Which type of electromagnetic radiation has the longest wavelength?

A) ultraviolet
B) infrared
C) X-ray
D) visible

Answer:

Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because the product, acetyl-CoA can enter the TCA cycle.

Oxidation of odd-number fatty acids such as undecanoic acid yields acetyl-CoA + propionyl-CoA in their last pass. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle.

The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme, which is inhibited by avidin.  Palmitate oxidation however, does not involve carboxylation.

Explanation:

Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because their oxidation product, acetyl-CoA, can enter the TCA cycle where it is oxidized to CO₂.

Undecanoic acid is an odd-number fatty acid having 11 carbon atoms. Oxidation of odd-number fatty acids such as undecanoic acid yields a five -carbon fatty acyl substrate for their last pass through β-oxidation which is oxidized and cleaved into acetyl-CoA + propionyl-CoA. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle. Since oxidation is occuring in a liver extract, CO₂ has to be externally sourced in order for the carboxylation of propionyl-CoA to proceed and thus resulting in comlete oxidation of undecanoic acid.

The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme.  The role of biotin is to activate the CO₂ before its tranfer to the propionate moiety. The addition of the protein avidin prevents the complete oxidation of undecanoic acid by  binding tightly to biotin, hence inhibiting the activation and transfer of CO₂ to propionate.

Palmitate oxidation however, does not involve carboxylation, hence addition of avidin has no effect on its oxidation.

Answer:

the atomic mass of any elemet contains avogardo numberof atoms

In case of Gallium,

69.72 gram is atomic mass and it cotnains around 6.023*10^23 atoms of Gallium

but, 2000 punds = 907184.7 grams

907184.7 gram of gallium contains= 6.023*10^23* 907184/69.72

                                                          = 79 *10^26 atoms

Explanation:

Answer:

[tex]\large \boxed{\text{122 000 J}}[/tex]

Explanation:

1. Calculate the energy needed

[tex]\text{Energy} = \text{0.800 pt} \times \dfrac{\text{36 400 cal}}{\text{1 pt}} = \text{ 29 120 cal}[/tex]

2. Convert calories to joules

[tex]\text{Energy} = \text{29 120 cal} \times \dfrac{\text{4.184 J}}{\text{1 cal}} = \textbf{122 000 J}\\\\\text{The water will have absorbed $\large \boxed{\textbf{122 000 J}}$}[/tex]

Answer:

18 moles

Explanation:

Here the combustion of one mole of glucose ----> carbon dioxide + water, releases 2870 kilojoules / moles.

_______________________________________________________

With one contraction cycle requiring 55 kilojoules,

2870 / 55 ≈ 52.18

And with the efficiency being 35 percent,

52.1818..... * 0.35 = ( About ) 18 moles

Hope that helps!

Answer:

[tex]T2=276K[/tex]

Explanation:

Given:

Initial volume of the balloon V1 = 348 mL

Initial temperature of the balloon T1 = 255C

Final volume of the balloon V2 = 322 mL

Final temperature of the balloon T2 =

To calculate T1 in kelvin

T1= 25+273=298K

Based on Charles law, which states that the volume of a given mass of a ideal gas is directly proportional to the temperature provided that the pressure is constant. It can be applied using the below formula

[tex](V1/T1)=(V2/T2)[/tex]

T2=( V2*T1)/V1

T2=(322*298)/348

[tex]T2=276K[/tex]

Hence, the temperature of the freezer is 276 K

Answer: 276 kelvins

Explanation:

Answer: their amounts vary throughout the atmosphere

Explanation:

There is very little that travels over the atmosphere

Vary=very little

Hope that helps

Answer:

63.518

Explanation:

The following data were obtained from the question:

Mass of Isotope A = 62.9 amu

Abundance of isotope A (A%) = 69.1%

Mass of isotope B = 64.9 amu

Abundance of isotope B (B%) = 30.9%

Atomic weight of the element =..?

The atomic weight of the element can be obtained as follow:

Atomic weight = [(Mass of A x A%)/100] + [(Mass of B x B%) /100]

Atomic weight = [(62.9 x 69.1)/100] + [(64.9 x 30.9)/100]

Atomic weight = 43.4639 + 20.0541

Atomic weight = 63.518

Therefore, the atomic weight of the element is 63.518.

