Sodium hydroxide neutralises hydrochloric acid as shown in the equation:
NaOH(aq) + HCl(aq) i NaCl(aq) + H2O(l)
(4)
dm3
(3)
The student found that 27.20 cm3 of 0.100 moles per dm3 sodium hydroxide neutralised 5.00 cm3 of hydrochloric acid.
Calculate the concentration of the hydrochloric acid in moles per dm3.
Give your answer to three significant figures.

Answers

Answer 1

Explanation:

Mole ratio of NaOH : HCl = 1: 1

Moles of NaOH used = 0.1 mol/1000 cm3 × 27.20 cm3

= 2.72 × 10^-3 mol

Therefore moles of HCl used is also 2.72 × 10^-3 mol

So concentration of HCl can be found by dividing the no.of moles of HCl by the volume of HCl as follows

2.72 × 10^-3 mol/ 5cm3

1000cm3 = 1dm3

Therefore,

1cm3 = 1/1000 dm3

5cm3 = 5/1000 dm3

HCl conc. = 2.72 × 10^-3 mol/ 5×10^-3 dm3

= 0.544 moldm-3


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How could you tell if a substance has undergone a physical change or a chemical one?

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Answer: Chemical changes occur when a substance combines with another to form a new substance, called chemical synthesis or, alternatively, chemical decomposition into two or more different substances. These processes are called chemical reactions and, in general, are not reversible except by further chemical reactions.

A physical change is are changes affecting the form of a chemical substance, but not its chemical composition. Physical changes are used to separate mixtures into their component compounds, but can not usually be used to separate compounds into chemical elements or simpler compounds.

Why we use two different methods for detection of cogulase enzyme ? Or what other reason or what basic different between them?

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Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?

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Answer:

Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?

Answer: Tristan is most like the Golgi Body

Explanation:

4. A system contracts from an initial volume of 15.0 L to a final volume of 10.0 L under a constant external pressure of 0.800 atm. The value of w, in J, is?

Answers

The value of the work done in joule is 405.3 J.

We know that the work done is obtained using the relation;

w = PΔV

Where;

w = work done

P = pressure

V = volume

Now, substituting values,

w = 0.800( 15.0 -  10.0)

w = 4 atm L

Since;

1 L atm = 101.325 J

4 atm L  = 4 atm L × 101.325 J/1 L atm

= 405.3 J

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How many mL of 0.150 M HF solution are required to produce 0.0370 moles of HF

Answers

Answer:

247 ml

Explanation:

How many mL of 0.150 M HF solution are required to produce 0.0370 moles of HF 0.150 moles/ liter = 0.150/1000 moles/ml =0.000150 moles/ml

0.000150 x ? = 0.0370 moles HF

? = 0.0370/0.000150 = 247 ml

check

247 ml = 247/1000 L = 0.247

(0.247) x (0.150) =0.370 check

A chemical property is a change in _____.

density

physical state

hardness

composition of matter

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Answer:

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A beaker containing a green liquid is left uncovered in a laboratory for one week., After the liquid evaporates, the beaker contains a dry green solid. Was the original liquid in the beaker an element, a compound, or a mixture?

Answers

The original liquid is regarded as a mixture.

A mixture is regarded as a material which comprises of two or more

substances which are combined physically. An example is the mixture of

dye and water.

A compound on the other hand is combined chemically and not physically

which when  exposed to the same type of situation either remains in the

beaker or nothing at all is found. Since the water evaporates and a green

solid was present then it means the material was a solution in which

water(solvent) and the green solid(solute) was physically combined and not

chemically combined together.

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he hybridization of carbon in diamond is _________. Enter your answer with no superscripts or subscripts, i.e., ab3.

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Diamond is composed of hexagonal rings in which sp3 hybridized carbon atoms are linked together.

Hybridization refers to the mixing of atomic orbitals to yield hybrid orbitals that are suitable for bonding. The energy of orbitals that combine to form hybrid orbitals must be close enough for such combination to take place.

Diamond is composed of hexagonal rings in which sp3 hbridized carbon atoms are linked together. Each carbon atom in diamond is tetrahedral.

