At what velocity (m/s) must a 20.0g object be moving in order to possess a kinetic energy of 1.00J

​

Explanation:

here's the answer to the question

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:

gamma ray

Explanation:

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:

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:

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.

Answer:

ITS ANSWER IS

OPTION B. ATOMIC NUMBER

HI HAVE A NICE DAY

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: 8 for both

Explanation:

Answer:

I don't know What can I do.

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:

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:

yes it is corrwect iyt is absolitle correct

Explanation:

Answer:

17

Explanation:

Step 1: Calculate the needed concentrations

[A]i = 1.00 mol/5.00 L = 0.200 M

[B]i = 1.80 mol/5.00 L = 0.360 M

[B]e = 1.00 mol/5.00 L = 0.200 M

Step 2: Make an ICE chart

        A(aq) + 2 B(aq) ⇄ C(aq)

I       0.200    0.360        0

C        -x           -2x         +x

E     0.200-x  0.360-2x   x

Then,

[B]e = 0.360-2x = 0.200

x = 0.0800

The concentrations at equilibrium are:

[A]e = 0.200-0.0800 = 0.120 M

[B]e = 0.200 M

[C]e = 0.0800 M

Step 3: Calculate the concentration equilibrium constant (K)

K = [C] / [A] × [B]²

K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17

Answer:

endet nach selam nw

4gh7

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:

"They established the laws of planetary motion"

Explanation:

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

After the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.

First, we will write the balanced equation for the reaction

H₂SO₄ + BaCl₂  → BaSO₄ + 2HCl

This means 1 mole of BaCl₂ is needed to react completely with 1 mole of H₂SO₄ to give 1 mole of BaSO₄ and 2 moles of HCl

From the question, 50.0g of sulfuric acid is mixed with 40.0 grams of barium chloride. To determine the quantity of each substance remaining after the complete reaction, we will first determine the number of moles present in each of the reactant.

For H₂SO₄

mass = 50.0g

Molar mass = 98.079 g/mol

From the formula

Number of moles = Mass / Molar mass

∴ Number of moles of H₂SO₄ = 50.0g / 98.079 g/mol

Number of moles of H₂SO₄ = 0.5098 mol

For BaCl₂

mass = 40.0 g

Molar mass = 208.23 g/mol

∴ Number of moles of BaCl₂ = 40.0g / 208.23 g/mol

Number of moles of BaCl₂ = 0.1921 mol

Since the number of moles of H₂SO₄ is more than that of BaCl₂, then H₂SO₄ is the excess reagent and BaCl₂ is the limiting reagent (that is, it will be used up completely during the reaction)

From the equation, 1 mole of H₂SO₄ is needed to completely react with 1 mole of BaCl₂

∴ 0.1921 mol of H₂SO₄ will be needed to completely react with 0.1921 mol of BaCl₂.

Therefore, after the reaction is complete, 0 mole (i.e 0 grams) of BaCl₂ will remain and (0.5098 mole - 0.1921 mole) of H₂SO₄ will remain.

Number of moles H₂SO₄ that will remain = 0.5098 mole - 0.1921 mole = 0.3177 moles

Now, we will convert this to grams

From the formula

Mass = Number of moles × Molar mass

Mass of H₂SO₄ that will remain = 0.3177 moles × 98.079 g/mol

Mass of H₂SO₄ that will remain = 31.1597 g

Mass of H₂SO₄ that will remain ≅ 31.16 g

Hence, after the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.

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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

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:

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

Answer: it’s checkbox 2&3

Answer:

1. more

2. less

3. more

4. less

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:

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:

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:

The phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.

Explanation:

The structure of nucleic acid polymers is built up from monomers of nucleotides.

A nucleotide consists of a sugar backbone which is either a ribose or deoxyribose sugar, a nitogenous base which is either a purine or pyrimidine, and a phosphate group. The nitrogenous base is attached to the carbon number 1 or C-1 of the sugar backbone by a covalent bond. The phosphate group on the other hand is covalently attached to the carbon number 5 or 5' carbon of the sugar backbone.

When polymers of nucleic acids are formed, the phosphate at the 5' carbon of the sugar backbone is covalently linked in a phosphodiester bond to the 3' carbon of the sugar backbone in another nucleotide molecule, thus extending the strands of the nucleic acid molecule.

Nucleases are enzymes that break down the phosphodiseter bonds in nucleic acids resulting in nucleotide monomers. After complete digestion ofmthe nucleic acid polymer by nucleases, the phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.

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:

Answer:

B. 0.0000005461m

I used the method of moving the decimal.

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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