Answer:

The net ionic equation is  [tex]ZnCO_3 _{(s)} + 4 OH^{-}_{(aq)} \to [Zn(OH)_4]^{2-} _{(aq)} + CO_3^{2-} _{(aq)}[/tex]

The equilibrium constant is  [tex]K = 4.06 *10^{4}[/tex]

Explanation:

From the question we are that

      The  [tex]K_f = 2.9 *10^{15 }[/tex]

The ionic equation is chemical represented as

    Step 1

         [tex]ZnCO_3 _{(s)}[/tex]  ⇔   [tex]Zn^{2+} _{aq} + CO_3^{2-} _{aq}[/tex]   The  solubility product constant for stage is     [tex]K_{sp} = 1.4*10^{-11}[/tex]

 Step 2

        [tex]Zn^{2+} _{(aq)} + 4 0H^{-} _{(aq)}[/tex]    ⇔  [tex][Zn(OH_4)]^{2-} _{(aq)}[/tex]  The formation constant for this step is given as [tex]K_f = 2.9 *10^{15 }[/tex]

 The net reaction is  

           [tex]ZnCO_3 _{(s)} + 4 OH^{-}_{(aq)} \to [Zn(OH)_4]^{2-} _{(aq)} + CO_3^{2-} _{(aq)}[/tex]

The equilibrium constant is mathematically evaluated as

         [tex]K = K_{sp} * K_f[/tex]

substituting values

         [tex]K = 1.4*10^{-11} * 2.9 *10^{15}[/tex]

        [tex]K = 4.06 *10^{4}[/tex]

Answer:

-226.0kJ = ΔH°f COCl₂(g)

Explanation:

Using Hess' law, it is possible to obtain the enthalpy of formation of a substance from the enthalpy change of a reaction and the other enthalpies of formation involved in the reaction.

For the reaction:

CO(g) + Cl₂(g) → COCl₂(g)

Hess's law is:

ΔHr = -115.5kJ = ΔH°f COCl₂(g) - (ΔH°f CO(g) + ΔH°f Cl₂(g))

ΔH°f CO(g) is -110.5kJ/mol

ΔH°f Cl₂(g) is 0 kJ/mol

Replacing in Hess's law:

-115.5kJ = ΔH°f COCl₂(g) - (-110.5kJ/mol + 0kJ/mol)

-115.5kJ = ΔH°f COCl₂(g) + 110.5kJ

Answer:

1) C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)

2) four molecules of CO2 will be produced and six molecules of water

3)9 molecules of water are formed when 9 molecules of oxygen are consumed.

4) 45 molecules of oxygen

Explanation:

The balanced chemical reaction equation is shown here and must guide our work. When ethanol is burned in air, it reacts as shown;

C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)

Hence, if we use 2 molecules of ethanol, the balanced reaction equation will look like this;

2C2H5OH(l)+6O2(g)⟶4CO2(g)+6H2O(l)

Hence four molecules of CO2 are formed and six molecules of water are formed

From the balanced stoichiometric equation;

3 molecules of oxygen yields 3 molecules of water

Therefore, 9 molecules of oxygen will yield 9 × 3/3 = 9 molecules of water

Therefore, 9 molecules of water are formed when 9 molecules of oxygen are consumed.

From the reaction equation;

1 molecule of ethanol reacts with 3 molecules of oxygen

Therefore 15 molecules of ethanol will react with 15 × 3/1 = 45 molecules of oxygen

Answer:

C(s) + 3 H₂(g) + 1/2 O₂(g) ⇒C₂H₅OH(l)

Explanation:

Ethanol (C₂H₅OH) is an alcohol and it is formed by carbon (C), H (hydrogen) and O (oxygen) atoms. These elements in their standard states are:

C: C(s), it is solid, could be graphite, diamond, among others.

H: H₂(g), it is a diatomic gas.

O: O₂(g), it is a diatomic gas.

