What is the definition of weight

Answer:

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Explanation:

Answer:

Chlorine

Explanation:

Hello.

In this case, for the given elements, we are able to know that oxygen has 6 valence electrons, chlorine 7, nitrogen 5 and neon 8, therefore neon is not able to react as it already has 8 valence electrons. Besides, the element having the most valence electrons is chlorine and its reaction with hydrogen forms hydrogen chloride as shown below:

[tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]

Therefore, the required element is chlorine.

Regards.

Chlorine element would have the most valence electrons and also be able to react with hydrogen

The atomic number of hydrogen is 1 and its electronic configuration is [tex]1s^{1}[/tex] .

The atomic number of oxygen is 8 and its electronic configuration is [tex]1s^{2} 2s^{2} 2p^{4}[/tex]. The valence electron in oxgen is 6.

The atomic number of chlorine is 17 and its electronic configuration is [tex]1s^{2} 2s^{2} 2p^{6}3s^{2} 3p^{5}[/tex]. The valence electron in chlorine is 7.

The atomic number of neon is 10 and its electronic configuration is [tex]1s^{2} 2s^{2} 2p^{6}[/tex]. The valence electron in neon is 8.

The atomic number of nitrogen is 7 and its electronic configuration is [tex]1s^{2} 2s^{2} 2p^{3}[/tex]. The valence electron in nitrogen is 5.

Except Neon, all three atoms will react with hydrogen as the configuration of neon is a stable electronic configuration.

Chlorine has the greatest number of valence electrons after neon that is 7. So, chlorine would have the most valence electrons and also be able to react with hydrogen as follows:-

[tex]H_2+Cl_2\rightarrow2HCl[/tex]

Hence, the correct answer is chlorine.

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Answer:

number of electrons, n

Gibbs free energy change, ΔG

temperature, T

coefficients from balanced redox equation (various)

Explanation:

The standard electrode potential of a cell can be obtained from;

∆G°= -nFE°cell

Also;

E°cell= RTlnK/nF

Where;

∆G°= standard free energy of the cell

n= number of electrons transferred

F= Faraday constant

E°cell= standard cell potential

R= universal gas constant

T= temperature

K= equilibrium constant

Answer:

Acetone.

Explanation:

Hello,

In this case, we can distinguish between polar solvent and nonpolar solvent by the nature of the bonds present in the compound. Thus, since the bonds C-Cl, C-C, C-H and C-O are nonpolar, which are contained in the toluene, carbon tetrachloride, diethyl ether and hexane, they are discarded as polar.

Nevertheless, since the carbonyl group contained in the acetone is a polar because of the formed positive and negative charges, it is actually the polar solvent, acting as an exception. This is substantiated by the fact the acetone is soluble in water whereas the other substances not,

Regards.

Answer:

THE VOLUME OF THE NITROGEN GAS AT 2.5  MOLES , 1.75 ATM AND 475 K IS 55.64 L

Explanation:

Using the ideal gas equation

PV = nRT

P = 1.75 atm

n = 2.5 moles

T = 475 K

R = 0.082 L atm/mol K

V = unknown

Substituting the variables into the equation we have:

V = nRT / P

V = 2.5 * 0.082 * 475 / 1.75

V = 97.375 / 1.75

V = 55.64 L

The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L

Answer:

B. cocamide DEA

C. folic acid

D. iron

G. lauryl glucoside

Answer:

The correct answer is option E) "none of the above".

Explanation:

This question is missing the following information "both the atmospheric pressure and temperature decreased"; as well as the last option, which is the correct one "none of the above". In this hypothetical scenario, a ballon has reach a height were the atmospheric pressure and the the temperature decreased. However, the change in the atmospheric pressure and the temperature would have no effect on the balloon's volume. The internal balloon's temperature and pressure tends to stabilize with the external conditions, therefore, the volume its not affected by the changes described in the question.

Answer:

E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)

Explanation:

Which ONE of the following is an oxidation–reduction reaction?

A) PbCO₃(s) + 2 HNO₃(aq) ⇒ Pb(NO₃)₂(aq) + CO₂(g) + H₂O(l). NO. All the elements keep the same oxidation numbers.

B) Na₂O(s) + H₂O(l) ⇒ 2 NaOH(aq). NO. All the elements keep the same oxidation numbers.

C) SO₃(g) + H₂O(l) ⇒ H₂SO₄(aq). NO. All the elements keep the same oxidation numbers.

D) CO₂(g) + H₂O(l) ⇒ H₂CO₃(aq). NO. All the elements keep the same oxidation numbers.

E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g). YES. C is reduced and H is oxidized.

Answer:

D. The lone pair forces bonding atoms away from itself.

Explanation:

The lone pairs are found in the outermost shell thereby making sharing of electrons easier. Lone pairs which are found in a covalent bond creates a bond angle which makes it a determinant of the bond angle.

