴M:

Ksp = 4X³

Ksp = 4(3.72x10⁻⁴)³

Answer: The Clean Air Act was important because it emphasized cost-effective methods to protect the air; encouraged people to study the effects of dirty air on human health; and created a regulation that makes any activities that pollute the air illegal.

Explanation:

Answer:

Clean Air Act (CAA), U.S. federal law, passed in 1970 and later amended, to prevent air pollution and thereby protect the ozone layer and promote public health. The Clean Air Act (CAA) gave the federal Environmental Protection Agency (EPA) the power it needed to take effective action to fight environmental pollution.

Answer:

Explanation:

KHT is a salt which ionises in water as follows

KHT ⇄ K⁺ + HT⁻

Solubility product Kw= [ K⁺ ] [ HT⁻ ]

product of concentration of K⁺ and HT⁻ in water

In KCl solution , the solubility product of KHT will be decreased .

In KCl solution , there is already presence of K⁺  ion in the solution . So

in the equation  

[ K⁺ ] [ HT⁻ ]  = constant

when K⁺ increases [ HT⁻ ] decreases . Hence less of KHT dissociates due to which its  solubility decreases . It is called common ion effect . It is so because here the presence of common ion that is K⁺ in both salt to be dissolved and in solvent , results in decrease of solubility of the salt .

Answer:

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

3.52 g of PbCl2

3.76 g of KCl

Explanation:

The equation of the reaction is;

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

Number of moles of Pb(NO3)2 =mass/molar mass 5.06g/331.2 g/mol = 0.0153 moles

Number of moles of KCl= mass/ molar mass= 6.03g/74.5513 g/mol= 0.081 moles

Next we obtain the limiting reactant; the limiting reactant yields the least number of moles of products.

For Pb(NO3)2;

1 mole of Pb(NO3)2 yields 1 mole of PbCl2

Therefore 0.0153 moles of Pb(NO3)2 yields 0.0153 moles of PbCl2

For KCl;

2 moles of KCl yields 1 mole of PbCl2

0.081 moles of KCl yields 0.081 moles ×1/2 = 0.041 moles of PbCl2

Therefore Pb(NO3)2 is the limiting reactant.

Theoretical Mass of precipitate obtained = 0.0153 moles of PbCl2 × 278.1 g/mol = 4.25 g of PbCl2

% yield = actual yield/theoretical yield ×100

Actual yield = % yield × theoretical yield /100

Actual yield= 82.9 ×4.25/100

Actual yield = 3.52 g of PbCl2

If 1 mole of Pb(NO3) reacts with 2 moles of KCl

0.0153 moles of Pb(NO3)2 reacts with 0.0153 moles × 2 = 0.0306 moles of KCl

Amount of excess KCl= 0.081 moles - 0.0306 moles = 0.0504 moles of KCl

Mass of excess KCl = 0.0504 moles of KCl × 74.5513 g/mol = 3.76 g of KCl

Answer:

Approximately [tex]10.88[/tex].

Explanation:

[tex]\rm OBr^{-}[/tex] can act as a weak Bronsted-Lowry base:

[tex]\rm OBr^{-}\; (aq) + H_2O\; (l) \rightleftharpoons HOBr\; (l) + OH^{-}\; (aq)[/tex].

(Side note: the state symbol of [tex]\rm HOBr[/tex] in this equation is [tex]\rm (l)[/tex] (meaning liquid) because [tex]\rm HOBr[/tex] is a weak acid.)

However, the equilibrium constant of this reaction, [tex]K_\text{eq}[/tex], isn't directly given. The idea is to find [tex]K_\text{eq}[/tex] using the [tex]\rm pH[/tex] value at the half-equivalence point. Keep in mind that this system is at equilibrium all the time during the titration. If temperature stays the same, then the same [tex]K_\text{eq}[/tex] value could also be used to find the [tex]\rm pH[/tex] of the solution before the acid was added.

At equilibrium:

[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].

At the half-equivalence point of this titration, exactly half of the base, [tex]\rm OBr^{-}[/tex], has been converted to its conjugate acid, [tex]\rm HOBr[/tex]. Therefore, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should both be equal to one-half the initial concentration of [tex]\rm OBr^{-}[/tex].

As a result, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should be the same. The expression for [tex]K_\text{eq}[/tex] can thus be simplified:

[tex]\begin{aligned}& K_\text{eq} \\&= \frac{\left(\text{half-equivalence $[\rm HOBr\; (l)]$}\right)\cdot \left(\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\right)}{\text{half-equivalence $[\rm OBr^{-}\; (l)]$}}\\ &=\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\end{aligned}[/tex].

In other words, the [tex]K_\text{eq}[/tex] of this system is equal to the [tex]\rm OH^{-}[/tex] concentration at the half-equivalence point. Assume that [tex]\rm p\mathnormal{K}_\text{w}[/tex] the self-ionization constant of water, is [tex]14[/tex]. The concentration of [tex]\rm OH^{-}[/tex] can be found from the [tex]\rm pH[/tex] value:

[tex]\begin{aligned}& \text{half-equivalence $[\rm OH^{-}\; (aq)]$} \\ &= 10^{\rm pH - p\mathnormal{K}_\text{w}}\;\rm mol \cdot L^{-1} \\ &= 10^{7.75 - 14}\; \rm mol \cdot L^{-1}\\ &= 10^{-6.25}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

Therefore, [tex]\begin{aligned} K_\text{eq} &= 10^{-6.25}\end{aligned}[/tex].