So, we can write the equation for the formation of  C₂H₅OH from C(s), H₂ and O₂ as follows:

C(s) + H₂(g) + O₂(g) ⇒C₂H₅OH(l)

Finally, we have to balance the equation by adding the estequiometrical coefficients:

C(s) + 3 H₂(g) + 1/2 O₂(g) ⇒C₂H₅OH(l)

2C(s)+3[tex]H_{2} [/tex](g)+[tex]\frac{1}{2} [/tex][tex]O_{2} [/tex](g)→[tex]C_{2} [/tex][tex]H_{5} [/tex]OH(l)

Explanation:

Standard state of carbon: C(s)

Standard state of oxygen: [tex]O_{2} [/tex](g)

Standard State of Hydrogen: [tex]H_{2} [/tex](g)

Then balance the equation C2H5OH(l) to get 2C(s)+3[tex]H_{2} [/tex](g)+[tex]\frac{1}{2} [/tex][tex]O_{2} [/tex](g)→[tex]C_{2} [/tex][tex]H_{5} [/tex]OH(l).

Answer:

900g of POCl₃

Explanation:

Hello,

To solve this question, we'll require the equation of reaction.

P₄O₁₀ + 6PCl₅ → 10POCl₃

Molar mass of P₄O₁₀ = 283.886 g/mol

Molar mass of PCl₅ = 208.24 g/mol

Molar mass of POCl₃ = 153.33 g/mol

But Number of moles = mass / molar mass

Mass = molar mass × number of moles

Mass of POCl₃ = 153.33 × 10 = 1533.3g

Mass of PCl₅ = 208.24 × 6 = 1249.44g

Mass of P₄O₁₀ = 283.886 × 1 = 283.886g

From the equation of reaction,

283.886g of P₄O₁₀ + 1249.44g of PCl₅ produces 1533.33g of POCl₃

I.e 1533.33g of reactants produces 1533.33g of product (law of conservation of mass)

Therefore, (225g of P₄O₁₀ + 675g of PCl₅) = 900g will give x g of POCl₃.

1533.33g of reactants = 1533.33g of products

900g of reactants = x g of products

x = (900 × 1533.33) / 1533.33

x = 900g of POCl₃

Answer:

0.129 M

Explanation:

0.100 M HCl = 0.100 mol/L solution HCl

5.00 mL = 0.00500 L solution HCl

0.100 mol/L HCl * 0.00500 L = 0.000500 mol HCl

                             HCl ------> H+ + Cl-

                           1 mol                   1 mol

                    0.000500 mol           0.000500 mol

0.200 M NaCl = 0.200 mol/L solution NaCl

2.00 mL = 0.00200 L solution NaCl

0.200 mol/L NaCl*0.00200 L = 0.000400 mol NaCl

                              NaCl ------> Na+ + Cl-

                            1 mol                        1 mol

                     0.000400 mol               0.000400 mol

Chloride ion altogether (0.000500 mol + 0.000400 mol) =0.000900 mol

Solution altogether (0.00500 L+0.00200 L) = 0.00700L

Molarity (Cl-)= solute/solution = 0.000900 mol/0.00700L = 0.129 mol/L=

= 0.129 M

Answer:

(a) [tex]W_{isoentropic}=8.125\frac{kJ}{mol}[/tex]

(b) [tex]W_{polytropic}=7.579\frac{kJ}{mol}[/tex]

(c) [tex]W_{isothermal}=5.743\frac{kJ}{mol}[/tex]

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant [tex]k[/tex] should be computed for air as an ideal gas by:

[tex]\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\[/tex]

[tex]0.2856=1-\frac{1}{k}\\\\k=1.4[/tex]

Next, we compute the final temperature:

[tex]T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K[/tex]

Thus, the work is computed by:

[tex]W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}[/tex]

(b) In this case, since [tex]n[/tex] is given, we compute the final temperature as well:

[tex]T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K[/tex]

And the isentropic work:

[tex]W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}[/tex]

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

[tex]W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}[/tex]

Regards.

