Below is the structure of what should be the peptide Thr-Glu-Arg-Met. However, one of these residues has been drawn incorrectly. Identify this residue
Thr

Glu

Arg

Met

Below Is The Structure Of What Should Be The Peptide Thr-Glu-Arg-Met. However, One Of These Residues

Answers

Answer 1

Answer:

the answer is glu and I know

Answer 2

The residue that has been drawn incorrectly is option (B) i.e. glu .

What is glutamate residue?

Glutathione consists of a glutamate residue linked to cysteine via its γ-carboxyl rather than the α-carboxyl group and followed by a conventional peptide bond between cysteine and glycine.

It plays an important role in maintaining the proper ratio of oxidized to reduced forms of metabolically important thiols such as coenzyme A.

It also provides reducing equivalents that detoxify reactive oxygen species such as peroxides.

The tripeptide is  regenerated through the concerted action enzymes in the so-called γ-glutamyl cycle.

The residue that has been drawn incorrectly is glutamate residue ,hence option (B) is correct.

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Related Questions


Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?
A. HBr, Na2S, Mg(OH)2, Na2CO3.
B. H2SO4, NaOH, NaCl, HF.
C. HNO3, H2SO4, KOH, K2SiO3.
D. Ca(OH)2, KOH, CH3COOH, NaCl.

Answers

Answer:

Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?

A. HBr, Na2S, Mg(OH)2, Na2CO3.

B. H2SO4, NaOH, NaCl, HF.

C. HNO3, H2SO4, KOH, K2SiO3.

D. Ca(OH)2, KOH, CH3COOH, NaCl.

Where do reactions in a solid generally take place?

A. At the center of the solid.
B. All throughout the solid
C. Only on opposite sides of the solid due to repelling forces.
D. On the surface of the solid.

Answers

Answer:

It's D.  On the surface of the solid.

Explanation:

If the reactant is a solid, the surface area of the solid will impact how fast the reaction goes. This is because the two types of molecule can only bump into each other at the liquid solid interface, i.e. on the surface of the solid. So the larger the surface area of the solid, the faster the reaction will be.

Arrange the following compounds in order of increasing reactivity (least reactive first.) to electrophilic aromatic substitution:.

Bromobenzene Nitrobenzene Benzene Phenol

a. Bromobenzene < Nitrobenzene < Benzene < Phenol
b. Nitrobenzene < Bromobenzene < Benzene < Phenol
c. Phenol < Benzene < Bromobenzene < Nitrobenzene
d. Nitrobenzene < Benzene < Bromobenzene < Phenol

Answers

Answer:

Nitrobenzene < Bromobenzene < Benzene < Phenol

Explanation:

Aromatic compounds undergo electrophilic aromatic substitution reaction in the presence of relevant electrophiles. Certain substituents tend to increase or decrease the tendency of an aromatic compound towards electrophilic aromatic substitution reaction.

Substituents that increase the electron density around the ring such as in phenol tends to make the ring more reactive towards electrophilic substitution. Halogens such as bromine has a -I inductive effect as well as a +M mesomeric effect.

However the -I(electron withdrawing effect) of the halogens supersedes the +M electron donation due to mesomeric effect.

Putting all these together, the order of increasing reactivity of the compounds towards electrophilic aromatic substitution is;

Nitrobenzene < Bromobenzene < Benzene < Phenol

Oxygen is composed of three isotopes: oxygen-16, oxygen-17 and oxygen-18 and has an average atomic mass of 15.9982 amu. Oxygen-17 has a mass of 16.988 amu and makes up 0.032% of oxygen. Oxygen-16 has a mass of 15.972 amu and oxygen-18 has a mass of 17.970 amu. What is the percent abundance of oxygen-18?

Answers

Answer:

The percent abundance of oxygen-18 is 1.9066%.

Explanation:

The average atomic mass of oxygen is given by:

[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]

Where:

m: is the atomic mass

%: is the percent abundance

Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:  

[tex] 1 = \%_{16} + \%_{17} + \%_{18} [/tex]

[tex] 1 = x + 3.2 \cdot 10^{-4} + \%_{18} [/tex]

[tex] \%_{18} = 1 - x - 3.2 \cdot 10^{-4} [/tex]

Hence, the percent abundance of O-18 is:  

[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]  

[tex]15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)[/tex]

[tex] x = 0.980614 \times 100 = 98.0614 \% [/tex]                                                              

Hence, the percent abundance of oxygen-18 is:

[tex]\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%[/tex]                      

Therefore, the percent abundance of oxygen-18 is 1.9066%.

