. The pI is called ________________. The rule of calculating pI of an amino acid is that first, write the dissociation equation from fully protonated form to fully deprotonated form, label the charge of each form; second, identify the zwitterionic form (zero charge) and find the closest pKs (left and right side in the dissociation equation); third, average these two pKs. Write the dissociation equations for amino acids, glutamate, histidine, and calculate their pIs.

Answer:

1= Volume

= Length x breath x height

= 13.90 x 2.9 x 0.081

=3.26511

2= Density = Mass ÷ volume

= 11.3 ÷ 3.26511

= 3.461 (3d.p)

idk the rest because you haven't shown a picture of the rest

Answer:

1. 3.3 cm³; 2. 3.5 g/cm³; 3. barium; 4. 4%

Explanation:

Experimental data:

Mass          = 11.3    g

Length      = 13.90 cm

Width        =  2.9    cm

Thickness = 0.081 cm

Calculations:

1. Volume of bar

V = lwh = 13.90 cm × 2.9 cm × 0.081 cm = 3.3 cm³

2. Experimental density

[tex]\text{Density} = \dfrac{m}{V} = \dfrac{\text{11.3 g}}{\text{3.27 cm}^{3}} = \textbf{3.5 cm}^{\mathbf{3}}[/tex]

3. Identity of metal

The three most likely metals are scandium (3.00 g/cm³), barium (3.59 g/cm³), and yttrium (4.47 g/cm³)

The metal is probably barium.

4. Percent difference

[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{ True - Calculated}\lvert}{ \text{True}} \times 100 \,\%\\\\& = & \dfrac{\lvert 3.59 - 3.5\lvert}{3.59} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.1\lvert}{3.59} \times 100 \, \%\\ \\& = & 0.04 \times 100 \, \%\\& = & \mathbf{4 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{4 \, \%} }$}[/tex]

Answer:

Amorphous solids .

Explanation:

They have no particular order or pattern.Each particle is in a particular spot, but the particles are in no organized pattern.

Answer: The solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.

Explanation:

The given data is as follows.

Volume of sample water = 46 ml

Temperature = [tex]21^oC[/tex]

After vaporization, washes and then drying the weight of mineral X = 0.87 g

This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.

          1 ml of water = [tex]\frac{0.87 g}{46.0 ml}[/tex]

                                = [tex]1.891 \times 10^{-2}[/tex] g/ml

Therefore, we can conclude that solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.

Complete Question

The diagram for this question is shown on the second uploaded image

Answer:

The organic product obtained is  shown  on the first uploaded image

Explanation:

The process that lead to this product formation is known as oxidative cleavage   which is a reaction that involves the cleavage of a carbon to carbon bond at the same time this carbon which formed the carbon bond are oxidized i.e oxygen is been added to them

Answer: Fossil fuels

Explanation:

Fossil fuels such as petroleum, oil,  and natural gas, are non-renewable energy resources which are formed from the remains of  prehistoric ancient  plants and animals beneath layers of rock of the earth surface.

By analyzing and studying fossil fuels using Radiocarbon analyses by archaeologists, earth scientists  etc, Information about the history of the earth can be obtained from the decomposition of dead organisms present in fossil fuels.

Answer: 125 g

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles[/tex]

The balanced reaction is:

[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]

According to stoichiometry :

1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]

Thus 1.30 moles of [tex]B_2H_6[/tex] will require=[tex]\frac{3}{1}\times 1.30=3.90moles[/tex]  of [tex]O_2[/tex]

Mass of [tex]O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g[/tex]

Thus 125 g of [tex]O_2[/tex] will be needed to burn 36.1 g of [tex]B_2H_6[/tex]

Answer:

10 g of CO2

Explanation:

Equation of the reaction:

CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2

Fom the above balanced equation,

1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2

Molar mass of Octane = 114 g/mol

Molar mass of oxygen gas = 32 g/mol

Molar mass of CO2 = 44 g/mol

Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.

From the given mass of reactants;

3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.

Therefore oxygen is the limiting reactant.