Answer:Mass of CO2 = 0.60g

Explanation:

Given the chemical rection

Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g

No of moles = mass / molar mass

molar mass Li2CO3 = Molecular mass  calculation: 6.941 x 2 + 12.0107 + 15.9994 x 3 =  

= 73.8909 g/mol

therefore Number of moles Li2CO3 = 1.0g / 73.89 g/mol

= 0.0135 moles Li2CO3

From our given Balanced equation,  shows that  

Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g

1 mole Li2CO3 produces 1 mole CO2

therefore 0.0135 mol Li2CO3 will produce  0.0135 moles of CO2

Also

No of moles = mass / molar mass

Mass = No of moles x molar mass

molar mass of CO2=12.0107 + 15.9994 x 2=44.0095 g/mol

Mass of CO2= 0.0135 X 44.0095 g/mol =0.594≈0.60g

Answer:

Total numbe of protons and neutrons in a single atom of that element

Explanation:

Hello,

I'll answer the question by filling in the blank spaces

"The atomic mass of an element is equal to the total number of proton and neutron in a particular atom of the element. The atomic mass of an element is equal to the atomic weight. Its mass number one-twelfth of the mass of carbon-12 atom a weighted mass of all naturally occurring isotopes of the elements. Its atomic mass is the average mass of all the naturally occurring isotopes of the element."

The atomic mass of an element is the total number of protons and neutrons in a single atom of that element.

The atomic mass of an element is equal to a weighted average mass of all of the naturally occurring isotopes of the element. The correct answer is option 2.

Isotopes are elements with the same number of protons (atomic number) but differing numbers of neutrons (mass number).

Most elements exist in nature as a mixture of isotopes, each with a different mass number and abundance. The atomic mass of an element is computed by adding the masses of all isotopes, multiplying by their relative abundance, and dividing by the total abundance of all isotopes.

This gives a weighted average mass that corresponds to the normal mass of an element's atom in nature.

Therefore, the correct answer is option 2. to a weighted average mass of all of the naturally occurring isotopes of the element.

Learn more about isotopes here:

https://brainly.com/question/27475737

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Here is the complete question.

Be sure to answer all parts. Three 8−L flasks, fixed with pressure gauges and small valves, each contain 4 g of gas at 276 K. Flask A contains He, flask B contains CH4, and flask C contains H2. Rank the flask contents in terms of:  the following properties. (Use the notation >, <, or =, for example B=C>A.)

(a) pressure

(b) average molecular kinetic energy

(c) diffusion rate after the valve is opened

(d) total kinetic energy of the molecules

Answer:

Explanation:

Given that:

Three flask A,B, C:

contains a volume of 8-L

mass m = 4g    &;

Temperature = 276 K

Flask A = He

Flask B = H₂

Flask C = CH₄

a) From the ideal gas equation:

PV = nRT

where;

n = number of moles = mass (m)/molar mass (mm)

Then:

PV = m/mm RT

If  T ,m and V are constant for the three flasks ; then

P ∝ 1/mm

As such ; the smaller the molar mass the larger the pressure.

Now; since the molecular weight of CH₄ is greater than He and H₂ and also between He and H₂,  He has an higher molecular weight .

Then the order of pressure in the flask is :

[tex]\mathbf{P_B >P_A>P_C}[/tex]

where :

[tex]P_A[/tex] = pressure in the flask A

[tex]P_B[/tex] = pressure in the flask B

[tex]P_C[/tex]= Pressure in the flask C

b)

average molecular kinetic energy

We all know that  the average molecular kinetic energy  varies directly proportional to the temperature.

Thus; the given temperature = 276 K

∴ The order of the average molecular kinetic energy is [tex]\mathbf{K.E_A =K.E_B =K.E_C}[/tex]

c)

The rate of diffusion of gas is inversely proportional to the square root of it density . Here the density is given in relation to their molar mass.

So;

rate of diffusion ∝ [tex]\dfrac{1}{\sqrt{mm} }[/tex]

where;

[tex]D_A[/tex] = rate of diffusion in flask A

[tex]D_B[/tex] = rate of diffusion in flask B

[tex]D_C[/tex] = rate of diffusion in flask C

Thus; the order of the rate of diffusion = [tex]D_B[/tex]  > [tex]D_A[/tex] > [tex]D_C[/tex]

d)  total kinetic energy of the molecules .