I hope it helps you!                                                      

385 x 42.13 x 0.079 is (consider significant figures):

Answers

385 x 42.13 x 0.079 = 1281.38395

What Volume of silver metal will weigh exactly 2500.0g. The density of silver

Answers

Answer:

cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3

What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.
1.8 M KCl

Answers

Answer:

Solution given:

1 mole of KCl[tex]\rightarrow [/tex]22.4l

1 mole of KCl[tex]\rightarrow [/tex]74.55g

we have

0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g

74.55g of KCl[tex]\rightarrow [/tex]22.4l

10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre

volume of each solution contains 0.14 mol of KCl contain 3.11litre.

[tex]\:[/tex]

1 mole of KCl → 22.4l

1 mole of KCl → 74.55g

we have

0.14 mole of KCl → 74.55*0.14=10.347g

74.55g of KCl  → 22.4l

10.347 g of KCl → 22.4/74.55*10.347=3.11litre

volume of each solution contains 0.14 mol of KCl contain 3.11litre.

convert 100kcals to kilojoules​

Answers

Answer:

Explanation:

418.4kj is the correct answer

An ice cube, measured at 260 Kelvin, is dropped into a cup of tea that is 350 Kelvin. The temperature of the tea is recorded every 30 seconds and shows the temperature dropping for 4 minutes. After 4 minutes the temperature stays steady at 300 Kelvin. What is this called?
A. Thermal equilibrium
B. Specific heat capacity
C. Latent heat
D. Temperature transfer

Answers

Answer:

Specific Heat Capacity

It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of by-product formed. What is the by-product

Answers

Answer:

Biphenyl

Explanation:

The reaction of bromo benzene with magnesium-ether solution yields a Grignard reagent.

The byproduct of this reaction is biphenyl. It is formed when two unreacted bromobenzene molecules are coupled together.

Hence, It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of biphenyl by-product formed.

A substance is tested and has a pH of 7.0. How would you classify it?

Answers

You can classify it as neutral.

Determine what product will be produced at the negative electrode for the following reaction:
2KCl(aq) + 2H20(1) -> H2(g) + Cl2(g) + 2KOH(aq)
A. H2
B. Cl2
с. КОН
D. K

Answers

Answer:

Choice A. [tex]\rm H_{2}[/tex] would be produced at the negative electrode.

Explanation:

Ionic equation for this reaction:

[tex]2\, {\rm K^{+}} + 2\, {\rm Cl^{-}} + {2\, \rm H_{2} O} \to {\rm H_{2}} + {\rm Cl_{2}} + 2\, {\rm K^{+}} + {\rm 2\, OH^{-}}[/tex].

Net ionic equation:

[tex]2\, {\rm Cl^{-}} + 2\, \rm H_{2} O} \to {\rm H_{2}} + {\rm Cl_{2}} + 2\, {\rm OH^{-}}[/tex].

Half-equations:

[tex]2\, {\rm Cl^{-}} \to {\rm Cl_{2}} + 2\, {e^{-}}[/tex].

(Electrons travel from the solution to an electrode.)

[tex]2\, {\rm \overset{+1}{H}_{2} O} + 2\, {e^{-}} \to \overset{0}{\rm H}_{2} + 2\, {\rm O\overset{+1}{H}\!^{-}}[/tex].

(An electrode supply electrons to the solution to reduce some of the [tex]\rm H[/tex] atoms from [tex]\rm H_{2}O[/tex].)

In a DC circuit, electrons always enter the circuit from the negative terminal of the power supply and return to the power supply at the positive terminal.

The negative electrode is connected to the negative terminal of the power supply. Electrons from the power supply would flow into the solution through this electrode.  

This continuous supply of electrons at the negative electrode would drive a reduction half-reaction. In this question, that corresponds to the reduction of water: [tex]2\, {\rm \overset{+1}{H}_{2} O} + 2\, {\rm e^{-}} \to \overset{0}{\rm H}_{2} + 2\, {\rm O\overset{+1}{H}\!^{-}}[/tex]. Hence, [tex]\rm H_{2}[/tex] would be produced at the negative electrode.

Convert the concentration of 0.700 M Na2SO4 to g/mol

Answers

To convert from mass concentration to molar concentration we use the formula;

Mass concentration = molar concentration * molar mass

Molar concentration of Na2SO4 = 0.700 M

Molar mass of Na2SO4  = 2(23) + 32 + 4(16) = 142 gmol-1

Hence;

Mass concentration = 0.700 M * 142 gmol-1

Mass concentration = 99.4 g/mol

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What is the basic unit of chemistry?
O A. The bond
O B. The atom
O C. The sun
O D. The cell

Answers

Answer:

B. The atom

Explanation:

Cells are the most basic unit of structure and the smallest unit of matter is the atom.

Answer:

B. The atom

Explanation:

yeee it was right

A Grignard reagent is prepared by reacting trans-1-bromo-1-butene with magnesium. What are the products of the reaction when this reagent is reacted with: a. Ethanol

Answers

Solution :

A Grignard compound or a Grignard reagent is defined as a chemical compound having a generic formula of  R−Mg−X.

Here, X = halogen

          R = organic group

The Grignard reagents are obtained by treating the organic halide  with a magnesium metal.

In the context, when  trans-1-bromo-1-butene is reacted with magnesium, a Grignard reagent is produced.

When this Grignard reagent is reacted with an ethanol, the following product is obtained in the attachment :

Write a balanced chemical equation for the reaction that occurs
when:
(a) titanium metal reacts with O21g2;
(b) silver(I) oxide decomposes into silver metal and oxygen gas when heated;
(c) propanol, C3H7OH1l2 burns in air;
(d) methyl tert-butyl ether, C5H12O1l2, burns in air.

Answers

Answer:

Explanation:

A balanced chemical equation refers to the reaction taking place whereby the number of atoms associated in the reactants side is equivalent to the number of atoms on the products side.

From the given information, the balanced equations are as follows:

[tex]\mathbf{(a) \ \ \ Ti(s) + O_{2(g)} \to TiO_{2(s)}}[/tex]

[tex]\mathbf{(b) \ \ \ 2Ag_{2}O \to 4Ag_{(s)} + O_{2(g)}}[/tex]

[tex]\mathbf{(c) \ \ \ 2C_3H_7OH + 9O_2 \to 6CO_2+8H_2O}[/tex]

[tex]\mathbf{(d) \ \ \ 2C_5 H_{12}O \to 10 CO_2 + 12 H_2O}[/tex]

11. An isotope Q has 18 neutrons a mass number of 34. (a) (i) What is an isotope? An isotope is one of two or C (b) Write its electron arrangement. Mass number=34 Number of neutrons=18 Number of Protons = 34-15-16 (c) To which period and group does Q belong? Protors - Electons - Atomic number Period - Group (d) How does Q form its ion?

Answers

An isotope is an element with the same atomic number but different mass number due to differences in number of neutrons.

electron configuration is 2,8,6.

Belongs to group 6 and period group 3.

It forms an ion by accepting 2 electrons

Four atoms and/or ions are sketched below in accordance with their relative atomic and/or ionic radii. Which of the following sets of species are compatible with the sketch?
Explain. (a) C,Ca2+,Cl−,Br−;
(b) Sr4, Cl,Br−,Na+

(d) Al,Ra2+,Zr2+

(c) Y,K,Ca,Na+, Mg2+;

e) Fe,Rb,Co,Cs


Answers

Answer:

Hence the correct option is an option (b) Sr4, Cl,Br−,Na+.

Explanation:    

Bromine and chlorine belong to an equivalent group. As we go down the group the dimensions increases which too there's a charge on the bromine atom. therefore the size of the Br- is going to be larger in comparison to the chlorine atom.

Sr atom is within the second group, and also it's below the above-mentioned atoms.so Sr is going to be the larger one among all the atoms.

Sodium and chlorine belong to an equivalent period .size decrease from left to right. but due to the charge on sodium its size decreases and there's an opportunity that Na+ size could be adequate for Cl.      

Here we finally assume that two atoms are of an equivalent size (Na+ and Cl) which are less in size compared to the opposite two(Sr and Br-) during which one is greater (Sr)and the opposite is smaller(Br-).

You have been contracted to determine how different salts affect the pH of water. Which of the solids in the following set should you test to investigate for the effects of cations on pH?

a. AlBr3
b. Rb2SO3
c. MgCl2
d. RbBrO
e. CH3NH3Br

Answers

Answer:

Hence the solids that should test to investigate the effects of cations on pH is

[tex]AlBr_{3}[/tex] (Cation is Al 3+)  

[tex]MgCl_{2}[/tex]  ( Cation is Mg 2+)  

[tex]CH_{3} NH_{3} Br[/tex] ( Cation is NH2+).