15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.

Mass of CO2 produced will be

(352 * 15.6)/544 = 10 g of CO2

Answer: if im not wrong the relations are that the electrochemistry can detect the cancer and any other sickness

just like it does with chemical phenomena

=)

Answer:

Mass of Al2S3 remaining is 17.212 g

Explanation:

Equation of the reaction is given below:

Al2S3 + 6H2O -----> 2Al(OH)3 + 3H2S

From the balanced equation above

6 mole of H20 reacts with 1 mole of Al2S3

i.e. 6 * 18.02 g of H2O reacts with 1 * 150.71 g of Al2S3

= 108.12 g of H2O reacts with 150.71 g of Al2S3

Therefore 2.0 g of water will react with 2.0 * (150.71/108.12) g of Al2S3

= 2.788 g of Al2S3

Mass of Al2S3 remaining = 20.0 g - 2.788 g = 17.212 g

According to the properly balanced chemical equation, the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction is 17.22 grams.

Given the following data:

To calculate the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction:

First of all, we would write a properly balanced chemical equation for this chemical reaction.

                        [tex]Al_2S_3 + 6H_2O ---> 2Al(OH)_3 + 3H_2S[/tex]

By stoichiometry:

1 mole of [tex]AL_2S_3[/tex] reacts with 6 moles of [tex]H_2O[/tex]

Next, we would calculate the mass of each compound.

For [tex]AL_2S_3[/tex]:

[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 1 \times 150.17[/tex]

Mass = 150.17 grams

For [tex]H_2O[/tex]:

[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 6 \times 18.02[/tex]

Mass = 108.12 grams

108.12 grams of [tex]H_2O[/tex] = 150.17 grams of [tex]AL_2S_3[/tex]

2.00 grams of [tex]H_2O[/tex] = X grams of [tex]AL_2S_3[/tex]

Cross-multiplying, we have:

[tex]108.12 \times X = 150.17 \times 2\\\\108.12X = 300.34\\\\X = \frac{300.34}{108.12}[/tex]

X = 2.78 grams of [tex]AL_2S_3[/tex]

Remaining mass = [tex]20.00 - 2.78[/tex]

Remaining mass = 17.22 grams of [tex]AL_2S_3[/tex]

Read more: https://brainly.com/question/13750908

Answer:

[CH₂Cl₂] = 7.07x10⁻² M

[CH₄] = 0.319 M

[CCl₄] = 0.164 M  

Explanation:

The equilibrium reaction is the following:

2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)  

The equilibrium constant of the above reaction is:

[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]

When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:

[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]

[tex]C_{CH_{4}} = 0.328 M[/tex]      

Now, the concentrations at the equilibrium are:

2CH₂Cl₂(g)   ⇄   CH₄(g)  +  CCl₄(g)

5.35x10⁻² - 2x   0.328 + x   0.173 + x    

[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]

[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]

Solving the above equation for x:  

x₁ = 0.076 and x₂ = -0.0086

Hence, the concentration of the three gases once equilibrium has been reestablished is:

[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M

[CH₄] = 0.328 + (-0.0086) = 0.319 M

[CCl₄] = 0.173 + (-0.0086) = 0.164 M  

We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.

I hope it helps you!

Answer:

Explanation:

use gas law eqation

P1 * V1  / T1 = P2 * V2 /T2

600*V1/227 = 300*800/23

V1 = 300*800*227 / 23*600 = ............ can you solve this and get the answer?

Answer:

Explanation:

Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.

In the living E. coli cells,

[ATP] = 7.9 mM;

[ADP] = 1.04 mM,

[glucose] = 2 mM,

[glucose 6-phosphate] = 1 mM.

Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

The reaction is given as

Glucose + ATP → glucose 6-phosphate + ADP

Now reaction quotient for given equation above is

[tex]q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}[/tex]

[tex]q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}[/tex]

so,

[tex]q<<K_e_q[/tex] ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable  until q = Keq

Answer:

The correct answer is 5.447 × 10⁻⁵ vacancies per atom.