The kinetic energy deals with how the speed of particles of a  substance determines how fast the substances will diffuse in a given set of condition.

The the order of the total kinetic energy depends on the molecular speed

Thus; the order of the total kinetic energy  for the three flask is as follows:

[tex]\mathbf{ K.E_B>K.E_A>K.E_C}[/tex]

Complete question:

On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and pressure.

Answer:

The  number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.

Explanation:

Given;

density of dry air, ρ = 1.1970 kg/m³

temperature of the air, T = 35.5°C  = 273 + 35.5 = 308.5 K

air volume, V = 1 m³

Apply ideal gas law for dry to calculate the air pressure;

[tex]P = \rho R_dT[/tex]

where;

P is the air pressure

ρ is the air density

Rd is gas constant for dry air = 287 J/kg/K

P = 1.197 x 287 x 308.5 = 105,981.78 Pa

(a) Now, determine the number of moles contained by an ideal gas at this temperature and pressure, by applying ideal gas law;

PV = nRT

where;

P is the pressure of the gas (Pa)

V is the volume of the gas (m³)

n is number of gas moles

R is gas constant = 8.314 m³.Pa / mol.K

T is temperature (K)

n = (PV) / (RT)

n = (105,981.78 x 1) / (8.314 x 308.5)

n = 41.32 moles

Therefore, the  number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.

The number of moles of an ideal gas at this temperature and pressure is 41.5 moles.

Given that;

Density of dry air = 1.1970 kg/m3

Pressure of dry air = ?

Temperature of dry air = 35.5°C + 273 = 308.5 K

Hence;

P = Density × gas constant of dry air × Temperature

P = 1.1970 kg/m3 × 287.1 J/Kg/K × 308.5 K

P = 106019 Pa or 1.05 atm

Using the ideal gas equation;

PV = nRT

n = PV/RT

n = 1.05 atm × 1000 L/0.082 atmL/K.mol × 308.5 K

n = 41.5 moles

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Answer:

Answers are in the explanation.

Explanation:

A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa. Having this in mind:

0.31 M ammonium bromide + 0.39 M ammonia . Is a good buffer system because ammonia is a weak base and its conjugate base, ammonium ion is in the solution.

0.31 M nitrous acid + 0.25 M potassium nitrite . Is a good buffer system because nitrous acid is the weak acid and nitrite ion its conjugate base.

0.21 M perchloric acid + 0.21 M potassium perchlorate . Perchloric acid is a strong acid. Thus, Is not a good buffer system.

0.16 M potassium cyanide + 0.21 M hydrocyanic acid . Hydrocyanic acid is a weak acid and cyanide ion is its conjugate base. Is a good buffer system.

0.14 M hypochlorous acid + 0.21 M sodium hypochlorite . Hypochlorous acid is a weak acid and hypochlorite ion its conjugate base. Is a good buffer system.

0.13 M nitrous acid + 0.12 M potassium nitrite . Is a good buffer system as I explained yet.

0.15 M potassium hydroxide + 0.22 M potassium bromide . Potassium hydroxide is a strong base. Is not a good buffer system.

0.23 M hydrobromic acid + 0.20 M potassium bromide . HBr is a strong acid. Is not a good buffer system.

0.34 M calcium iodide + 0.29 M potassium iodide . CaI and KI are both salts, Is not a good buffer system.

0.33 M ammonia + 0.30 M sodium hydroxide . Ammonia is a weak base but its conjugate base ammonium ion is not in solution. Is not a good buffer system.

0.20 M nitrous acid + 0.18 M potassium nitrite . Is a good buffer system.

0.30 M ammonia + 0.34 M ammonium bromide . Ammonia and ammonium in solution, Good buffer system.

0.29 M hydrobromic acid + 0.22 M sodium bromide . HBr is a strong acid, is not a good buffer system.

0.17 M calcium hydroxide + 0.28 M calcium bromide . CaOH is a strong base, is not a good buffer system.

0.34 M potassium iodide + 0.27 M potassium bromide. KI and KBr are both salts, is not a good buffer system.

Answer:

A

Explanation:

Answer:

The answer is in the explanation

Explanation:

Acetic acid, CH₃COOH, is a weak acid that will produce a buffer when its conjugate base, CH₃COO⁻, acetate ion, is added to the solution.