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The mixing of the two different orbitals to form a compound is called hybridization. For example mixing of s and p orbits.

The correct answer is sp3.

The arrangement of the elements in a different manner to form a new compound is called allotropes. For example, diamond and graphite are the allotropes of carbon.

The valence electrons are in p orbitals and p orbit mixed after the s orbitals.

Therefore, the correct answer is sp3

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Thin-layer chromatography explain ?????​

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Answer:

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. ... After the sample has been applied on the plate, a solvent or solvent mixture (known as the mobile phase) is drawn up the plate via capillary action.

Pleeeeasee someone who’s good at chemistry?! 10 grade

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Answer:

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the solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm. what is the solubility of nitrogen gas in a deepsea divers blood at a depth of 200 feet and pressure of 7.00 atm

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The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

We want to relate the solubility of a gas with its partial pressure.

We can do so using Henry's law.

What does Henry's law state?

Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

C = k × P

where,

C is the concentration of a dissolved gas. k is the Henry's Law constant. P partial pressure of the gas.

The solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm.

Since the solvent is basically water, we can understand that the concentration of nitrogen gas is 1.90 mL/dL at 1.00 atm.

We can use this information to calculate Henry's Law constant.

k = C/P = (1.90 mL/dL)/1.00 atm = 1.90 mL/dL.atm

We want to calculate the solubility of nitrogen gas at a pressure of 7.00 atm.

We will use Henry's law.

C = k × P = (1.90 mL/dL.atm) × 7.00 atm = 13.3 mL/dL

The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

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Al2(SO3)3
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Answer:

from the words below underline six example of rhetorical patterns

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Answer:

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Humans detect smell by inhaling air that contains odor molecules, hence, bind to receptors inside the nose, relaying messages to the brain. Most scents are composed of many odorants, for example, a chocolate is made up of hundreds of distinct odor molecules

Answer:

Atoms themselves do not have a smell. Molecules do.

Explanation:

What happens is that particular molecules bind to receptors in our nose, activating nerves, that then send signals to our brain, which we then interpret and perceive as “smells”. Different molecules bind to different receptors, which is why we can perceive a number of different smells

3.00 L of a gas is collected at 35.0 C and 0.93 atm. What is the volume at STP

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2.47L

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Answer:

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Suppose that in an equilibrium mixture of HCl, Cl2, and H2, the concentration of H2 is 1.0 x 10-11 mol-L-1and that of Cl2 is 2.0 x 10-10 mol-L-1. What is the equilibrium molar concentration of HCl at 500 K, given Kc = 4.0 x 1018 for H2(g) +Cl2(g) ⇆ 2HCl(g).

Answers

Considering the definition of Kc, the equilibrium molar concentration of HCl at 500 K is 0.0894 [tex]\frac{mol}{L}[/tex].

The balanced reaction is:

H₂(g) +Cl₂(g) ⇆ 2 HCl(g)

Equilibrium is a state of a reactant system in which no changes are observed as time passes, despite the fact that the substances present continue to react with each other. In other words, reactants become products and products become reactants and they do so at the same rate.

In other words, chemical equilibrium is established when there are two opposite reactions that take place simultaneously at the same speed.

The concentration of reactants and products at equilibrium is related by the equilibrium constant Kc. Its value in a chemical reaction depends on the temperature and the expression of a generic reaction aA + bB ⇄ cC is

[tex]K_{c} =\frac{[C]^{c} x[D]^{d} }{[A]^{a} x[B]^{b} }[/tex]

That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case, the constant Kc can be expressed as:

[tex]K_{c} =\frac{[HCl]^{2} }{[H_{2} ]x[Cl_{2} ] }[/tex]

You know that in an equilibrium mixture of HCl, Cl₂, and H₂:

the concentration of H₂ is 1.0×10⁻¹¹ [tex]\frac{mol}{L}[/tex]the concentration of Cl₂ is 2.0×10⁻¹⁰ [tex]\frac{mol}{L}[/tex]Kc=4×10¹⁸

Replacing in the expression for Kc:

[tex]4x10^{18} =\frac{[HCl]^{2} }{[1x10^{-11} ]x[2x10^{-10} ] }[/tex]

Solving:

[tex]4x10^{18} =\frac{[HCl]^{2} }{2x10^{-21} }[/tex]

[tex]4x10^{18} x 2x10^{-21}=[HCl]^{2}[/tex]

[tex]8x10^{-3} =[HCl]^{2}[/tex]

[tex]\sqrt[2]{8x10^{-3}} =[HCl][/tex]

0.0894 [tex]\frac{mol}{L}[/tex]= [HCl]

Finally, the equilibrium molar concentration of HCl at 500 K is 0.0894 [tex]\frac{mol}{L}[/tex].

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What identifies the number of protons in the nucleus of an atom?

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Answer: Atomic number

Explanation:

I hope this helps you!

Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if an additional 0.182mol of methane is added to the container under constant temperature and pressure? Give your answer in three significant figures.

Answers

The final volume of the methane gas in the container is 6.67 L.

The given parameters;

initial volume of gas in the container, V₁ = 2.65 Linitial number of moles of gas, n₁ = 0.12 moladditional concentration, n = 0.182 mol

The total number of moles of gas in the container is calculated as follows;

[tex]n_t = 0.12 + 0.182 = 0.302 \ mol[/tex]

The final volume of gas in the container is calculated as follows;

[tex]PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L[/tex]

Thus, the final volume of the methane gas in the container is 6.67 L.

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Calculate the Ka of your acetic acid solution. Discuss this calculation. Based on the value of Ka, is acetic acid a strong acid or a weak acid

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Based on our knowledge of strong and weak acids, we can confirm that the Ka value for acetic acid will be relatively low since it is a weak acid.

Acids can be strong or weak. This is determined by its tendency to break apart into ions or stay together to form molecules. Although somewhat counter-intuitive, strong acids are those that are most likely to break apart and therefore contain a high number of ions within their solutions.

Weak acids, on the other hand, are those that tend to stay together in the form of molecules and therefore possess very low ion counts in their solutions. The acid dissociation constant, Kₐ, is used to measure whether an acid is weak or strong and how much so. In the case of Acetic acid, the ka measurement will offer a low value, indicating a weak acid.

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A mixture of solids containing a ketone, a carboxylic acid, and an amine, are dissolved in DCM. What is the best way to begin an extraction to separate the amine from the mixture

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There are different ways of extraction. The best way to begin an extraction to separate the amine from the mixture is to extract with dilute NaOH.

An acid-base extraction is often used in the extraction of carboxylic acids from the organic layer and thereafter into the aqueous layer.

NaOH is known to be the most common compound that is used to convert a carboxylic acid into its more water-soluble ionic carboxylate form.

But if  the mixture has a compound that you want, and that can react with NaOH, another milder base such as sodium bicarbonate is preferably used.

See full question below

A mixture of solids containing a ketone, a carboxylic acid, and an amine, are dissolved in DCM. What is the best way to begin an extraction in order to separate the carboxylic acid from the mixture?

A) Extract with dilute NaOH

B) Extract with dilute HCl

C) Extract with dichloromethane

D) Extract with water

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Convert 1.36x10 to standard form​

Answers

Answer:

13.6 is the correct answer written in standard form.

Explanation:

1.36, move the decimal once to the right to get 13.6

Answer:

13.6

Explanation:

The standard form is 13.6

Calculate the second ionization energy of the metal M (?Hion2� in kJ/mol) using the following data:


Lattice enthalpy of MO(s), ?Hl� = -2383 kJ/mol

Bond dissociation enthalpy of O2(g) = +498 kJ/mol

First electron affinity of O = -141 kJ/mol

Second electron affinity of O = +744 kJ/mol

Enthalpy of sublimation of M = + 130 kJ/mol

First ionization energy of M = + 267 kJ/mol

Standard enthalpy of formation of MO(s), ?Hf� = -307 kJ/mol

Answers

From the information provided in the question, the second ionization energy of the metal is  578 kJ/mol.