The lone pairs which are negatively charged repulses the bond pairs thereby creating a distortion in the shape of the molecule.

The lone pairs creates the distortion by forcing bonding atoms away from itself.

Answer: 3Ag⁺ (aq) + PO₄³⁻ (aq) → Ag₃PO₄ (s)

Explanation:

The question requires us to write down the net ionic equation for the precipitation of silver phosphate from aqueous solution:________

Note that we we told to:

(Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds.)

Answer:

From the solubility rule,

Phosphate (PO₄³⁻) and Carbonates are insoluble except for NH₄⁺ and Group1A compounds.

According to the solubility rule, the precipitation of silver phosphates and that of the net ionic equations is given thus;

3AgNO³ (aq) + K₃PO₄ (aq) → Ag₃PO₄ (s) + 3KNO₃ (aq)

while the Net ionic equation is given;

3Ag⁺ (aq) + PO₄³⁻ (aq) → Ag₃PO₄ (s)

cheers i hope this helped !!

Answer:

This question is incomplete

Explanation:

This question is incomplete as the actual values are not provided however you can solve the question yourself if you have the actual value using the formulas below.

The formula for Molarity is

Molarity = number of moles (n) ÷ volume (in dm³ or Litre)

If number of moles is not given, to calculate for number of moles

number of moles (n) = given mass of NaOH ÷ molar mass of NaOH

Please NOTE that the volume MUST be converted to dm³ (or litre) if not provided in dm³.

The unit for Molarity is mol/dm³ or mol/L or M (molar concentration).

Average molarity = (Trial 1 + Trial 2 + Trial 3)/ 3

A 50.3 g piece of aluminum (specific heat = 0.930 J/g･°C) will change its temperature from 23.0°C to 67.0°C while absorbing 2.06 × 10³ J of heat.

When a material absorbs heat, its temperature increases. We can calculate the amount of heat (Q) absorbed using the following expression.

[tex]Q = c \times m \times \Delta T[/tex]

where,

[tex]Q = c \times m \times \Delta T\\Q = \frac{0.930J}{g. \° C } \times 50.3 g \times (67.0 \° C - 23.0 \° C) = 2.06 \times 10^{3} J[/tex]

A 50.3 g piece of aluminum (specific heat = 0.930 J/g･°C) will change its temperature from 23.0°C to 67.0°C while absorbing 2.06 × 10³ J of heat.

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Answer:

0.000237mL

Explanation:

0.237 x 10^-6L = 0.000000237L = mL

0.000237mL

Explanation:

Efficiency of a Carnot heat engine depends on :

(D) The temperature difference between hot and cold reservoirs

Answer:

pKa bromocresol green is ≅4.5

Explanation:

The bromocresol green is a chemical indicator used in titrations with equivalence point at pH's between 3 and 5.

Is an indicator that, in acidic region is yellow, and in basicic region is blue. The intermediate color is green (at pH≅ 4.5).

As at the intermediate color of the indicator pKa = pH,

Answer:

Mitosis results in the formation of two haploid gametes which can then combine to form a diploid daughter cell

Explanation:

Meiosis creates different cells

Answer:

During the process of meiosis, haploid cells are formed. After fertilization, the diploid number of chromosomes is restored.

Explanation:

Answer:

Yes, the reaction is supported by spectrochemical series.

Explanation:

When copper oxidizes with concentrated nitric acid, it produces copper ions; and the nitric acid is then reduced to nitrogen dioxide, a very poisonous gas which is brown in color and with an irritating odor.

[tex]$ Cu + 4HNO_{3} \rightarrow Cu(NO_{3})_{2} + 2NO_2+2H_2O$[/tex]

The copper ions ([tex]$Cu^{2+} $[/tex]) product which is initially coordinated to form nitrate ions from nitric acid and thus first gives the solution a green color.

Now when the solution is further diluted with water, the water molecules then displaces the nitrate ions  around the copper ions and causes the solution to change its color to blue. The reaction is -

[tex]$ 3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2+ 2NO + 4H_2O $[/tex]

Yes, the results are supported by spectrochemical series.

Answer:

[tex]\rho =1g/mL[/tex]

Explanation:

Hello,

In this case, since the density is defined as the ratio between the mass and the volume as shown below:

[tex]\rho =\frac{m}{V}[/tex]

We can compute the density of water for the given 43 g that occupy the volume of 43 mL:

[tex]\rho =\frac{43g}{43mL}=1g/mL[/tex]

Regards.

Answer:

a. H⁺(aq)+ OH⁻(aq) →  H₂O(l)

Explanation:

First, we will write the molecular equation for the reaction between H₂SO₄ and KOH.

H₂SO₄(aq) + 2 KOH(aq) →  K₂SO₄(aq) + 2 H₂O(l)

The complete ionic equation includes all the ions and the molecular species.

2 H⁺(aq) + SO₄²⁻(aq) + 2 K⁺(aq) + 2 OH⁻(aq) →  2 K⁺(aq) + SO₄²⁻(aq) + 2 H₂O(l)

The net ionic equation includes only the ions that participate in the reaction and the molecular species.

H⁺(aq)+ OH⁻(aq) →  H₂O(l)

Answer:

The standard cell potential for this galvanic cell is 0.27 V.

Explanation:

The standard redox potentials, E° of the Pb and Cd are:

Pb²⁺(aq) + 2e⁻ → Pb   E° = -0.13 V

Cd²⁺(aq) + 2e⁻ → Cd   E° = -0.40 V    

The standard cell potential for this galvanic cell can be calculated as follows:

[tex] E_{cell}^{0} = E^{0}_{c} - E^{0}_{a} [/tex]     (1)

Where:

c: is for cathode    

a: is for anode  

As we can see in the standard redox potentials of Pb and Cd, the Pb is going to be reduced (cathode) and the Cd is going to be oxidated (anode).

By replacing the standard redox potentials of Pb and Cd into equation (1) we have:

[tex] E_{cell}^{0} = E^{0}_{c} - E^{0}_{a} = -0.13 V - (-0.40 V) = 0.27 V [/tex]

Therefore, the standard cell potential for this galvanic cell is 0.27 V.

I hope it helps you!                

Answer:

[tex]\Delta _rH=-136.27kJ/mol[/tex]

Explanation:

Hello,

In this case, for the given reaction, the enthalpy of reaction is computed in terms of the enthalpies of formation as:

[tex]\Delta _rH=\Delta _fH_{C_2H_6}-\Delta _fH_{C_2H_4}-\Delta _fH_{H_2}[/tex]

Whereas hydrogen, ethene and ethane enthalpies of formation are 0 kJ/mol, 52.47 kJ/mol and -83.8kJ/mol respectively. Therefore, we compute:

[tex]\Delta _rH=-83.8kJ/mol-52.47kJ/mol-0kJ/mol\\\\\Delta _rH=-136.27kJ/mol[/tex]

Best regards.

Answer:

An unknown metal between 10-30 g is transferred to a test tube, place the tube in water that is boiling for ten minutes until it reaches a thermal equilibrium.

Explanation:

Answer:

Explanation:

Soak your wooden skewers in water for at least 30 minutes before using them to cook with. That way the skewers are water logged and won't catch on fire while you're cooking your soon-to-be delicious kebabs. Remember, only you can prevent kebab fires.

Answer:

[tex](C_2H_5)_3N~+~H_2O~->~(C_2H_5)_3NH^+~+~OH^-[/tex]

Explanation:

For this question, we have to remember that definition of acid and base in the Bronsted-Lowry theory:

Acid

A substance with the ability to produce a hydronium ion ([tex]H^+[/tex]).

[tex]HA~->~H^+~+~A^-[/tex]

Base

A substance with the ability to accepts a hydronium ion ([tex]H^+[/tex]).

[tex]B~+~H^+->BH^+[/tex]

If we check the reaction mechanism (figure 1). We can see that the lone pair of electrons in the "N" atom will remove an "H" from the water molecule producing a positive charge in the nitrogen and a hydroxyl group ([tex]OH^-[/tex]).

With all this in mind, the net ionic equation would be:

[tex](C_2H_5)_3N~+~H_2O~->~(C_2H_5)_3NH^+~+~OH^-[/tex]

I hope it helps!

Answer:

Option D. 400 mmHg

Explanation:

The following data were obtained from the question:

Mole of He (nHe) = 0.04 mole

Mole of Ne (nNe) = 0.06 mole

Total pressure = 10³ mmHg

Partial pressure of He =.?

Next, we shall determine the total number of mole in the reaction vessel.

This can be obtained as follow:

Mole of He (nHe) = 0.04 mole

Mole of Ne (nNe) = 0.06 mole

Total mole =?

Total mole = nHe + nNe

Total mole = 0.04 + 0.06

Total mole = 0.1

Next, we shall determine the mole fraction of He.

This can be obtained as follow:

Mole fraction = mole of gas /total mole

Mole of He (nHe) = 0.04 mole

Total mole = 0.1

Mole fraction of He =.?

Mole fraction of He = nHe/total mole

Mole fraction of He = 0.04/0.1

Mole fraction of He = 0.4

Finally, we shall determine the partial pressure of He as follow:

Partial pressure = mole fraction x total pressure

Mole fraction of He = 0.4

Total pressure = 10³ mmHg

Partial pressure of He =.?

Partial pressure of He = 0.4 x 10³

Partial pressure of He = 400 mmHg.

Therefore, the partial pressure of He is 400 mmHg.

Answer:

B. Ni

D. Sn

Explanation:

Electrode Potential is the potential difference set up between an element and a solution of its ion. It is a measure of the tendency of an element to form ions.

The electrode potentials vary from one metal ion or metal system to another and the value depends on:

concentration of ions in the solution

the temperature at which the measurement is made , and

the overall energy change.

When two half-cells are joined together through a salt bridge, the e.m.f (electromotive force) of the cell formed is the algebraic difference between the two electrode potentials.

However, the set up in which chemical energy is converted to electrical energy is known as an Electrochemical cell. It consists of two half cells ;

an oxidation half-cell reaction  

a reduction  half cell reaction.

From the information given:

the standard reduction potential for each metal under standard conditions in the electrochemical series is as follows :

[tex]E^0 _{Pb} = -0.126 \ V[/tex]

[tex]E^0 _{Cr} = - 0.74 \ V[/tex]

[tex]E^0 _{Ni} =- 0.23 \ V[/tex]

[tex]E^0 _{Zn} =- 0.76 \ V[/tex]

[tex]E^0 _{Sn} = -0.13 \ V[/tex]

[tex]E^0 _{Cd} = - 0.40 \ V[/tex]

We will realize that  Ni and Sn have reduction values in between Pb and Cd.

Thus , Ni can be oxidized by Pb2+  solution  but not with a Cd2+ solution

The metal(s) that can be oxidized with a [tex]Pb^{2+}[/tex] solution

B. Ni

D. Sn

It is the potential contrast set up between a component and an answer of its particle. It is a proportion of the inclination of a component to shape particles. The cathode possibilities shift from one metal particle or metal framework to another and the worth relies upon:

Centralization of particles in the arrangement, the temperature at which the estimation is made , and the general energy change.

Whenever two half-cells are combined through a salt scaffold, the e.m.f (electromotive power) of the cell shaped is the mathematical contrast between the two terminal possibilities.

However, the set up in which chemical energy is converted to electrical energy is known as an Electrochemical cell. It consists of two half cells ;

an oxidation half-cell reaction  and a reduction  half cell reaction.

On seeing the values of electrode potential from electrochemical series we observe that  Ni and Sn have reduction values in between Pb and Cd.

Thus , Ni and Sn can be oxidized by  [tex]Pb^{2+}[/tex] solution but not with a [tex]Cd^{2+}[/tex]solution.

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Answer:

Option a: positron emission.

Explanation:

In the transformation we have:

⁶⁷Ga  →  ⁶⁷Zn  

The reaction is:

[tex]^{67}_{Z}X \rightarrow ^{67}_{Z -1}Y[/tex]

For Ga to become Zn, the atom nucleus has to lose a proton, so in the given options, the reaction that involves the transformation of a proton is the option a, positron emission.

In a positron emission, a proton becomes into a neutron and a positron:

[tex]^{A}_{Z}X \rightarrow ^{A}_{Z-1}Y + ^{0}_{+1}e[/tex]

Therefore, the correct answer is option a: positron emission.

I hope it helps you!

Answer:

See the answer below

Explanation:

The pH scale is important in science because it gives an indication of how acidic or basic a solution is. The scale ranges from 0 - 14 with 0 being the most acidic and 14 being the most basic while a pH of 7 is a neutral pH.

The pH scale is widely applicable in several scientific applications such as in medicine/health, agricultural processes, industrial processes, environmental monitoring, research and development, etc.

In medicine, the pH of the stomach is monitored in order to make some diagnosis. The normal pH of the human stomach ranges from 1.5 - 3.5 and a major deviation from this range can give an indication of wrong health.

In agriculture, the pH condition of the soil on which crops are grown is quite important. While some crops require slightly acidic soil, some will only do well in alkaline soil. Hence, the pH condition of the soil must be monitored to ensure the optimal yield of crops.

Several industrial processes require the monitoring of pH in order to ensure product's quality or monitor some important reactions. In food industries, for example, monitoring the pH of reactions is necessary in order to prevent contamination by pathogens or ensure a good organoleptic quality of the final product. It is also necessary to monitor the pH of industrial wastewaters in order to avoid polluting the environment.

Monitoring pH is also important for environmental monitoring, The pH of various water bodies or soil can give an indication of the level of pollution in the water or the soil.

The pH can be defined as the concentration of the hydrogen ions in the sample. The determination of pH helps in the designing of the study and investigating the reactions.

Some of the examples for the scientific application of pH has been:

For more information about the pH importance, refer to the link:

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Answer:

The correct answer is c. second order

Explanation:

In a rate law, second order means that the chemical rate is related with the reactant concentration raised to 2. For example, if the chemical equation is:

A ⇒ B

A is the reactant and it is a second order reaction, so the rate law will be:

r = [A]²