How much heat will be absorbed by a 50.3 g piece of aluminum (specific heat = 0.930 J/g･°C) as it changes temperature from 23.0°C to 67.0°C?

Answer:

0.000237mL

Explanation:

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

0.000237mL

Answer:

The correct answer is  1.977 % m/v ≅ 2% m/v

Explanation:

We have:

0.617 M = 0.617 moles methanol/ 1 L solution

We need:

%m/v= grams of methanol/100 mL solution

So, first we convert the moles of methanol to grams by using the MM (32.04 g/mol). Then, we multiply by 0,1 to convert the volume in liters to 100 mL by using the ratio: 100 mL= 0.1 L:

0.617 mol / 1 L x 32.04 g/mol 0.1 L/100 mL= 1.977 g/100 mL= %m/v

The concentration in [tex]\%_{m/v}[/tex] of a 0.617 M aqueous solution of methanol is 1.98%.

To find the [tex]%_{m/v}[/tex] concentration of methanol we need to use the following equation:

[tex] \%_{m/v} = \frac{m_{s}}{V_{sol}} \times 100 [/tex]   (1)

Where:

[tex] m_{s}[/tex]: is the mass of methanol in grams

[tex] V_{sol} [/tex]: is the volume of the solution in milliliters

The molar concentration of methanol is:

[tex] C = 0.617 M = 0.617 \:\frac{mol}{L} [/tex]

From this concentration, we can find the mass of methanol

[tex] m = n*MM [/tex]   (2)

Where:

n: is the number of moles = C*V

MM: is the molar mass = 32.04 g/mol

Then, the mass of methanol is (eq 2):

[tex] m = n*MM = C*V*MM = 0.617 mol/L*1 L*32.04 g/mol = 19.77 g [/tex]

Knowing that 1 L = 1000 mL, the [tex]%_{m/v}[/tex] concentration is (eq 1):

[tex] \%_{m/v} = \frac{m_{s}}{V_{sol}} \times 100 = \frac{19.77 g}{1000 mL} \times 100 = 1.98 \% [/tex]

Therefore, the concentration in [tex]\%_{m/v}[/tex] is 1.98%.

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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:

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:

OLA  IQESOXPJMXXXXOX{OADUOOAOOOOOSCDCSXWAXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

Explanation:

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:

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:

[tex]\%m/m=0.15\%[/tex]

Explanation:

Hello,

In this case, we are asked to compute the by mass percent representing the toxicity of ethylene glycol in the body mass. In such a way, since the by mass percent is computed as follows:

[tex]\%m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%[/tex]

Whereas the solute is the ethylene glycol in the body mass, we obtain:

[tex]\%m/m=\frac{1.5g}{1.5g+1000g} *100\%\\\\\%m/m=0.15\%[/tex]

Best regards.

The percentage of ethylene glycol that is fatal would be 0.15%.

The toxic quantity is 1.5 g of ethylene glycol for 1000 g of body mass.

The percentage toxic quantity in relation to the body mass can be calculated as:

Percentage = mass of toxic quantity/body mass x 100%

                    = 1.5/1000 x 100%

                       = 0.15 %

Thus, the fatal level of ethylene glycol is 0.15% of body mass.

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Answer:

142.36 kg

Explanation:

volume of water in the lake = surface area x depth

= 18.5 x ( 1760 x 3 )² x 39 ft³

= 2.011 x 10¹⁰ ft³

= 2.011 x 10¹⁰ x 28.3168 liter .

= 56.945 x 10¹⁰ liter .

concentration of mercury = .25 x 10⁻⁶ g / liter

= 25 x 10⁻⁸ g / liter

= 25 x 10⁻¹¹ kg / liter

mass of mercury in the water of lake

= 25 x 10⁻¹¹ x 56.945 x 10¹⁰ kg

= 142.36 kg .

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:

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. 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:

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:

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:

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]\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:

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:

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:

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Answer:

The reaction is exothermic

Explanation:

If we look at this reaction system closely, we will discover that the energy of reactants is greater than that of the product. The excess energy possessed by the reactants must be given off at the end of the reaction.

Since ∆Hrxn = ∆Hproducts - ∆Hreactants and ∆Hreactants > ∆Hproducts hence ∆Hrxn is negative and the reaction is exothermic, hence the answer.

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:

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.

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Explanation:

Efficiency of a Carnot heat engine depends on :

(D) The temperature difference between hot and cold reservoirs

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:

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; If a chemical has a pH of 3, how could you change its pH value to be more basic? Adding water to a chemical will dilute the acid, thus lowering the pH value to more basic.

If a chemical has a pH of 3, that means it is strong acid.To alter its pH value to be more basic, we have add strong base in excess.

First, neutralization reaction occur. After adding excess strong base, the solution becomes basic and pH become more basic.

The reaction between strong acid and base to form salt and water is called neutralization reaction.

Example: HCl + NaOH → NaCl + H2O

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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:

H2C=CH-CH2OH

Explanation:

Aldehydes are reduced to primary alcohols.

The first step in the reaction sequence is that the nucleophilic Hydrogen atom in the hydride reagent attacks the electrophilic Carbon in the polar carbonyl group of the aldehyde, there will now be a shift of electrons from the carbonyl to the Oxygen atom creating an intermediate metal alkoxide complex.

The reaction sequence is completed by a simple addition of a proton to the alkoxide oxygen which creates the primary alcohol product from the intermediate complex.

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:

B. cocamide DEA

C. folic acid

D. iron

G. lauryl glucoside