Given the following balanced reaction: 2Na(s) + F2(g) --> 2NaF(s)
a) How many moles of NaF will be made from 2.6 moles of F2?

b) How many moles of NaF will be made from 4.8 moles of Na?

The correct answer is C. square planar

According to the Valence Shell Electron Pair Repulsion Theory(VSEPR), The shape of a molecule depends on the number of electron pairs in the molecule.

VSEPR theory was first coined by Gillespie  and Nyhlom in 1957 as an improvement over the Sidgwick - Powell theory.

According to this theory, the shape of a molecule is determined by the number of electron pairs that surround the valence shell of the central atom in the molecule.  The electron pairs are positioned as far apart in space as possible to minimize repulsion of electron pairs.

However, the presence of lone pairs distorts the shape anticipated for the molecule on the basis of VSEPR.

For a molecule having six electron pairs, an octahedral geometry is expected(electron domain geometry). However, the presence of two lone pairs which are positioned at opposite side of the four single bonds leads to an observed square planar molecular geometry.

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Answer:

square planar

Explanation:

Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.

At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.

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Answer:

Na2S(aq)+Cd(No3)2(aq)=CdS(s)+2NaNo3(aq)

Answer: it’s checkbox 2&3

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

Answer:

ITS ANSWER IS

OPTION B. ATOMIC NUMBER

HI HAVE A NICE DAY

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:

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:

gggggggggg

Explanation:

gggggggg

The tripeptide formed from threonine, glycine and alanine is neutral at the pH of 7.3. The carboxylic end is negative charged by donating its proton to form the NH₃⁺ group.

Peptides are protein units formed from two or more amino acids bonded through peptide bonds. There are essential and non-essential amino acids. Essential amino acids have to be uptake from food and non-essential amino acids are synthesized inside the body.

Threonine is an essential amino acid with a CH₃CHOH side group. Glycine has the simplest side group hydrogen and alanine has  CH₃ side chain. Both glycine and alanine are non-essential amino acids.

Each amino acids are represented with a three letter code or one letter symbol. Thus threonine is T,  G for glycine and A for alanine. At a pH of 7.3 the peptide formed from these amino-acids contains a negatively charged carboxylic end.

A positively charged amino end made by protonation from the acid group make the overall charge zero. The structure of the peptide is given in the uploaded image.

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Answer:

25.88 g/mol

Explanation:

Graham's law is a famous law which states that the diffusion rate or the effusion rate of any gas varies inversely to the square root of the molecular weight the gas.

So from Graham's law, we have,

[tex]$\frac{\text{time}}{M^{1/2}}=\text{constant}$[/tex]

Using the sample of Kr gas having M = 83.8

[tex]$\frac{8.15}{(83.8)^{0.5}}= \frac{4.53}{M^{0.5}}$[/tex]

[tex]$M^{0.5}= 5.088$[/tex]

M = 25.88 g/mol

Answer:

See explanation

Explanation:

The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equation;

k= Ae^-Ea/RT

Where;

k= rate constant

A= pre-exponential factor

Ea=activation energy

R= gas constant

T= temperature

We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have a very high activation energy are markedly slow.

Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.

The study of chemistry and bonds is called chemistry. There are two types of elements metal and nonmetals.

The correct answer is mentioned below.

The equation is as follows:-

[tex]k= Ae^{-Ea/RT[/tex] Where;

We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have very high activation energy are markedly slow. Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.

Hence, the correct answer is mentioned above.

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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:

Pure water has an acidity of about 7 on the pH scale. -is a chemical property of pure water. Pure water has an acidity of about 7 on the pH scale

Answer: không màu , không mùi không vị

Explanation:

Answer:

1. more

2. less

3. more

4. less

Explanation:

Answer:

B. 0.0000005461m

I used the method of moving the decimal.

Answer:

1176 grams

Explanation:

nH2SO4 =2*6=12 mol

mH2SO4=12*98=1176 grams

Answer:

solution given:

molarity of H₂SO₄=6 M

volume=2L

no of mole =6M*2=12mole

we have

mass =mole* actual mass=12*98=1176g

the mass is 1176g.

The question is incomplete. The complete question is :

Hydrogen [tex](H_2)[/tex] gas and oxygen [tex](O_2)[/tex] gas react to form water vapor [tex](H_2O)[/tex]. Suppose you have 11.0 mol of [tex]H_2[/tex] and 13.0 mol of [tex]O_2[/tex] in a reactor. Calculate the largest amount of [tex]H_2O[/tex] that could be produced. Round your answer to the nearest 0.1 mol .

Solution :

The balanced reaction for reaction is :

[tex]$2H_2(g) \ \ \ \ + \ \ \ \ \ O_2(g)\ \ \ \rightarrow \ \ \ \ 2H_2O(g)$[/tex]

11.0                      13.0

11/2                       13/1     (dividing by the co-efficient)

6.5 mol               13 mol    (minimum is limiting reagent as it is completely consumed during the reaction)

Therefore, [tex]H_2[/tex] is limiting reagent. It's stoichiometry decides the product formation amount from equation above it is clear that number of moles for [tex]H_2O[/tex] will be produced = number of moles of [tex]H_2[/tex]

                                     = 11.0 mol

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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Answer:

I don't know What can I do.

Answer:

Part A

HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)

Part B

ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)

Explanation:

The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;

RCOOH + NaOH ----> RCOONa + H2O

We have to note the fact that the net ionic reaction still remains;

H^+(aq) + OH^-(aq) ---> H2O(l)

In both cases, the reaction can occur and they actually do occur as written.

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:

Transition temperature is the temperature at which a substance changes from one state to another.

Allotropy is the existence of an element in many forms.

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:

"They established the laws of planetary motion"

Explanation:

Mr. Kepler was the astronomer who came up with the "Laws of Planetary Motion."

Answer:

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:

endet nach selam nw

4gh7

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: 8 for both

Explanation:

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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