a. You have a stock solution of 14.8 M NH3. How many milliliters of this solution should you dilute to make 1000.0 mL of 0.250 M NH3?
b. If you take a 10.0 mL portion of the stock solution and dilute it to a total volume of 0.500 L, what will be the concentration of the final solution?

Answer:

AgNO3 + NaOH = AgOH + NaNO3.

Explanation:

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.

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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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.

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.

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.

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]

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]

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.

Answer:

[tex]\boxed {\boxed {\sf 0.14 \ mol \ CaCl_2}}[/tex]

Explanation:

We are asked to calculate the number of moles of 15 grams of calcium chloride (CaCl₂).

To convert from grams to moles, we use the molar mass, or the mass of 1 mole of a substance. Molar masses are found on the Periodic Table because they are equivalent to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up the individual elements in the compound: calcium and chloride.

Notice the chemical formula has a subscript of 2 after Cl or chlorine. There are 2 moles of chlorine in every 1 mole of calcium chloride. We must multiply chlorine's molar mass by 2 before adding calcium's molar mass.

We will convert using dimensional analysis, so we must create a ratio using the molar mass.

[tex]\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]

We are converting 15 grams of calcium chloride to moles, so we must multiply the ratio by this value.

[tex]15 \ g \ CaCl_2 *\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]

Flip the ratio so the units of grams of calcium chloride cancel.

[tex]15 \ g \ CaCl_2 *\frac { 1 \ mol \ CaCl_2}{110.98 \ g \ CaCl_2}[/tex]

[tex]15 *\frac { 1 \ mol \ CaCl_2}{110.98}[/tex]

[tex]\frac { 15}{110.98} \ mol \ CaCl_2[/tex]

[tex]0.1351594882\ mol \ CaCl_2[/tex]

The original measurement of grams (15) has 2 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 3 up to a 4.

[tex]0.14 \ mol \ CaCl_2[/tex]

15 grams of calcium chloride is approximately 0.14 moles of calcium chloride.

Answer:

B. 0.0000005461m

I used the method of moving the decimal.

Answer: 1090°C

Explanation: According to combined gas laws

(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2

where P1 = initial pressure of gas = 80.0 kPa

V1 = initial volume of gas = 10.0 L

T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K

P2 = final pressure of gas = 107 kPa

V2 = final volume of gas = 20.0 L

T2 = final temperature of gas

Substituting the values,

(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2

T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)

T2 = 513 K × (1.3375) × (2)

T2 = 1372.275 K

T2 = (1372.275 - 273) °C

T2 = 1099 °C

1090 degree Celsius

hope it helps

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.

Answer:

Al + MnO4- + 2H2O → Al(OH)4- + MnO2

Explanation:

First of all, we out down the skeleton equation;

Al + MnO4- → MnO2 + Al(OH)4-

Secondly, we write the oxidation and reduction equation in basic medium;

Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-

Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-

Thirdly, we add the two half reactions together to obtain:

Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-

Lastly, cancel out species that occur on both sides of the reaction equation;

Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O

The simplified equation now becomes;

Al + MnO4- + 2H2O → Al(OH)4- + MnO2

The volume occupied by the given amount of sulfur dioxide will be 84 L.

please show working my dear citizen

The number of replaceable electrons in an atom is called its valency.

Examples

Thanks !

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Answer:

the combining capacity if an atom is know as valency.

the property of an element that determines the number of other atimd with an aton if the element can combine.

Answer:

D. -80m/s^2

Explanation:

V = u + at

5 = 65 + a (0.75)

0.75a = -60

a = -60/0.75

a = -80m/s^2

Therefore, is decelerating at 80m/s^2

Answer:

[tex]\boxed {\boxed {\sf D. \ 80 \ m/s^2 \ down}}[/tex]

Explanation:

We are asked to find the acceleration of a skydiver. Acceleration is the change in velocity over the change in time, so the formula for calculating acceleration is:

[tex]a= \frac{v_f-v_i}{t}[/tex]

The skydiver was initially traveling 65 meters per second, then he slowed down to a final velocity of 5 meters per second. He slowed down in 0.75 seconds.

[tex]\bullet \ v_f = 5 \ m/s \\\bullet \ v_i= 65 \ m/s \\\bullet \ t= 0.75 \ s[/tex]

Substitute the values into the formula.

[tex]a= \frac{ 5 \ m/s - 65 \ m/s}{0.75 \ s}[/tex]

Solve the numerator.

[tex]a= \frac{-60 \ m/s}{0.75 \ s}[/tex]

Divide.

[tex]a= -80 \ m/s^2[/tex]

The acceleration of the skydiver is -80 meters per second squared or 80 meters per second squared down. The skydiver is slowing down or decelerating, so the acceleration is negative or down.

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:

Answer:

3

Explanation:

The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.

As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.

Hence, M shell contains s,p and d subshells.

water is known as the mixture

Answer:

iv. Iron

water is not a mixture

hydrogen is the simplest element

air is pure

Answer:

yes it is corrwect iyt is absolitle correct

Explanation:

Chemistry 11 Solutions

978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85

Amount in moles, n, of the NaCl(s):

NaCl

2.5 g

m

n

M

58.44 g

2

4.2778 10 m l

ol

o

/m

u

Molar concentration, c, of the NaCl(aq):

–2 4.2778 × 10 mol

0.100

0.42778 mol/L

0.43 mol

L

/L

n

c

V

The molar concentration of the saline solution is 0.43 mol/L.

Check Your Solution

The units are correct and the answer correctly shows two significant digits. The

dilution of the original concentrated solution is correct and the change to mol/L

seems reasonable.

Section 8.4 Preparing Solutions in the Laboratory

Solutions for Practice Problems

Student Edition page 386

51. Practice Problem (page 386)

Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,

(NH4)2SO4(aq).

What volume of the stock solution do you need to use to prepare each of the

following solutions?

a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)

b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)

c. 250 mL of 0.300 mol/L NH4

+

(aq)

What Is Required?

You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution

needed to prepare each given dilute solution.

The dilution gives the relationship between the molarity and the volume of the solution. The volume of stock solution with a molarity of 1.50 mol/L is 50 mL.

Dilution is said to be the addition of more volume to the concentrated solution to make it less in molar concentration. This tells about the inverse and indirect relationship between the volume and the molar concentration of the solution.

Given,

Initial volume = V₁

Initial molar concentration (M₁) = 1.50 mol/L

Final volume (V₂) = 125 mL = 0.125 L

Final molar concentration (M₂)= 0.60 mol/L

The dilution is calculated as:

M₁V₁ = M₂V₂

V₁ = M₂V₂ ÷ M₁

Substituting the values in the above formula as

V₁ = M₂V₂ ÷ M₁

V₁ = (0.60 mol/L × 0.125 L) ÷ 1.50 mol/ L

V₁ = 0.05 L

= 50 mL

Therefore, 50 mL of stock solution is needed to make a 0.60 mol/L solution.

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In the original solution you have the mixture of a weak acid (Hypochlorous acid) and its conjugate base (Sodium hypochlorite). That is a buffer.

The barium hydroxide will react with hypochlorous acid. If this reaction cause the complete reaction of hypochlorous acid, the buffer break its capacity and the pH change in several units. In this case:

The addition of barium hydroxide will raise the pH slightly because the buffer still working.

The initial moles of those species are:

Hypochlorous acid:

[tex]1L * \frac{0.370mol}{1L} = 0.370 moles[/tex]

Sodium hypochlorite:

[tex]1L * \frac{0.493mol}{1L} = 0.493 moles[/tex]

Now, a strong acid as barium hydroxide (Ba(OH)₂) reacts with a weak acid as hypochlorous acid (HClO) as follows:

Ba(OH)₂ + 2HClO → Ba(ClO)₂ + 2H₂O

For a complete reaction of 0.092 moles of barium hydroxide are required:

[tex]0.092 moles Ba(OH)_2*\frac{2mol HClO}{1molBa(OH)_2} = 0.184 moles HClO[/tex]

As there are 0.370 moles, the moles of HClO after the reaction are:

0.370 moles - 0.184 moles = 0.186 moles of HClO will remain

As you still have hypochlorite and hypochlorous acid you still have a buffer.

Thus, the pH will raise slightly because the amount of acid is decreasing and slightly because the buffer can keep the pH.

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Answer:

Both have solutions in the  graduated cylinder.

Explanation:

Recording the amount of alcohol used as the difference between two masses is the wrong method used for measuring tert-butyl alcohol for the experiment.  For measuring tert-butyl alcohol for this experiment, the student has to measure the two masses when both the graduated cylinders has solution of tert-butyl alcohol not when one of it is empty (having no tert-butyl alcohol ).

The wrong aspect is that the liquid didn't need to be weighed. Instead the volume should have been recorded with the aid of the graduated cylinder.

This is a cylinder with marked readings and is used to measure the volume of liquids in the laboratory.

The graduated cylinder will accurately measure the amount of  alcohol used due to it being volatile and the mass fluctuating during the measurement.

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Answer:

46.2%

Explanation:

Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles

Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.

Hence;

Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g

% yield = actual yield/theoretical yield × 100

% yield = 1.7 g/3.68 g × 100

% yield = 46.2%

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⁺

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.

Answer:

a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. i. NO₂⁻ is the oxidizing agent

ii. NO₃⁻ is the reducing agent.

Explanation:

a. Balance the following skeleton reaction

The reaction is

NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)

The half reactions are

NO₂ (g) → NO₃⁻ (aq)  (1) and

NO₂ (g) → NO₂⁻  (aq) (2)

We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)

We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms

NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)

The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.

So, NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

We now add equation (4) and (5)

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

+  NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + e⁻  (4)

2NO₂ (g) + H₂O (l)  → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq)  

We now add two hydroxide ions to both sides of the equation.

So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + 2OH⁻ (aq)

The hydrogen ion and the hydroxide ion become a water molecule

2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H₂O (l)

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

So, the required reaction is

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. Identify the oxidizing agent and reducing agent

Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = 0

x - 4 = 0

x = 4

Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = -1

x - 4 = -1

x = 4 - 1

x = 3

Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = -1

x + 3(-2) = -1

x - 6 = -1

x = 6 - 1

x = 5

i. The oxidizing agent

The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus  NO₂⁻ is the oxidizing agent

ii. The reducing agent

The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and  NO₃⁻ is the reducing agent.

Answer:

c. C3H9N2O2

Explanation:

The empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a molecule. To solve this question we need to convert the mass of each atom to moles. With the moles we can find the ratio as follows:

Moles N -Molar mass: 14.01g/mol-

0.420g N * (1mol/14.01g) = 0.0300 moles N

Moles O -Molar mass: 16g/mol-

0.480g O * (1mol/16g) = 0.0300 moles O

Moles C -Molar mass: 12.01g/mol-

0.540g C * (1mol/12.01g) = 0.0450 moles C

Moles H -Molar mass: 1.0g/mol-

0.135g H * (1mol/1g) = 0.135moles H

Dividing in the moles of N (Lower number of moles) the ratio of atoms is:

N = 0.0300 moles N / 0.0300 moles N = 1

O = 0.0300 moles O / 0.0300 moles N = 1

C = 0.0450 moles C / 0.0300 moles N = 1.5

H = 0.135 moles H / 0.0300 moles N = 4.5

As the empirical formula requires whole numbers, multiplying each ratio twice:

N = 2, O = 2, C = 3 and H = 9

And the empirical formula is:

Answer:

endet nach selam nw

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A cell is the structural and fundamental unit of life. The study of cells from its basic structure to the functions of every cell organelle is called Cell Biology. Robert Hooke was the first Biologist who discovered cells

two types of cell

1) Prokaryotes

2) Eukaryotes

Characteristics of Cells

1) Cells provide structure and support to the body of an organism.

2) The cell interior is organised into different individual organelles surrounded by a separate membrane.

3) The nucleus (major organelle) holds genetic information necessary for reproduction and cell growth

[tex]hope \: its \: helpful \: to \: you \: please \: mark \: me \: a \: brainliest[/tex]

A cell is defined as the fundamental, structural and functional unit of all life.

have a great day

God bless you

Answer:

a. Decrease

b. Increase

c. Increase

d. No effect

Explanation:

Glycolysis is present in muscle cells which converts glucose to pyruvate, water and NADH. It produces two molecules of ATP. Cellular respiration produces more molecules of ATP from pyruvate in mitochondria. Glycolysis increases in pyruvate kinase.

a. Loss of binding site for fructose 1,6-bisphosphate in pyruvate kinase: Decrease

b. Loss of allosteric binding site for ATP in pyruvate kinase: No effect

c. Loss of allosteric binding site for AMP in phosphofructokinase: Increase

d. Loss of regulatory binding site for ATP in phosphofructokinase: Increase

A. An important substrate in the glycolysis pathway is fructose 1,6-bisphosphate. It stimulates pyruvate kinase, an essential enzyme in glycolysis. The amount of pyruvate kinase that is activated will decrease if the fructose 1,6-bisphosphate binding site in pyruvate kinase is eliminated. As a result the rate of glycolysis in the muscle cells will probably decrease.

B. The allosteric ATP binding site of pyruvate kinase controls how active the enzyme is. However, pyruvate kinase is not significantly regulated by ATP in muscle cells. Therefore, it is unlikely that deletion of the ATP-binding allosteric site in pyruvate kinase would have no effect on the rate of glycolysis in muscle cells.

C. The rate-limiting enzyme in glycolysis, phosphofructokinase, is activated from all forms by AMP. It increases the rate of glycolysis by stimulating the activity of phosphofructokinase. If the allosteric binding site for AMP is eliminated, phosphofructokinase activation will be reduced. As a result, the rate of glycolysis in muscle cells will decrease.

D. Phosphofructokinase is inhibited allosterically by ATP. It regulates the rate of glycolysis by a feedback mechanism. High ATP concentrations cause phosphofructokinase to bind to its regulatory site, limiting its activity and delaying glycolysis. If the regulatory binding site for ATP is eliminated, the inhibitory action of ATP on phosphofructokinase would be lost. As a result, muscle cells will glycolysis at a faster rate.

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Answer:

lots more of the carbon 12 than the others

havent calculated it percentage-wise but you can see its very close to 12 meaning it is of far greater abundance that carbon 13 and 14

Explanation:

Answer:

The degree of compactness of a substance

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