Explanation:

The solids in the following should you test to investigate the effects of cations on pH.  

[tex]AlBr_{3}[/tex] contains (Cation is Al 3+)  

[tex]MgCl_{2}[/tex] contains ( Cation is Mg 2+)  

[tex]CH_{3} NH_{3} Br[/tex] contains( Cation is NH2+ )

The atoms or the molecules containing the positive charge that gets attracted to the cathode are called cations. The compounds a. [tex]\rm AlBr_{3}[/tex], c. [tex]\rm MgCl_{2}[/tex] and e. [tex]\rm CH_{3}NH_{3}Br[/tex] should be investigated.

What are cations and pH?

Cations are the positive charge containing molecules and atoms that have more protons in their nucleus than the number of electrons in their shells. They are formed when they lose one or more electrons to another atom.

The addition or release of the electrons of the cations and anions affects the pH system as absorption of the cation decreases the pH and absorption of the anions increases the pH.

Hence, [tex]\rm Al^{3+}[/tex], [tex]\rm Mg^{2+}[/tex] and [tex]\rm NH^{2+}[/tex] are the cation that should be investigated. The addition of the cations will reduce the pH of the reaction.

Therefore, absorption of the cation reduces the pH.

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What must happen to uranium before it can be used as a fuel source?

Answers

Answer: Uranium enrichment. Uranium is used to fuel nuclear reactors; however, uranium must be enriched before it can be used as fuel. Enriching uranium increases the amount of uranium-235 (U235) that can sustain the nuclear reaction needed to release energy and produce electricity at a nuclear power plant.

A hot pot of water is set on the counter to cool. After a few minutes it has lost 495 J of heat energy. How much heat energy has the surrounding air gained?

_____unit_____

Answers

Answer:

495 J

Explanation:

When the hot pot was set on the counter to cool, heat energy was lost from the pot. Note that according to the first law of thermodynamics, heat is neither created nor destroyed.

This implies that, the heat energy lost from the pot must be gained by the surrounding air. Therefore, if 495 J of energy is lost from the pot, then 495 J of energy is gained by the surrounding air.

Conversion Problem (show all work):
1. A patient required 3.0 pints of blood during surgery. How many liters does this correspond
to? Show all work. Use conversion factors available in the text or the exam packet. (4)

Answers

1.42liters, which is equivalent to 3pints, of blood is required for the surgery

Pints is a unit of measurement for volume in the United States. However, it can be converted to litres using the following equation:

1 US pint = 0.473 liters

Hence, according to this question which states that a patient required 3.0 pints of blood during surgery. This means that the patient required:

3 × 0.473

= 1.419 liters of blood for the surgery

1.42liters, which is equivalent to 3pints, of blood is required for the surgery

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A solution is made by dissolving 5.84 grams of NaCl in enough distilled water to give a final volume of 1.00 L. What is the molarity of the solution
Group of answer choices

0.0250 M

0.400 M

0.100 M

1.00 M

Answers

Answer:

Explanation:

1. A solution is made by dissolving 5.84g of NaCl is enough distilled water to a give a final volume of 1.00L. What is the molarity of the solution? a. 0.100 M b. 1.00 M c. 0.0250 M d. 0.400 M 2. A 0.9% NaCl (w/w) solution in water is a. is made by mixing 0.9 moles of NaCl in a 100 moles of water b. made and has the same final volume as 0.9% solution in ethyl alcohol c. a solution that boils at or above 100°C d. All the above (don't choose this one) 3. In an exergonic process, the system a. gains energy b. loses energy c. either gains or loses energy d. no energy change at all

Answer:

[tex]\boxed {\boxed {\sf 0.100 \ M }}[/tex]

Explanation:

Molarity is a measure of concentration in moles per liter.

[tex]molarity = \frac{moles \ of \ solute}{liters \ of \ solution}}[/tex]

The solution has 5.84 grams of sodium chloride or NaCl and a volume of 1.00 liters.

1. Moles of Solute

We are given the mass of solute in grams, so we must convert to moles. This requires the molar mass, or the mass of 1 mole of a substance. These values are found on the Periodic Table as the atomic masses, but the units are grams per mole, not atomic mass units.

We have the compound sodium chloride, so look up the molar masses of the individual elements: sodium and chlorine.

Na: 22.9897693 g/mol Cl: 35.45 g/mol

The chemical formula (NaCl) contains no subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the 2 molar masses to find the compound's molar mass.

NaCl: 22.9897693 + 35.45 = 58.4397693 g/mol

There are 58.4397693 grams of sodium chloride in 1 mole. We will use dimensional analysis and create a ratio using this information.

[tex]\frac {58.4397693 \ g\ \ NaCl} {1 \ mol \ NaCl}[/tex]

We are converting 5.84 grams to moles, so we multiply by that value.

[tex]5.84 \ g \ NaCl *\frac {58.4397693 \ g\ NaCl} {1 \ mol \ NaCl}[/tex]

Flip the ratio. It remains equivalent and the units of grams of sodium chloride cancel.

[tex]5.84 \ g \ NaCl *\frac {1 \ mol \ NaCl}{58.4397693 \ g\ NaCl}[/tex]

[tex]5.84 *\frac {1 \ mol \ NaCl}{58.4397693 }[/tex]

[tex]0.09993194823 \ mol \ NaCl[/tex]

2. Molarity

We can use the number of moles we just calculated to find the molarity. Remember there is 1 liter of solution.

[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]

[tex]molarity= \frac{ 0.09993194823 \ mol \ NaCl}{1 \ L}[/tex]

[tex]molarity= 0.09993194823 \ mol \ NaCl/L[/tex]

3. Units and Significant Figures

The original measurements of mass and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 9 in the ten-thousandths place tells us to round the 9 to a 0, but then we must also the next 9 to a 0, and the 0 to a 1.

[tex]molarity \approx 0.100 \ mol \ NaCl/L[/tex]

1 mole per liter is 1 molar or M. We can convert the units.

[tex]molarity \approx 0.100 \ M \ NaCl[/tex]

The molarity of the solution is 0.100 M.

The half life of radium-226 is 1600 years. If you have 200 grams of radium today how many grams would be present in 8000 years?

Answers

Answer:

Half life is the time taken by a radio active isotope to reduce by half of its original amount. Radium-226 has a half life of 1602 years meaning that it would take 1602 years for a mass of radium to reduce by half.

Number of half lives in 9612 years = 9612/1602 = 6 half lives

New mass = Original mass x (1/2)n where n is the number of half lives.

Therefore, New mass= 500 x (1/2)∧6

                                 = 500 x 0.015625

                                 = 7.8125 g

Hence the mass of radium after 9612 years will be 7.8125 grams.

Explanation:

Answer:

[tex]\boxed {\boxed {\sf 6.25 \ grams}}[/tex]

Explanation:

We are asked to find the mass of a sample of radium-226 after half-life decay. We will use the following formula:

[tex]A= A_o *\frac{1}{2}^{\frac{t}{h}}[/tex]

In this formula, [tex]A_o[/tex] is the initial amount, t is the time, and h is the half-life.

For this problem, the initial amount is 200 grams of radium-226, the time is 8,000 years, and the half-life is 1,600 years.

[tex]\bullet \ A_o= 200 \ g \\\\bullet \ t= 8,000 \ \\\bullet \ h= 1,600[/tex]

Substitute the values into the formula.

[tex]A= 200 \ g * \frac{1}{2} ^{\frac{8.000}{1,600}[/tex]

Solve the fraction in the exponent.

[tex]A= 200 \ g * \frac{1}{2}^{5}[/tex]

Solve the exponent.

[tex]A= 200 \ g *0.03125[/tex]

[tex]A= 6.25 \ g[/tex]

In addition, we can solve this another way. First, we find the number of half-lives by dividing the total time by the half-life.

8,000/1,600= 5 half-lives

Every half-life, 1/2 of the mass decays. Divide the initial mass in half, then that result in half, and so on 5 times.

1.  200 g/2= 100 g2. 100 g / 2 = 50 g3. 50 g / 2 = 25 g 4. 25 g / 2 = 12.5 g5. 12.5 g / 6.25 g

After 8,000 years, 6.25 grams of radium-226 remains.

How many moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+?

Answers

Answer:

0.1 mol

Explanation:

6/(15*3+15)

0.1 mol moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+

What is mole?

The mole, symbol mol, exists as the SI base unit of the amount of substance. The quantity amount of substance exists as a measure of how many elementary entities of a provided substance exist in an object or sample.A mole corresponds to the mass of a substance that includes 6.023 x 1023 particles of the substance. The mole exists the SI unit for the amount of a substance. Its symbol stands mol.

The compound trimethylamine, (CH3 )3N, exists as a  weak base when dissolved in water.

A mole exist expressed as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole exists as a convenient unit to utilize because of the great number of atoms, molecules, or others in any substance.

To find the amount of the substance (CH3)3NH+ to calculate its molar mass:

M((CH3)3NH+) = (12+3)*3 + 14+1 = 60 g/mol

n((CH3)3NH+) = m/M

m((CH3)3NH+) = 6g

Thus,

n((CH3)3NH+) = 6g/60 g/mol = 0.1 mol

Hence,

n((CH3)3NH+) = 0.1 mol

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15.27
The following equilibria were attained at 823 K:
COO(s) + H2() Co(s) + H2O(g) K = 67

COO(s) + CO(8) = Co(s) + CO2(8) K = 490

Based on these equilibria, calculate the equilibrium con-
stant for
H2(g) + CO2(g) = CO(g) + H2O(g) at 823 K.

Answers

The equilibrium constant for the reaction is K = 0.137

We obtain the equilibrium constant considering the following equilibria and their constants:

COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

COO(s) + CO(g) → Co(s) + CO₂(g)   K₂ = 490

We write the first reaction in the forward direction because we need H₂(g) in the reactants side:

(1)     COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):

(2)   Co(s) + CO₂(g) → COO(s) + CO(g)   K₂ = 1/490

From the addition of (1) and (2), we obtain:

COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

+

Co(s) + CO₂(g) → COO(s) + CO(g)   K₂ = 1/490

-------------------------------------------------

H₂(g) +  CO₂(g) → CO(g) + H₂O(g)

Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.

Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:

K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137

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The density of toluene (C7H8) is 0.867 and that of thiophene (C4H4S) is 1.065 g/ml. A solution is made by dissolving 10.00g thiophene in 250.00ml of toluene. a)Calculate the molarity of the solution
b)Assuming the volume are addictive ,calculate the molarity of the solution

Answers

Answer:

Calcular la molaridad de una solución que se preparó disolviendo 14 g de KOH en suficiente  

agua para obtener 250 mL de solución. (masa molar del KOH = 56 g/mol).

Resolución: de acuerdo a la definición de “molaridad” debemos calcular primero, el número de mol de soluto (KOH) que  

se han disuelto en el volumen dado, es decir, “se transforma g de soluto a mol de soluto” por medio de la masa molar,  

así:

56 g de KOH 14 g de KOH

----------------- = ------------------- X = 0,25 mol de KOH

1 mol X

Ahora, de acuerdo con la definición de molaridad, el número de mol debe estar contenido en 1000 mL (o 1 L) de  

solución, que es el volumen estándar para esta unidad de concentración, lo que se determina con el siguiente planteamiento:

0,25 mol X

----------------------- = ------------------------- X = 1 mol de KOH

250 mL de solución 1000 mL de solución

Explanation:

HBr can be added to an alkene in the presence of peroxides (ROOR). What function does the peroxide serve in this reaction

Answers

Answer:

Radical chain initiator

Explanation:

The peroxide here serves as a radical chain initiator. In the field of chemistry the radical initiatives are those substances that are used in industrial processes like polymer synthesis. These initiatives have weak bonds generally and they're mostly used to create free radicals. These radicals are atoms that have odd numbers of electrons. Peroxide is an example of such.

how can we convert plastic garbage energy into electric energy​

Answers

Answer:

Unfortunately, we don`t know how to convert plastic material into electricity yet. I suppose an idea is for someone to invent a machine similar to biomass, where dead plants created energy, but here it`s plastic. The only issue is that it could release deadly chemicals.

Sorry if this isn`t much help, but there isn`t really an answer.   :/

Answer:

Plastics are among the most valuable waste materials – although with the way people discard them, you probably wouldn’t know it. It’s possible to convert all plastics directly into useful forms of energy and chemicals for industry, using a process called “cold plasma pyrolysis”.

Hope this helps you ❤️

MaRk mE aS braiNliest ❤️

What is the balanced form of the following equation?
Br2 + S2O32- + H2O → Br1- + SO42- + H+

Answers

Answer:

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

Explanation:

We will balance the redox reaction through the ion-electron method.

Step 1: Identify both half-reactions

Reduction: Br₂ ⇒ Br⁻

Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻

Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate

Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺

Step 3: Perform the charge balance, adding electrons where appropriate

2 e⁻ + Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻

Step 4: Make the number of electrons gained and lost equal

5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)

1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)

Step 5: Add both half-reactions

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

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