In that question how( b) are enantiomers? I want the steps pleas ?

Answer: True

Explanation: For example, changing direction can change velocity

Answer:

B. Conceiving a child who is a bone marrow match to a living child.

Explanation:

I don't know it just is.

Answer:

C

Explanation:

Answer:

[tex]T_2=637\°C[/tex]

Explanation:

Hello there!

In this case, considering this problem as pressure constant, since the change is exhibited in temperature and volume only, it is possible for us to use the Charles' law as shown below:

[tex]\frac{V_2}{T_2}=\frac{V_1}{T_1}[/tex]

Thus, by solving for the final temperature, we obtain:

[tex]T_2=\frac{T_1V_2}{V_1}\\\\T_2=\frac{273K*1L}{0.30L}\\\\T_2=910K-273\\\\T_2=637\°C[/tex]

Best regards!

Answer:

B.

Explanation:

The metal forms metallic bonds which are more greatly affected by gravity, increasing the mass.

Answer:

Hi

Explanation:

We assume you are converting between moles H2O and gram. You can view more details on each measurement unit: molecular weight of H2O or grams This compound is also known as Water or Dihydrogen Monoxide. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles H2O, or 18.01528 grams.

1 mole is equal to 6.023 × 10 ²³ molecules. 63 grams  are in 3.5 moles of H2O.

The term mole is defined as the amount of substance of a system which contains as many elementary entities.

One mole of any substance is equal to 6.023 × 10²³ units of that substance such as atoms, molecules, or ions. The number 6.023 × 10²³ is called as Avogadro's number or Avogadro's constant.

The mole concept can be used to convert between mass and number of particles.

We expect you are converting between moles H2O and gram. The molecular weight of H2O or gram's This compound is also known as Water or Dihydrogen Monoxide. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles H2O, or 18.01528 grams.

Thus, 63 grams  are in 3.5 moles of H2O.

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Answer:

substances is never entirely a one way

Explanation:

no possibility of obtaining complete combustion at hig

temperature.

always loss of heat from the temperature

its temperature is lower than the constant volume process because some of the energy is utilized to change the volume of the system

Four reasons why the temperature of flame is lower than adiabatic flame temperature is that there is  no possibility of obtaining complete combustion at high temperature due to high thermal energy.There is always loss of heat from temperature,its temperature is lower than constant volume process because some of  energy is utilized to change the volume of the system.

Thermal energy is defined as a type of energy which is contained within a system which is responsible for temperature rise.Heat is a type of thermal energy.It is concerned with the first law of thermodynamics.

Thermal energy arises from friction and drag.It includes the internal energy or enthalpy of a body of matter and radiation.It is related to internal energy and heat .It arises when a substance whose molecules or atoms are vibrating faster.

These vibrating molecules and atoms collide and as a result of which heat is generated in a substance , more the collision of particles , higher is the thermal energy.

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Answer:

outside the nucleus

Explanation:

The primary porosity of a sediment or rock consists of the spaces between the grains that make up that material. The more tightly packed the grains are, the lower the porosity.

Answer:

15.2 grams of calcium chloride are produced and HCl is the limiting reactant.

Explanation:

Hello there!

In this case, according to the described scenario, it is possible to realize that the reaction between hydrochloric acid and calcium hydroxide is:

[tex]2HCl+Ca(OH)_2\rightarrow CaCl_2+2H_2O[/tex]

Whereas there is a 2:1 mole ratio of the acid to the base. In such a way, with the given masses, we can compute how much calcium chloride product is produced due to the chemical reaction via stoichiometry:

[tex]m_{CaCl_2}^{by HCl}=10.0gHCl*\frac{1molHCl}{36.46gHCl}*\frac{1molCaCl_2}{2molHCl} *\frac{110.98gCaCl_2}{1molCaCl_2} =15.2gCaCl_2\\\\m_{CaCl_2}^{by Ca(OH)_2}=10.5gHCl*\frac{1molCa(OH)_2}{74.09gCa(OH)_2}*\frac{1molCaCl_2}{1molCa(OH)_2} *\frac{110.98gCaCl_2}{1molCaCl_2} =15.7gCaCl_2[/tex]

Whereas we infer that the correct amount is 15.2 g since HCl is the limiting reactant as it produces the fewest grams of the desired product. Consequently, the calcium hydroxide is the excess reactant here.

Regards!

Molar mass of C3H8 = C 3 (12.01 g/mol) = 36.03 (g/mol)

H 8 (1.008 g/mol) = 8.064 (g/mol)

44.09 (g/mol)

74.6 g propane x 1 mole propane x 6.022 x 10

23

molecules

44.09 g propane 1 mole propane

= 1.02 x 10

24 molecules propane

Answer:

B the weight of the passengers

Answer:

The correct approach is "1 M".

Explanation:

The given values are:

Volume of HCL,

V₁ = 45 ml

In prepared solution,

V₂ = 15 ml

Concentration,

C₁ = ?

C₂ = 3.0 M

As we know,

⇒  [tex]V_1C_1=V_2C_2[/tex]

or,

⇒      [tex]C_1=\frac{V_2C_2}{V_1}[/tex]

On substituting the values, we get

⇒           [tex]=\frac{15\times 3}{45}[/tex]

⇒           [tex]=\frac{45}{45}[/tex]

⇒           [tex]=1 \ M[/tex]

Answer:

Option A

Explanation:

As we know

C1V1 = C2V2

C1 = concentration of solution before dilution

V1 = Volume of solution before dilution

C2 = concentration of solution after dilution

V2 = Volume of solution after dilution

Substituting the given values in above equation, we get -

125 mL * 2.00 M  = X mL * 1.50 M

X mL = 125 mL * 2.00 M  / 1.50 M

X = 167 mL

Hence, option A is correct

Answer:

Reducing the amount of water you use, by having a 5-minute shower or not running the water when washing up the dishes, can help protect vital wetlands. Plant scientists are also working to help conserve by developing crop varieties that use less water.

Explanation:

Answer:

1. Most of the endangered organisms may represent a source of income.

2. The conservation or preservation may conflict with morals.

3. The role of the species or organisms might not be understood.

Explanation:

Biodiversity simply means the amount or number of living organisms that exist in the world.

The practice of protecting these existing living organisms for specific known purposes is regarded as the preservation or conservation of biodiversity.

This process or practice of preserving biodiversity is important to meet most of the human needs, such as generating income, source of food and fuel among others.

Answer:

I think its true I dont really know

Explanation:

true

Answer:

For a): The mass percent of naphthalene in the original sample is 31.00 %.

For b): The mass percent of 3-nitroaniline in the original sample is 32.03 %.

For c): The mass percent of benzoic acid in the original sample is 29.97 %.

For d): The total percent recovered is 93.00 %.

Explanation:

Percentage by mass is defined as the ratio of the mass of a substance to the mass of the solution multiplied by 100. The formula used for this is:  

[tex]\text{Percent by mass}=\frac{\text{Mass of substance}}{\text{Mass of a solution}} \times 100[/tex]               ......(1)

a):

Mass of naphthalene = 0.483 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\%\text{ mass of naphthalene}=\frac{0.483 g}{1.558}\times 100\\\\\%\text{ mass of naphthalene}=31.00 \%[/tex]

b):

Mass of 3-nitroaniline = 0.499 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\%\text{ mass of 3-nitroaniline}=\frac{0.499 g}{1.558}\times 100\\\\\%\text{ mass of 3-nitroaniline}=32.03 \%[/tex]

c):

Mass of benzoic acid = 0.467 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\%\text{ mass of benzoic acid}=\frac{0.467 g}{1.558}\times 100\\\\\%\text{ mass of benozic acid}=29.97 \%[/tex]

d):

Total mass recovered = [0.483 + 0.499 + 0.467] = 1.449 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\text{Total percent recovered}=\frac{1.449 g}{1.558}\times 100\\\\\text{Total percent recovered}=93.00\%[/tex]

Answer: n= 4, , l= 3, [tex]m_l[/tex] = 1,  permissible

Explanation:

Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1).For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as [tex]m_l[/tex]. The value of this quantum number ranges from [tex](-l\text{ to }+l).[/tex]

Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]. The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.

a) n=1 , l= 2, [tex]m_l[/tex] = 0,  not permissible as l can not greater than n.

b) n= 4, , l= 3, [tex]m_l[/tex] = 1,  permissible

c) n= 3, l= 3, [tex]m_l[/tex] = 1, not permissible as l can not equal than n.

d) n= 4 , l= 3, [tex]m_l[/tex] = 4,  not permissible as [tex]m_l[/tex] can not greater than l.

Answer: Atomic mass (Cu) = (x)(63.0 amu) + (1 – x)(65.0 amu) = 63.5 amu 63.0x + 65 – 65.0x = 63.5 –2x = –1.5 x = 0.75 The percent abundance of each isotope is 75.0 % (Cu-63) and 25.0 % (Cu-65).

Explanation: As you know, the average atomic mass of an element is determined by taking the weighted average of the atomic masses of its naturally occurring isotopes.

Simply put, an element's naturally occurring isotopes will contribute to the average atomic mass of the element proportionally to their abundance.

Answer:

34 mL

Explanation:

We'll begin by calculating the number of mole in 1.25 g of sodium carbonate, Na₂CO₃. This can be obtained as follow:

Mass of Na₂CO₃ = 1.25 g

Molar mass of Na₂CO₃ = (23×2) + 12 + (16×3)

= 46 + 12 + 48

= 106 g/mol

Mole of Na₂CO₃ =?

Mole = mass /molar mass

Mole of Na₂CO₃ = 1.25 / 106

Mole of Na₂CO₃ = 0.012 mole

Next, we shall determine the number of mole HCl needed to react with 0.012 mole of Na₂CO₃.

The equation for the reaction is given below:

Na₂CO₃ + 2HCl —> H₂CO₃ + 2NaCl

From the balanced equation above,

1 mole of Na₂CO₃ reacted with 2 moles of HCl.

Therefore, 0.012 mole of Na₂CO₃ will react with = 0.012 × 2 = 0.024 mole of HCl.

Next, we shall determine the volume of HCl required for the reaction. This is illustrated:

Mole of HCl = 0.024 mole

Molarity of HCl = 0.715 M

Volume of HCl =?

Molarity = mole /Volume

0.715 = 0.024 / volume of HCl

Cross multiply

0.715 × volume of HCl = 0.024

Divide both side by 0.715

Volume of HCl = 0.024 / 0.715

Volume of HCl = 0.034 L

Finally, we shall convert 0.034 L to mL

This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.034 L = 0.034 L × 1000 mL / 1 L

0.034 L = 34 mL

Therefore, 34 mL of HCl is needed for the reaction.

The amount of HCl required for counterbalancing 1.25 g of Na2CO3(Sodium Carbonate) would be:

- [tex]34 ml[/tex]

Given that,

Mass of Na2CO3 [tex]= 1.25 g[/tex]

To find the Moles of Na2CO3, we will find the molar mass of Na2CO3,

Molar Mass of or Na2CO3 [tex]= 106 g/mol[/tex]

So,

Moles of Na2CO3 [tex]= mass /molar mass[/tex]

[tex]= 1.25/106[/tex]

[tex]= 0.012 mol[/tex]

To determine the quantity of HCl required to display the reaction with 0.012 mol of Na2CO3

[tex]Na_{2} CO_{2} + 2HCl[/tex] → [tex]H_{2}CO_{3} + 2NaCl[/tex]

While balancing the equation, we know that [tex]0.012[/tex] × [tex]2 = 0.024 mole of HCl[/tex] is necessary to process the reaction.

Now,

As we know,

HCl moles [tex]= 0.024[/tex]

Molarity of HCl [tex]= 0.715 M[/tex]

∵ Quantity of HCl required = Moles/Molarity

[tex]= 0.024 / 0.715[/tex]

[tex]= 0.034 l[/tex] [tex]or 34ml[/tex]

Thus, 34 ml is the correct answer.

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Answer:

0.550 M HCl

Explanation:

M1V1 = M2V2

M1 = 0.455 M NaOH

V1 = 14.50 mL NaOH

M2 = ?

V2 = 12.0 mL HCl

Solve for M2 --> M2 = M1V1/V2

M2 = (0.455 M)(14.50 mL) / (12.0 mL) = 0.550 M HCl

Answer:

The appropriate answer is "0.549 M".

Explanation:

The given values are:

N₁ = 14.50 mL

V₁ = 0.455 M

N₂ = 12 mL

Let

V₂ = C = ?

As we know,

⇒  [tex]N_1\times V_1=N_2\times V_2[/tex]

On substituting the values, we get

⇒  [tex]14.50\times 0.455 = 12\times C[/tex]

⇒            [tex]6.5975=12\times C[/tex]

⇒                   [tex]C=\frac{6.5975}{12}[/tex]

⇒                       [tex]=0.549 \ M[/tex]

Answer:

1.51448 atms

Explanation:

Answer:

A. Number of energy levels

Answer:

3.15 × 10⁻⁶ mol H₂/L.s

1.05 × 10⁻⁶ mol N₂/L.s

Explanation:

Step 1: Write the balanced equation

2 NH₃ ⇒ 3 H₂ + N₂

Step 2: Calculate the rate of production of H₂

The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s

Step 3: Calculate the rate of production of N₂

The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

Step 1:  When we Write the balanced equation

Step 2: Calculate the rate of production of H₂

Step 3: Calculate the rate of production of N₂

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Answer:

Molecular formula => C₅H₅O₅

Explanation:

From the question given above, the following data were obtained:

Empirical formula = CHO

molar mass of compound = 145 g/mol

Molecular formula =?

The molecular formula of the compound can be obtained as follow:

Molecular formula = Empirical formula × n

Molecular formula = [CHO]ₙ

[CHO]ₙ = 145

[12 + 1 + 16]n = 145

29n = 145

Divide both side by 29

n = 145 / 29

n = 5

Molecular formula = [CHO]ₙ

Molecular formula = [CHO]₅

Molecular formula => C₅H₅O₅

Answer:

II. 0.10 M HF + 0.10 M NaF.

Explanation:

A buffer system may be formed in one of two ways:

Which of the following mixtures could be considered a buffer?

I. 0.10 M HCl + 0.10 M NaCl. No, because HCl is a strong acid.

II. 0.10 M HF + 0.10 M NaF. Yes, since HF is a weak acid and F⁻ (coming from NaF)  is its conjugate base.

III. 0.10 M HBr + 0.10 M NaBr. No, because HBr is a strong acid.

Answer:

85.952 ml [tex]Zn^2^+[/tex]  needed to titrate the excess complexing reagent .

Explanation:

Lets calculate

After addition of 80 ml of EDTA the solution becomes = 20 + 70 = 90 ml

As the number of moles of [tex]CoSO_4[/tex] =[tex]\frac{Given mass }{molar mass}[/tex]

                                                       =[tex]\frac{1.817}{155}[/tex]

                                                          =0.01172

Molarity = [tex]\frac{no. of moles}{volume of solution}[/tex]

           =[tex]\frac{0.01172}{20}[/tex]

        =0.000586 moles

Excess of EDTA = concentration of EDTA - concentration of CoSO4

                            = 0.009005 - 0.000586

                           = 0.008419 M

As M1V1 ( Excess of EDTA ) = M2V2 [tex](Zn^2^+)[/tex]

           [tex]0.008419\times100ml=0.009795\times V2[/tex]

           [tex]V2=\frac{0.008419\times100}{0.009795}[/tex]

             V2 =85.952 ml

Therefore , 85.952 ml [tex]Zn^2^+[/tex] needed to titrate the excess complexing reagent .