Explanation:

Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,  

Fv = Nv/N------ (i)  

Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,  

N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,  

N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol

N = 5.14073 × 10²⁸ atoms/m³

Now putting the values of Nv and N in the equation (i) we get,  

Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3

Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.  

Answer:

ΔH of the reaction is -802.3kJ.

Explanation:

Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.

Using the reactions:

(1) Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ

(2) Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k J

(3) H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ

The sum of (2) - (1) produce:

CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ

And the sum of this reaction with 2×(3) produce:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =

-802.3kJ

Answer:

D: CH2=CH-CH(CH3)-CH2-CH3 (R & S enantiomers)

E: CH3-CH2-(CH3)-CH2-CH3

(Please see the figures enclosed )

Explanation:

D is a racemic mixture (R & S) of 3-metyl-pent-1-ene, so it is optically inactive (although each of two enantiomers is optically active, the mixture is optically  inactive. The reason is that two enantiomers are present in an equal amount).

E is optically inactive, so its structure has to be symmetric.

Answer:

  1140 mmHg

Explanation:

1 atmosphere is 760 mmHg, so 1.5 atmospheres is ...

  1.5×760 mmHg = 1140 mmHg

Answer:

Explanation:

From the given information ;the objective is to determine how the instructor made the 0.25L elution buffer

0.25 L elution buffer = 250 mL elution butter

The breaking buffer that we use this week contains

10mM Tris    =   0.01 M

150mM NaCl  =   0.15 M

300mM imidazole.  = 0.3 M

The stock concentration  of Tris in 1M

Therefore ; by using the formula: [tex]M_1V_1 = M_2 V_2[/tex]; we can determine the volume in the preparation; so;

[tex]1*V_1 = 0.0 1 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0.0 1 \ M * 250 \ mL}{1 }[/tex]

[tex]V_1 = 2.5 \ mL[/tex]

In NaCl, The amount of stock concentration is 5 M

so; using the same formula; we have:

[tex]5*V_1 = 0.15 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0. 15 \ M * 250 \ mL}{5 }[/tex]

[tex]V_1 = 7.5 \ mL[/tex]

From Imidazole ; the amount of stock concentration is

[tex]1*V_1 = 0.3 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0. 3 \ M * 250 \ mL}{1 }[/tex]

[tex]V_1 = 75 \ mL[/tex]

Thus; we can have a table as shown as :

Stock concentration        volume to be added        Final concentration

1 M of Tris                              2.5 mL                            10 mM

5 M of  NaCl                          7.5 mL                             150 mM

1 M of Imidazole                    75  mL                            300  mM

In conclusion. the addition of all the volume make up the 250 mL elution buffer that is equivalent to 0.25 L.

Answer: The molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]

Explanation:

Formula used for Elevation in boiling point :

[tex]\Delta T_b=k_b\times m[/tex]

or,

[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]

where,

[tex]T_b-T^o_b =(125.2-120.7)^0C=4.5^0C[/tex]

[tex]k_b[/tex] = boiling point constant  = ?

m = molality

[tex]w_2[/tex] = mass of solute (urea) = 55.4 g

[tex]w_1[/tex] = mass of solvent  X =  500 g

[tex]M_2[/tex] = molar mass of solute (urea) = 60 g/mol

Now put all the given values in the above formula, we get:

[tex]4.5^oC=k_b\times \frac{55.4g\times 1000}{60\times 500g}[/tex]

[tex]k_b=2.4^0C/m[/tex]

Thus the molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]

Answer:

FOR EVERY 10 DEGREE CELSIUS INCREASE IN TEMPERATURE, THE ACTIVATION ENERGY THAT SHOWS THIS IS 52.4 KJ/MOL

Explanation:

From Arrhenius equation, the relationship between the rate constant and the temperature is as shown below:

k = Ae^ -Ea/RT

At initial temperature T1, the initial rate constant is (k1)

At final temperature T2, the final rate constant is k2

For the reaction rate to be doubled, we must double the rate constant which shows that the ratio of k2 / k1 must be equal to 2.

That is, k2 / k1 = 2 (rate is doubled)

Equating this into the Arrhenius equation, we have:

k2 / k1 = Ae^ (-Ea / R ) (1/ T2 - 1/T1)

2 = e^ (-Ea / R) (1 / T2 = 1 / T1)

Taking the natural logarithm of both sides:

ln 2 = - (Ea / R) (1 / T2 - 1 / T1)

Making Ea the subject of the formula, we obtain:

Ea = - (ln 2 R / (1 / T2- 1 / T1))

Let T1 = 25 C = 25 + 273 K = 298 K

T2 = 35 C = 35 + 273 K = 308 K

R = 8.314

So,

Ea = - (ln 2 * 8.314 / ( 1/308 - 1 / 298))

Ea = - (0.693 * 8.314 / 0.00324 - 0.00335)

Ea = - 5.7616 / -0.00011

Ea = 52 378,18 J / mol

So therefore, the activation energy Ea is 52.4 kJ/mol.

Answer:

That its small pointed. Pink(Himalayan salt)or white(normal salt)

Explanation:

Summa dees questions are so stupid, deys makin me salty.

Answer:

For our assumed experiment; the expected  volume of Hcl acid needed to neutralize the bicarbonate is 0.13 mL

Explanation:

We are going attempt this question experimentally.

We know that benzoic acid originate from the relationship between  benzene and a carboxylic group. So basically , the functional group of a carboxylic acid (-COOH) joins with a benzene ring(C₆H₆) to form a simple aromatic carboxylic acid known as Benzoic acid. (C₇H₆O₂)

However, it is possible to isolate benzoic acid from  a bicarbonate solution in the presence of an acidified concentrated hydrochloric acid.

Let assume that ;

0.20 g of benzoic acid was reacted with 2 mL of a 20% solution of NaHCO₃, the amount of the  excess NaHCO₃ can be determined by subtracting the amount of benzoic acid from the amount of NaHCO₃.

Let first calculate the number of moles in 0.20 g of benzoic acid

we know that the standard  molar mass of benzoic acid is 122.12 g/mol

number of moles of benzoic acid = mass of benzoic acid/molar mass of benzoic acid =

number of moles of benzoic acid = 0.20/ 122.12

number of moles of benzoic acid = 0.0016 mol

number of moles of bicarbonate  solution = mass of bicarbonate solution/ molar mass of bicarbonate solution

number of moles of bicarbonate  solution =  0.2/84.00654 g/mol

number of moles of bicarbonate  solution =  0.00238 mol

∴

(0.00238 - 0.0016) mol

= 7.8 × 10⁻⁴ mol

Let assume that the concentrated HCl is 12  M

Also. HCl and NaHCO₃ react together at the ratio of 1:1; thus the  volume of Hcl acid needed to neutralize the bicarbonate is:

[tex]= ( 7.8 * 10^{-4} \ \ mol )* ( \dfrac{2\ L}{ 12 \ M})*( \dfrac{10^3 \mL}{1 \ L})[/tex]

= 0.13 mL

Thus; for our assumed experiment; the expected  volume of Hcl acid needed to neutralize the bicarbonate is 0.13 mL

Answer:

B. the molecules in liquid are loosely packed and scattered thus, they cannot be arranged

Answer:

6-metyl-2-heptyne

Explanation:

C-C-C-C-C-C-C hept

   2

C-C≡C-C-C-C-C  2-heptyne

                   C

                    | 6

C-C≡C-C-C-C-C

6-metyl-2-heptyne

The IUPAC name for the above structure is 6 methyl, hept-2-yne.

IUPAC stands for international Union of pure and applied chemistry. It is the body in charge of naming organic chemical compounds.

The naming is is based on a molecule's longest chain of carbons connected by single/double/triple bonds, whether in a continuous chain or in a ring etc.

According to this question, a structure is given. The following applies;

Learn more about IUPAC at: https://brainly.com/question/33646537

#SPJ6

Answer:

[tex]B_2O_3[/tex]

Explanation:

First, we have to find the reaction:

[tex]B_2O_3~+~C~+~Cl_2~->~BCl_3~+~CO[/tex]

The next step is to balance the reaction:

[tex]B_2O_3~+~3C~+~3Cl_2~->~2BCl_3~+~3CO[/tex]

Now, we have to calculate the molar mass for  each compound, so:

[tex]B_2O_3=~69.62~g/mol[/tex]

[tex]C=~12~g/mol[/tex]

[tex]Cl_2=~70.96~g/mol[/tex]

With these values, we can calculate the moles of each compound:

[tex]95.7~g~B_2O_3\frac{1~mol~B_2O_3}{69.62~g~B_2O_3}=1.37~mol~B_2O_3[/tex]

[tex]75.7~g~C\frac{1~mol~C}{112~g~C}=6.30~mol~C[/tex]

[tex]369~g~Cl_2\frac{1~mol~Cl_2}{70.96~g~C}=5.20~mol~Cl_2[/tex]

Now we can divide by the coefficient of each compound in the balanced equation:

[tex]\frac{1.37~mol~B_2O_3}{1}=~1.37[/tex]

[tex]\frac{6.30~mol~C}{3}=~2.1[/tex]

[tex]\frac{5.20~mol~Cl_2}{3}=~1.73[/tex]

The smallest values are for  [tex]B_2O_3[/tex], so this is our limiting reagent.

I hope it heps!

Answer:

$11.81

Explanation:

27 lb cost $16

27/16=$1.69 per pound

$1.69*7=$11.81 for 7 lbs

Answer:

The correct answer is 66.35 kilograms.

Explanation:

Based on the data given in the question, the energy consumed by the body of a human being is 50%. Based on the given data, the energy consumed in a day is 8000 kJ, 50 percent is the energy transfer efficiency. Thus, the consumption of total energy is 4000 kJ, and for the transformation of ADP to ATP, the energy involved is 30.6 kJ per mole.  

Hence, the total ATP produced in the process is,  

ATP = 4000 kJ / 30.6 kJ/mol

= 130.7189 mol.  

Thus, with the energy transfer efficiency of 50 percent, the total moles of ATP produced is 130.7 mol.  

The mass of ATP can be calculated by using the formula,  

moles = mass/molecular mass

The molecular mass of ATP is 507.18 g per mol

Now by putting the values we get,  

mass of ATP = 130.7189 mol * 507.18 g/mol

= 66298.011 g or 66.298 kg

It is mentioned that human comprise 50 g of ATP or 0.05 kg of ATP. Therefore, the sum of the available ATP will be.  

= Total production of ATP + Total ATP available

= 66.298 kg + 0.05 kg

= 66.348 kg

Hence, the sum of the ATP that is turned over by a resting human in a day is 66.35 kg.  

Answer:

The percentage yield is 50%

Answer:

The enthalpy of reaction per mole of HBr for this reaction = ΔrH = -40.62 kJ/mole.

Explanation:

2HBr(g) + Cl2(g) → 2HCl(g) + Br2(g)

When 23.9 g HBr(g) reacts with sufficient Cl2(g), 12.0 kJ of heat is evolved, calculate the value of ΔrH for the chemical reaction.

Note that ΔrH is the enthalpy per mole for the reaction.

Molar mass of HBr (g) = 80.91 g/mol.

Hence, 1 mole of HBr = 80.91 g

23.9 g of HBr led to the reaction giving off 12.0 kJ of heat

80.91 g of HBr will lead to the evolution of (80.91 × 12/23.9) = 40.62 kJ heat is given off.

Hence, 40.62 kJ of heat is given off per 80.91 g of HBr.

This directly translates to that 40.62 kJ of heat is given off per 1 mole of HBr

Hence, the heat given off per mole of HBr for this reaction is 40.62 kJ/mole.

But since the reaction liberates heat, it means the reaction is exothermic and the enthalpy change for the reaction (ΔHrxn) is negative.

Hence, ΔrH = -40.62 kJ/mole.

Hope this Helps!!!