From the question, we have the following information;

Lattice enthalpy of MO(s) = -2383 kJ/mol

Bond dissociation enthalpy of O2(g) = +498 kJ/mol

First electron affinity of O = -141 kJ/mol

Second electron affinity of O = +744 kJ/mol

First ionization energy of M = + 267 kJ/mol

Heat of sublimation of M = + 130 kJ/mol

Standard enthalpy of formation of MO(s) = -307 kJ/mol

Using Hess law of constant heat summation;

ΔHf = ΔHs + BE + ∑IE + ∑EA + U

ΔHs  = Heat of sublimation of metal

ΔHf = Heat of formation MO

BE = Bond energy of O2

∑EA  = sum of electron affinities of Oxygen

∑IE  = Sum of the ionization energies of M

U = Lattice energy of MO

Let the second ionization energy be x

Substituting values;

(-307) = 130 + 498 + (267 + x) + 603 + (-2383)

(-307) = -885 + x

-x =   -885 + 307

-x = -578

x = 578 kJ/mol

The second ionization energy of the metal is  578 kJ/mol.

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To what pH should you adjust a standard hydrogen electrode to get an electrode potential of -0.128 V ? (Assume that the partial pressure of hydrogen gas remains at 1 atm.) Express your answer using two decimal places.\

Answers

The pH of the standard hydrogen electrode that has electrode potential of -0.128 V  is 4.3.

The equation of the hydrogen electrode is;

2H^+(aq) + 2e ⇄ H2(g)

The standard electrode potential of hydrogen is 0.00 V

Using the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Now;

E°cell = 0.00 V

n = 2

Q = 1/[H^+]

-0.128 = 0.00 - 0.0592/2 log  1/[H^+]

-0.128 = 0.00 - 0.0296 log 1/[H^+]

 -0.128 =  - 0.0296 log 1/[H^+]

-0.128/  - 0.0296  =  log 1/[H^+]

1/[H^+] = Antilog (4.32)

[H^+] = 4.79 × 10^-5

Now;

pH = -log[H^+]

pH = -log (4.79 × 10^-5)

pH = 4.3

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A pan containing 40 grams of water was allowed to cool from a temperature of 91.0 °C. If the amount of heat released is 1,300 joules, what is the approximate final temperature of the water? 74 °C 78 °C 81 °C 83 °C

Answers

Answer:

d. 83 °c

Explanation:

Answer:

D. 83 c

Explanation:

took the test

Radio waves bounce off of _____________ before returning to Earth

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Answer:   ionosphere

Explanation:       First it bounces off a top layer of the atmosphere called the ionosphere, then it bounces back to the Earth (this is reflection. It then bounces up again to the ionosphere, and continues bouncing back again until it reaches the radio receiver. This is called a skywave, which works around 3 to 30 MHz.

an expression of Avogradros law

Answers

Answer:

The formula for Avagadro's law is V1/n1 = V2/n2, where V = volume and n = amount of gas (in moles).

Explanation:

Oil does not dissolve in water because

Answers

Explanation:

Oils are made up of long hydrocarbon chains which are non polar. Since water is a polar solvent oil doesn't dissolve in water well.( Polar solutes dissolve in polar solvents and non polar solutes dissolve in non polar solvents) The hydrocarbon chains are hydrophobic.

Using the Kf value of 1.2×109 calculate the concentration of Ni2+(aq) and Ni(NH3)62+ that are present at equilibrium after dissolving 1.47 gNiCl2 in 100.0 mL of NH3(aq) solution such that the equilibrium concentration of NH3 is equal to 0.20 M.
Express your answers in moles per liter to two significant figures separated by a comma.

Answers

This problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.

Firstly, we can write out the chemical equation to be considered:

[tex]Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}[/tex]

Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:

[tex][Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M[/tex]

Afterwards, we set up an equilibrium expression for this chemical reaction:

[tex]Kf=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}[/tex]

Which can be written in terms of the reaction extent, [tex]x[/tex]:

[tex]Kf=\frac{x}{(0.113-x)(0.2)^6}[/tex]

Now, for the calculation of [tex]x[/tex], we plug in Kf, and solve for it:

[tex]1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M[/tex]

Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.

Thus, the required concentrations at equilibrium are about:

[tex][Ni(NH_3)_6^{2+}]=0.113M[/tex]

[tex][Ni^{2+}]=0M[/tex]

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