Again, since [tex]\rm KOBr[/tex] is a soluble salt, all that [tex]0.200\; \rm M[/tex] of [tex]\rm KOBr[/tex] in this solution will be in the form of [tex]\rm K^{+}[/tex] and [tex]\rm OBr^{-}[/tex] ions. Before any hydrolysis takes place, the concentration of [tex]\rm OBr^{-}[/tex] should be equal to that of [tex]\rm KOBr[/tex]. Therefore:

[tex]\text{$[\rm OBr^{-}\; (aq)]$ before hydrolysis} = 0.200\; \rm M[/tex].

Let the equilibrium concentration of [tex][\rm OH^{-}\; (aq)][/tex] be [tex]x\; \rm M[/tex]. Create a RICE table for this reversible reaction:

[tex]\begin{array}{c|ccccccc} & \rm OBr^{-}\; (aq) &+&\rm H_2O\; (l)& \rightleftharpoons & \rm HOBr\; (l)& + & \rm OH^{-}\; (aq) \\ \textbf{I}& 0.200\; \rm M & & & & 0 \; \rm M & & 0\; \rm M \\ \textbf{C} & -x\; \rm M & & & & +x \; \rm M & & +x\; \rm M \\ \textbf{E}& (0.200 + x)\; \rm M & & & & x \; \rm M & & x\; \rm M \end{array}[/tex].

Assume that external factors (such as temperature) stays the same. The [tex]K_\text{eq}[/tex] found at the half-equivalence point should apply here, as well.

[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].

At equilibrium:

[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x}[/tex].

Assume that [tex]x[/tex] is much smaller than [tex]0.200[/tex], such that the denominator is approximately the same as [tex]0.200[/tex]:

[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x} \approx \frac{x^2}{0.200}[/tex].

That should be equal to the equilibrium constant, [tex]K_\text{eq}[/tex]. In other words:

[tex]\displaystyle \frac{x^2}{0.200} \approx K_\text{eq} = 10^{-6.25}[/tex].

Solve for [tex]x[/tex]:

[tex]x \approx 3.35\times 10^{-4}[/tex].

In other words, the [tex]\rm OH^{-}[/tex] before acid was added was approximately [tex]3.35\times 10^{-4}\; \rm M[/tex], which is the same as [tex]3.35\times 10^{-4}\; \rm mol \cdot L^{-1}[/tex]. Again, assume that [tex]\rm p\mathnormal{K}_\text{w} = 14[/tex]. Calculate the [tex]\rm pH[/tex] of that solution:

[tex]\begin{aligned}\rm pH &= \rm p\mathnormal{K}_\text{w} + \log [\mathrm{OH^{-}}] \approx 10.88\end{aligned}[/tex].

(Rounded to two decimal places.)

Answer:

I think the answer is " Night vision glasses detect Infrared" energy emitted from cooling objects.

Explanation:

Answer:

25.99mL is the volume internal volume of the flask

Explanation:

To complete the question:

The temperature of the water was measured to be 21ºC. Use this data to find the internal volume of the stoppered flask

The flask was filled with water, that means the internal volume of the flask is equal to the volume that the water occupies.

To find the volume of the water you need to find the mass and by the use of density of water at 21ºC (0.997992g/mL), you can find the volume of the flask, thus:

Mass water = Mass filled flask - Mass of clean flask

Mass water = 60.167g - 34.232g

Mass water = 25.935g of water.

To convert this mass to volume:

25.935g × (1mL / 0.997992g) =

Answer:

Acidic proteins are proteins that move faster than serum albumin on zone electrophoresis (starch or acrylamide gel) and which bind most strongly to the basic ion exchangers used in protein chromatography.

Basic protein is a late gene product associated with the viral DNA within the nucleocapsid. The harnessing of this promoter allows the expression of foreign genes at earlier times than those using the very late phase promoters of the polyhedron and p10 genes.

Answer:

Acidic proteins are proteins that move faster than serum albumin on zone electrophoresis (starch or acrylamide gel) and which bind most strongly to the basic ion exchangers used in protein chromatography.

Basic protein is a late gene product associated with the viral DNA within the nucleocapsid. The harnessing of this promoter allows the expression of foreign genes at earlier times than those using the very late phase promoters of the polyhedron and p10

Explanation:

Answer:

0.60 mol

Explanation:

There is some info missing. I think this is the original question.

Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5 mol of water.

Step 1: Given data

Moles of water required: 1.5 mol

Step 2: Write the balanced equation

C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)

Step 3: Calculate the moles of oxygen needed to produce 1.5 mol of water

The molar ratio of O₂ to H₂O is 2.5:1. The moles of oxygen needed to produce 1.5 mol of water are (1/2.5) × 1.5 mol = 0.60 mol

Answer:

C. To determine how efficient reactions are.

D. To determine how much reactant they need.

Explanation:

When you are doing a reaction, you are hoping for a percent yield to close of 100%. You make the reaction and determine how many product you obtain. If you know the percent yield of a reaction you can calculate the amount of reactant you need to obtain a determined amount of product.

Having this in mind:

A. To balance the reaction equation.  false. To calculate percent yield you need to balance the reaction before. You don't use percent yield to balance the reaction

B. To determine how much product they will need.  false. You determine how much product you obtain after the reaction. How much product you need is independent of percent yield

C. To determine how efficient reactions are.  true. A way to determine efficience of a reaction is with percent yield. An efficient reaction has a high percent yield.

D. To determine how much reactant they need. true. If you know percent yield of a reaction you can know how many reactant you must add to obtain  the amount of product you want.