That is because a buffer is the mixture of a weak acid and its conjugate base or vice versa.

When an acid (HX) is added to the solution, the acetate ion will react producing acetic acid, thus:

CH₃COO⁻ + HX → CH₃COOH + X⁻

For this reason, the pH doesn't change abruptly because H⁺ ions are not produced.

Now, if a  base (BOH) is added to the buffer, CH₃COOH will react producing acetate ion and water, thus:

CH₃COOH + BOH → CH₃COO⁻ + H₂O + B⁺.

In the same way, there are not produced free OH⁻ and the pH doesn't change significantly.

Answer:

Explanation:

Kindly note that I have attached the complete question as an attachment.

Here, we are told that elimination occurs by an E2 mechanism. What this means is that the hydrogen and the halogen must be above and below for the reaction to proceed.

The possible products are as follows;

Please check attachment for complete equations and diagrams of compounds too.

Answer:

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base or vice versa.

For the acetic acid buffer, CH₃CO₂H is the weak acid and its conjugate base is the ion without H⁺, that is CH₃CO₂⁻. The equilibrium equation in water knowing this is:

In the equilibrium, the acid is dissociated in the conjugate base and the hydronium ion.

The acetic acid/acetate buffer system is a common buffer used in the laboratory, the equilibrium equation for the acetic acid/acetate buffer system. The formula of acetic acid is CH3CO2H -

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

An acid buffer is a solution that contains roughly the same concentrations of a weak acid and its conjugate base.

The equation is:

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

where CH₃CO₂H - acetic acid

and, CH₃CO₂⁻ acetate ion

Thus, CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺ is the equilibrium equation for the acetic acid/acetate buffer system.

Learn more:

https://brainly.com/question/3435382

Answer:

Depending on the anions and cations present within a hydrolysis reaction, the solution can be more... ... This lesson will explain how this occurs. ... that could react with water and create products that affect the characteristics of the solution.

Answer:

Salts of weak bases and strong acids do hydrolyze, which gives it a pH less than 7. This is due to the fact that the anion will become a spectator ion and fail to attract the H+, while the cation from the weak base will donate a proton to the water forming a hydronium ion.

Explanation:

I hope this is the answer your looking for

Answer:

See the explanation

Explanation:

In this case, we have to keep in mind that in the monosubstituted product we only have to replace 1 hydrogen with another group. In this case, we are going to use the methyl group [tex]CH_3[/tex].

In the axial position, we have a more steric hindrance because we have two hydrogens near to the [tex]CH_3[/tex] group. If we have more steric hindrance the molecule would be more unstable. In the equatorial positions, we don't any interactions because the [tex]CH_3[/tex] group is pointing out. If we don't have any steric hindrance the molecule will be more stable, that's why the molecule will the equatorial position.

See figure 1

I hope it helps!

Answer:

The answer is A

Explanation:

Answer:

997.65cmH2O

Explanation:

Barometric pressure = 739 mmHg

density of Hg = 13.5 g/ml

density of water (H2O) = 1.00 g/ml

Calculate Barometric pressure in centimetres of water ( cmH20)

equate the barometric pressure of Hg and water

739 * 13.5 * 9.8 = x * 1 * 9.81

x ( barometric pressure of water in mmH2O ) = 739 *13.5 / 1 = 9976.5mmH2O

in cmH2O = 997.65cmH2O

Answer:

When copper is heated, it decomposes to form copper oxide and carbon dioxide. It is an endothermic reaction, which means that it absorbs heat. When heated, copper is easily bent or molded into shapes.

Explanation:

Answer:

Evolution of gas.

Formation of a precipitate.

Change in color.

Explanation:

Answer:

How are bees able to fly?

Explanation: