Which substrate is used in the last step of glycolysis

Answer:

4.00 is the pH of the mixture

Explanation:

The ethyl amine reacts with HNO3 as follows:

C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻

To solve this question we need to find the moles of ethyl amine and the moles of HNO3:

Moles C2H5NH2:

0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine

Moles HNO3:

0.201L * (0.025mol/L) = 0.005025 moles HNO3

That means HNO3 is in excess. The moles in excess are:

0.005025 moles HNO3 - 0.00500 moles ethyl amine =

2.5x10⁻⁵ moles HNO₃

In 50 + 201mL = 251mL = 0.251L:

2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]

As pH = -log [H+]

pH = -log 9.96x10⁻⁵M

pH = 4.00 is the pH of the mixture

Answer:

what do you need help with

Answer:

1.30 m

Explanation:

First we calculate the number of moles of solute in the solution, using the definition of molarity:

Then we calculate the mass of solvent, using the given volume and density:

Now we calculate the molality of the solution:

Based on the calculations, the Gibbs's free energy for this chemical reaction is equal to -1,680.906 kJ/mol.

Given the following data:

Gibbs's free energy simply refers to the quantity of energy that is associated with a particular chemical reaction.

Mathematically, the Gibbs's free energy for this chemical reaction can be calculated by using this formula:

ΔG° = ΔH° - ΔS°

Substituting the given parameters into the formula, we have;

ΔG° = -1652 × 10³ - (298 × 0.097)

ΔG° = -1652 × 10³ - 28.906

ΔG° = -1,680.906 kJ/mol.

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Answer:

verdadero/a

falso/b

verdadero/c

Explanation:

Answer:

Explanation:

Target organ toxins are chemicals that can cause adverse effects or disease states manifested in specific organs of the body. Toxins do not affect all organs in the body to the same extent due to their different cell structures.

Answer: The reaction, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.

Explanation:

A chemical reaction in which one element of a compound is replaced by another element participating in the reaction.

For example, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex]

Here, the element manganese is replaced by carbon atom. As only one element gets replaced so, it is a single-displacement reaction.

Thus, we can conclude that [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.

Answer:

ΔH = -2446J

Explanation:

Based on the reaction:

2 Na(s) + Cl2(g) → 2NaCl

We can find the enthalpy of this reaction using Hess's law:

The enthalpy of a reaction is equal to the sum of the enthalpy of products times their reaction quotient subtracting the enthalpy of reactants times their reaction quotient. For the reaction of the problem:

ΔH = 2ΔH(NaCl) - [2ΔH(Na) + ΔHCl2)]

ΔH(NaCl) = 670J

ΔH(Na) = 235cal * (4.184J/1cal) = 983J

ΔHCl2 = 435cal * (4.184J/1cal) = 1820J

ΔH = 2*670J - [2*983J + 1820J]

ΔH = 1340J - [3786J]

Answer:

the heat content of a system at constant pressure

Explanation:

Answer:

C

Explanation:

Think of 'exo' as exit and 'thermic' as relating to thermal energy/ heat. Thus, an exothermic release thermal energy as the reaction proceeds.

In an exothermic reaction, the total energy of the products is lesser than that of the reactants and ΔH (change in energy) is less than zero.

When heat is absorbed as the reaction proceeds, the chemical reaction is an endothermic reaction.

Answer:

dissolved in the solute

Explanation:

A solvent is the component that dissolves the solute and is present in larger amount. The type of solution is determined by the state of the solute and solvent. If you have NaCl, a solid, dissolved in water, a liquid, the type of solution is a solid/liquid solution.

The question is incomplete, the complete question is;

. A chemist determined by measurements that 0.015 moles of hydrogen chloride participate in a chemical reaction. Calculate the mass of hydrogen chloride that participates Round your answer to 2 significant digits. x S. ?

Answer:

0.54 g

Explanation:

Recall that;

Number of moles = mass/molar mass

Molar mass of HCl =36.5 g/mol

Mass= number of moles × molar mass

Mass= 0.015 moles × 36 g/mol

Mass= 0.54 g

Answer:

b. The anion affects the color of the solution more than the intensity of the color.

Explanation:

An ionic compound is one that is made up of ions held together by electrostatic forces. The ions are made up of positively charged ions known as cations and negatively charged ions known as anions. In compounds with a high ionic attribute, anions produce colorless compounds. But compounds that have less ionic attributes produce deeper colors like black and red. When in solutions, ionic compounds take on color and this is because of the anion that absorbs infrared light energy.

Answer:

36.55kJ/mol

Explanation:

The heat of solution is the change in heat when the KNO3 dissolves in water:

KNO3(aq) → K+(aq) + NO3-(aq)

As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.

To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:

Moles KNO3 -Molar mass: 101.1032g/mol-

10.6g * (1mol/101.1032g) = 0.1048 moles KNO3

Change in heat:

q = m*S*ΔT

Where q is heat in J,

m is the mass of the solution: 10.6g + 251.0g = 261.6g

S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-

And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C

q = 261.6g*4.184J/g°C*3.5°C

q = 3830.87J

Molar heat of solution:

3830.87J/0.1048 moles KNO3 =

36554J/mol =

Answer:

Which piece of equipment would be BEST to measure 5 mL of a liquid?

A 50 mL beaker filled halfway between 0 and 10 mL markings

A 50 mL Erlenmeyer flask filled halfway between 0 and 10 ml marking

A 10 mL graduated cylinder

A disposable dropper

A 100 mL graduated cylinder

Explanation:

To measure 5 mL of a liquid, the best equipment is

a 10 mL graduated cylinder.

The remaining apparatus do not give an accurate measurement because readings are not marked on them.

Answer:

the molar mass of the element

Answer:

11

1. 6.02×10 23

this is the answer Hope it helps you

Answer:

(NH4)2SO4 (aq) + Ba(C2H302)2 (aq) - BaSO4(s) + 2 NH4C2H3O2 (aq).

Explanation:

In balancing redox reaction equations, the rule of thumb is that the number of atoms on the left hand side of the reaction equation must be equal to the number of atoms of the same element on the left hand side of the reaction equation.

To ensure this, the correct coefficients are written in front of each of the reactants/products and the number of atoms of each element is counted to ensure that the equation is balanced.

If all these procedures are meticulously carried out, the balanced equation for the reaction between aqueous ammonium sulfate and aqueous barium acetate is:

(NH4)2SO4 (aq) + Ba(C2H302)2 (aq) - BaSO4(s) + 2 NH4C2H3O2 (aq).

This is an acronym of the word Systeme International in French. Its International System of Units (SI) is a metric system that would be universally acknowledged as a measurement standard, and the further discussion can be defined as follows:

Conversion [tex]^{\circ} \ to \ K[/tex]:

let,

[tex]\to 0^{\circ}\ C + 273.15 = 273.15\ K\\\\[/tex]

So,

[tex]\to -35.903+273.15=237.247\ K[/tex]

Therefore, the final answer is "237.247".

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Answer:

Does a reaction occur when aqueous solutions of copper(II) nitrate and calcium chloride are combined?

Explanation:

The given reaction is an example of a chemical double displacement reaction.

The mutual interchange of the ions takes place.

The balanced chemical equation of the reaction is:

[tex]Cu(NO_3)_2(aq)+CaCl_2(aq) -> Ca(NO_3)_2(aq)+CuCl_2(aq)[/tex]

All the products formed are also highly soluble.

So, all the ions are spectator ions.

Hence, all will cancel with each other and ions remain in the net ionic equation and hence, the reaction does not take place.

Answer:

Pharmaceutical laboratory helps in devloping and conducting research, vaccines. Various kinds of drugs and chemical substances used and are produced at a Pharmaceutical laboratory.

The pharmaceutical laboratories performs with various hazardous substances that results in exposure to various chemicals, biological substances and radiation. To avoid any injury or infection labs need to maintain all safety measures.

Spillage and relaseing chemical substances can be lethal during transportaions by safety measures for heling in for manufacturing of such therapeutic agents spillage and avoid wastage.

Maintaining good safety standards in the pharmaceuticals laboratory will help promote the health of technicians and workers which in turn will increase productivity and attain positive outcomes.

Answer:

liquefied petroleum gas

LPG is prepared by refining natural gas. it is made by refining crude oil or from extracted natural gas streams as they emerge from the ground.

a. Br, Cl, F

b. Na, K, Ba

c. He, Ar, Ne

d. Ca, Ba, Mg

Answer:

a. halogen : F ,Cl ,Br l ,At

b an alkali metal: Na,Li, Rb, Cs

c. a noble gas: He, Ne, Kr, Ar

d. an alkaline earth metal: Be,Mg,Ca, Sr

hope it helps

Answer:

a)10.87

b)9.66

c)9.15

d)7.71

e) 5.56

f) 3.43

Explanation:

tep 1: Data given

Volume of 0.030 M NH3 solution = 30 mL = 0.030 L

Molarity of the HCl solution = 0.025 M

Step 2: Adding 0 mL of HCl

The reaction:    NH3 + H2O ⇔ NH4+ + OH-

The initial concentration:  

[NH3] = 0.030M    [NH4+] = 0M    [OH-] = OM

The concentration at the equilibrium:

[NH3] = 0.030 - XM

[NH4+] = [OH-] = XM

Kb = ([NH4+][OH-])/[NH3]

1.8*10^-5 = x² / 0.030-x

1.8*10^-5 = x² / 0.030

x = 7.35 * 10^-4 = [OH-]

pOH = -log [7.35 * 10^-4]

pOH = 3.13

pH = 14-3.13 = 10.87

Step 3: After adding 10 mL of HCl

The reaction:

NH3 + HCl ⇔ NH4+ + Cl-

NH3 + H3O+ ⇔ NH4+ + H2O

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.010 L = 0.00025 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00025 =0.00065 moles

Moles HCl = 0

Moles NH4+ = 0.00025 moles

Concentration at the equilibrium:

[NH3]= 0.00065 moles / 0.040 L = 0.01625M

[NH4+] = 0.00625 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.00625/0.01625)

pOH = 4.34

pH = 9.66

Step 3: Adding 20 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.020 L = 0.00050 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00050 =0.00040 moles

Moles HCl = 0

Moles NH4+ = 0.00050 moles

Concentration at the equilibrium:

[NH3]= 0.00040 moles / 0.050 L = 0.008M

[NH4+] = 0.01 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.01/0.008)

pOH = 4.85

pH = 14 - 4.85 = 9.15

Step 4: Adding 35 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.035 L = 0.000875 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000875 =0.000025 moles

Moles HCl = 0

Moles NH4+ = 0.000875 moles

Concentration at the equilibrium:

[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M

[NH4+] = 0.000875 M / 0.065 L = 0.0135 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.0135/3.85*10^-4)

pOH = 6.29

pH = 14 - 6.29 = 7.71

Step 5: adding 36 mL HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.036 L = 0.0009 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.0009 =0 moles

Moles HCl = 0

Moles NH4+ = 0.0009 moles

[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M

Kw = Ka * Kb

Ka = 10^-14 / 1.8*10^-5

Ka = 5.6 * 10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka =5.6 * 10^-10 =  x² / 0.0136

x = 2.76 * 10^-6 = [H3O+]

pH = -log(2.76 * 10^-6)

pH = 5.56

Step 6: Adding 37 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.037 L = 0.000925 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000925 =0 moles

Moles HCl = 0.000025 moles

Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M

pH = -log 3.73*10^-4= 3.43

The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:

a) pH = 10.86

b) pH = 9.66

c) pH = 9.15

d) pH = 7.70

e) pH = 5.56

f) pH = 3.43          

Initially, the pH of the solution is given by the dissociation of NH₃ in water.  

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻     (1)

The constant of the above reaction is:

[tex] Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.76\cdot 10^{-5} [/tex]   (2)

At the equilibrium, we have:  

   NH₃    +    H₂O   ⇄   NH₄⁺    +    OH⁻     (3)  

0.030 M - x                      x               x

[tex] 1.76\cdot 10^{-5}*(0.030 - x) - x^{2} = 0 [/tex]

After solving for x and taking the positive value:

x = 7.18x10⁻⁴ = [OH⁻]  

Now, we can calculate the pH of the solution as follows:

[tex] pH = 14 - pOH = 14 + log(7.18\cdot 10^{-4}) = 10.86 [/tex]

Hence, the initial pH is 10.86.

After the addition of HCl, the following reaction takes place:

NH₃ + HCl ⇄ NH₄⁺ + Cl⁻  (4)  

We can calculate the pH of the solution from the equilibrium reaction (3).            

[tex] 1.76\cdot 10^{-5}(Cb - x) - (Ca + x)*x = 0 [/tex] (5)  

The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:

[tex] n_{b} = n_{i} - n_{HCl} [/tex]     (6)

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^{-4} moles [/tex]          

[tex] n_{a} = n_{HCl} [/tex]   (7)

[tex] n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles [/tex]

The concentrations are given by:

[tex] Cb = \frac{6.5\cdot 10^{-4} moles}{(0.030 L + 0.010 L)} = 0.0163 M [/tex]   (8)

[tex] Ca = \frac{2.5 \cdot 10^{-4} mole}{(0.030 L + 0.010 L)} = 6.25 \cdot 10^{-3} M [/tex]      (9)

After entering the values of Ca and Cb into equation (5) and solving for x, we have:  

[tex] 1.76\cdot 10^{-5}(0.0163 - x) - (6.25 \cdot 10^{-3} + x)*x = 0 [/tex]

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

[tex] pH = 14 + log(4.54\cdot 10^{-5}) = 9.66 [/tex]

Hence, the pH is 9.66.

We can find the pH of the solution from the reaction of equilibrium (3).

The concentrations are (eq 8 and 9):

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 8.0\cdot 10^{-3} M [/tex]    

[tex] Ca = \frac{0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 0.01 M [/tex]    

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(8.0\cdot 10^{-3} - x) - (0.01 + x)*x = 0 [/tex]

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:  

[tex] pH = 14 + log(1.40\cdot 10^{-5}) = 9.15 [/tex]

So, the pH is 9.15.

We can find the pH of the solution from reaction (3).

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 3.85\cdot 10^{-4} M [/tex]      

[tex] Ca = \frac{0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 0.0135 M [/tex]      

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(3.85\cdot 10^{-4} - x) - (0.0135 + x)*x = 0 [/tex]

x = 5.013x10⁻⁷ = [OH⁻]      

Then, the pH is:  

[tex] pH = 14 + log(5.013\cdot 10^{-7}) = 7.70 [/tex]  

So, the pH is 7.70.

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 [/tex]    

[tex] n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles [/tex]

Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.

At the equilibrium, we have:                

NH₄⁺    +    H₂O   ⇄   NH₃    +    H₃O⁺

Ca - x                             x               x

[tex] Ka = \frac{x^{2}}{Ca - x} [/tex]  

[tex] Ka(Ca - x) - x^{2} = 0 [/tex]   (10)          

We can find the acid constant as follows:

[tex] Kw = Ka*Kb [/tex]

Where Kw is the constant of water = 10⁻¹⁴

[tex] Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} [/tex]  

The concentration of NH₄⁺ is:

[tex] Ca = \frac{9.0 \cdot 10^{-4} moles}{(0.030 L + 0.036 L)} = 0.0136 M [/tex]      

After solving the equation (10) for x, we have:

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:  

[tex] pH = -log(H_{3}O^{+}) = -log(2.78\cdot 10^{-6}) = 5.56 [/tex]

Hence, the pH is 5.56.

Now, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).

[tex] C_{HCl} = \frac{0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L}{(0.030 L + 0.037 L)} = 3.73 \cdot 10^{-4} M = [H_{3}O^{+}] [/tex]      

[tex] pH = -log(H_{3}O^{+}) = -log(3.73 \cdot 10^{-4}) = 3.43 [/tex]

Therefore, the pH is 3.43.

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Answer:

C

Explanation:

an impure substance made through chemical process

Answer: It is true about the type of reaction that it is a single replacement reaction, and the cations in the two ionic compounds are different.

Explanation:

When one element in a compound is replaced by another element in a chemical reaction then it is called a single replacement reaction.

For example, [tex]K + NaCl \rightarrow KCl + Na[/tex]

Here, potassium metal is replacing the sodium metal in the sodium chloride compound.

As metals become cation by losing an electron in a chemical reaction.

Thus, we can conclude that it is true about the type of reaction that it is a single replacement reaction, and the cations in the two ionic compounds are different.

Answer: Its A

Explanation:

a single replacement reactions, and the ANIONS in the two ionic compounds are different

Answer:

n = 2.23 moles

Explanation:

Given the following data;

Standard temperature = 273 K

Standard pressure = 101.325 kPa

Volume = 50 liter

R = 8.314 J/mol·K  

To find the number of moles, we would use the ideal gas law formula;

PV = nRT

Where;

Making n the subject of formula, we have;

[tex] n = \frac {PV}{RT} [/tex]

Substituting into the formula, we have;

[tex] n = \frac {101.325*50}{8.314*273} [/tex]

[tex] n = \frac {5066.25}{2269.722} [/tex]

n = 2.23 moles

Therefore, 2.23 moles of carbon dioxide at STP will fit in a 50 liter container.

The internal energy of system is raised by 3 times

Answer:

The reaction is the combustion of methanol (CH3OH).

Write the balanced chemical equation.

Draw Lewis structures for each structure.

Explanation:

The balanced chemical equation for the combustion of methane is shown below:

[tex]2CH_3OH(g)+3O_2(g)->2CO_2(g)+ 4 H_2O(g)[/tex]

Lewis structures of the given molecules are shown below: