What is the mass number of an atom with 24 protons anſ 30 neutrons?

rate=0.2*3^1*3^2=0.2*3*9=5.4(mol/L)s so the correct answer is C.

Answer:

Mg(s) + 2Ag^+(aq) ---->Mg^2+(aq) + 2Ag(s)

Explanation:

The key to balancing redox reaction equations is this; ensure that the number of electrons lost in the oxidation half reaction equation is equal to the number of electrons gained in the reduction half reaction equation. The both half reaction equations can now be combined to give the overall reaction equation.

For the redox reaction under consideration;

Oxidation half equation;

Mg(s) ------> Mg^2+(aq) + 2e

Reduction half equation;

2Ag^+(aq) +2e ----> 2Ag(s)

Overall balanced redox reaction equation;

Mg(s) + 2Ag^+(aq) ---->Mg^2+(aq) + 2Ag(s)

Answer:

The correct answer is mass of Al3+ will be 3.23 grams and the mass of Co2+ will be 8.50 grams.

Explanation:

Based on the given information, 0.3731 L of 1.735 M of NaOH is added in a solution resulting in the precipitation of the ions as Al(OH)₃ and Co(OH)₂. Thus, the moles of NaOH will be molarity × V(L) = 1.735 × 0.3731 L = 0.647 moles.  

The mass of the precipitate given is 22.73 grams.  

Now let us assume that the mass of Al(OH)₃ will be x grams and the mass of Co(OH)₂ will be (22.73-x) grams

Therefore, the moles of Al(OH)₃ will be x grams/78 g/mol and as 3OH⁻ ions are needed so the moles will be 3x/78 mole.  

And, the moles of Co(OH)₂ will be (22.73-x)grams/92.94 g/mol and as 2OH⁻ ions are needed so the moles will be 45.46-2x/92.94 moles.

Now the equation will become,  

3x/78 + 45.46-2x/92.94 = 0.647 moles

0.03846 x + 0.489 - 0.02152 x = 0.647  

0.01694 x + 0.489 = 0.647

0.01694 x = 0.158

x = 0.158/0.01694

x = 9.327 grams

Hence, the mass of Al(OH)₃ is 9.327 grams, and the mass of Al³⁺ will be,  

= 9.327 gm/78 g/mol × 27 g/mol = 3.23 grams

Now the mass of Co(OH)₂ will be, (22.73 - 9.327) grams = 13.403 grams

the mass of Co²⁺ will be,  

= 13.403 grams / 92.94 g/mol × 58.94 g/mol = 8.50 grams

Answer:

Polarity in chemistry referred to physical properties of compounds related to solubility, melting and boiling properties.

Polarity of black pepper can be seen when black pepper is sprinkled on water. The balck pepper float on water and get displaced if touched.

It means black pepper is non-polar and have no difference in electronegativity between bonded atoms. Black pepper is so light in weight and non-polar, the surface tension of water keep it floating in the water.

Answer:

10.96 grams of compound C will be produced from 18.24 grams of compound A and excess compound B.

Explanation:

3A + 2B ⇒ 5C

By stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) the following amounts of reagent and products participate in the reaction:

The excess reagent will be that which is not completely depleted during the reaction.

The amount of product obtained from the reaction will always depend on the amount of limiting reagent in the reaction. Then, being B the excess reagent and therefore A the limiting reagent and knowing that compound A has a molar mass of 159.7 g/mole and compound C has a molar mass of 57.6 g/mole, by stoichiometry the following mass amounts of A and C participate in the reaction:

Then it is possible to apply the following rule of three: if by stoichiometry of the reaction 479.1 grams of A produce 288 grams of C, 18.24 grams of A, how much mass of C does it produce?

[tex]mass of C=\frac{18.24 grams of A*288 grams of C}{479.1 grams of A}[/tex]

mass of C= 10.96 grams

10.96 grams of compound C will be produced from 18.24 grams of compound A and excess compound B.

Answer:

The base is involved in the rate determining step of an E2 reaction mechanism

Explanation:

Let us get back to the basics. Looking at an E1 reaction, the rate determining step is unimolecular, that is;

Rate = k [Carbocation] since the rate determining step is the formation of a carbonation.

For an E2 reaction however, the reaction is bimolecular hence for the rate determining step we can write;

Rate = k[alkyl halide] [base]

The implication of this is that an excess of either the alkyl halide or base will facilitate an E2 reaction.

Hence, when excess base is used, E2 reaction is favoured since the base is involved in its rate determining step. In an E1 reaction, the base is not involved in the rate determining step hence an excess of the base has no effect on an E1 reaction.

Answer:

10.328 m

Explanation:

normal atmospheric pressure = 101325 Pa

density of water at 25 °C = 1.0 g/cm^3 = 1000 kg/m^3

pressure = pgh

where p = density

g = acceleration due to gravity = 9.81 m/s^2

h = height of column

imputing values, we have

101325 = 1000 x 9.81 x h

height of column h = 101325/9810 = 10.328 m

Answer:

WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.

Explanation:

Equation for the reaction:

K2SO4 + H20 ------->2 K+ + SO4^2-

When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.

1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-

At STP, 1 mole of K2SO4 will be the molar mass of the substance

Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol

Molar mass = 174 g/mol

So therefore;

1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion

When 63.7 g is used; we have:

174 g = 2 moles of K+

63.7 g = ( 63.7 * 2 / 174) moles of K+

= 0.73 moles of K+

Forr sulfate ion, we have:

174 g = 1 mole ofSO4^2-

63.7 g = (63.7 * 1 / 174) moles of SO4^2-

= 0.366 moles of SO4^2-

In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.

Answer:

The new solution is 1.4% m/V

Explanation:

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

We have 2.5 mL (V₁) of a concentrated solution and add it to 22.5 mL of distilled water. Assuming the volumes are additives, the volume of the new solution (V₂) is:

[tex]2.5 mL + 22.5 mL = 25.0 mL[/tex]

We want to prepare a dilute solution from a concentrated one, whose concentration is 14% m/V (C₁). We can calculate the concentration of the dilute solution (C₂) using the dilution rule.

[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{14\% m/V \times 2.5 mL}{25.0 mL} = 1.4 \% m/V[/tex]

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

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Answer:

3,4,1 and 6,5,2

Explanation:

In the electromagnetic spectrum the arrangement of the waves in increasing frequencies and decreasing wavelengths are as follows;

Radio waves

Microwaves

Infrared waves

Visible light rays

Ultraviolet rays

X-rays

Gamma rays

(a simple mnemonic is RMIVUXG)

Answer:

[tex]n_{Cl^-}=25molCl^-[/tex]

Explanation:

Hello,

In this case, since the given 5-M concentration of magnesium chloride is expressed as:

[tex]5\frac{molMgCl_2}{L}[/tex]

We can notice that one mole of salt contains two moles of chloride ions as the subscript of chlorine is two, in such a way, with the volume of solution we obtain the moles of chloride ions as shown below:

[tex]n_{Cl^-}=5\frac{molMgCl_2}{L}*\frac{2molCl^-}{1molMgCl_2} *2.5L\\\\n_{Cl^-}=25molCl^-[/tex]

Best regards.

Answer:

[tex]14\\8[/tex]O

Explanation:

The top number always represents the mass number.

The bottom number always represents the atomic number.

The element always goes after the numbers.

If charge is present, that comes after the element.

Answer:

680g of NaNO3.

Explanation:

The balanced equation for the reaction is given below:

2NaNO3 —> 2NaNO2 + O2

Next, we shall determine the mass of NaNO3 that decomposed and the mass of O2 produced from the balanced equation. This is illustrated below:

Molar mass of NaNO3 = 23 + 14 + (16x3) = 85g/mol

Mass of NaNO3 from the balanced equation = 2 x 85 = 170g

Molar mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 1 x 32 = 32g

From the balanced equation above,

170g of NaNO3 decomposed to produce 32g of O2.

Now, we can obtain the mass of NaNO3 needed to produce 128g of O2 as shown below:

From the balanced equation above,

170g of NaNO3 decomposed to produce 32g of O2.

Therefore, Xg of NaNO3 will decompose to produce 128g of O2 i.e

Xg of NaNO3 = (170 x 128)/32

Xg of NaNO3 = 680g

Therefore, 680g of NaNO3 are needed to produce 128g of O2.

Answer:

The correct answer is -  13.33 kJ of heat

Explanation:

To know which one is the limiting reagent, determine the number of moles of each reagent in order .

n(K) = mass/atomic weight = 1.41/39 = 0.036 moles

Density of ICl = Mass/Volume

3.24 = Mass/6.52

Mass of ICl = 21.12 g

n(ICl) = mass/molar mass = 21.12/162.35 = 0.130 moles

2 moles of K reacts with 1 mole of ICl

0.036 moles of K will react with = 0.036/2 = 0.018 moles of ICl

since the amount of moles of ICl is more than 0.018, it is in excess and hence K is the limiting reagent. Now, use the balance equation to determine the amount of heat liberated:

2 moles of K gives out -740.71 kJ of heat

1 mole of K will give out = -740.71/2 = 370.36 kJ of heat

0.036 moles of K will give out = 0.036 × 370.36 = 13.33 kJ of heat

Thus, the correct answer is -  13.33 kJ of heat

The amount of heat liberated at constant pressure is -13.33 kJ

The given parameters are:

Start by calculating the number (n) of moles of each reagent using:

[tex]n = \frac{Mass}{Atomic\ weight }[/tex]

For the potassium metal, we have:

[tex]n_k = \frac{1.41g}{39g/mole}[/tex] ---where 39 is the atomic weight of potassium

[tex]n_k = 0.036\ moles[/tex]

For the liquid iodine monochloride, we start by calculating its mass using:

[tex]Mass = Density \times Volume[/tex]

So, we have:

[tex]Mass = 3.24 \times 6.52[/tex]

[tex]Mass = 21.12g[/tex]

The number of moles is then calculated as:

[tex]n_I=\frac{21.12g}{162.35}[/tex]

[tex]n_I = 0.130\ moles[/tex]

The reaction equation 2K(s) + ICl(l) → KCl(s) + KI(s) means that:

2 moles of potassium reacts with 1 mole of liquid iodine monochloride.

So,  0.036 moles of potassium will react with the following moles of liquid iodine monochloride.

[tex]A = \frac{0.036}{2}[/tex]

[tex]A = 0.018\ moles[/tex]

i.e. 0.036 moles of potassium will react with of liquid iodine monochloride

By comparison: 0.018 moles is less than 0.036 moles

So, the amount of heat liberated at constant pressure is:

[tex]Amount = 0.036 \times -\frac{740.71}{2} kJ[/tex]

[tex]Amount = -13.33 kJ[/tex]

Hence, the amount of heat liberated at constant pressure is -13.33 kJ

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Answer:

Option B. 0.136 g

Explanation:

The balanced equation for the reaction is given below:

2AgNO3(aq) + 2NaOH(aq) —> Ag2O(s) + 2NaNO3(aq) + H2O(l)

Next, we shall determine the masses of AgNO3 and NaOH that reacted and the mass of Ag2O produced from the balanced equation. This is illustrated below:

Molar mass of AgNO3 = 108 + 14 + (16x3) = 170g/mol

Mass of AgNO3 from the balanced equation = 2 x 170 = 340g

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of Ag2O = (108x2) + 16 = 232g/mol

Mass of Ag2O from the balanced equation = 1 x 232 = 232g

Summary:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH to produce 232g of Ag2O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH.

Therefore, 0.2g of AgNO3 will react with = (0.2 x 80)/340 = 0.047g of NaOH.

From the calculations made above, only 0.047g out of 0.2g of NaOH given, reacted completely with 0.2g of AgNO3. Therefore, AgNO3 is the limiting reactant and NaOH is the excess reactant.

Now, we can calculate the mass of Ag2O produced from the reaction of 0.2g of AgNO3 and 0.2g of NaOH.

In this case, we shall use the limiting reactant because it will produce the maximum yield of Ag2O as all of it is used up in the reaction.

The limi reactant is AgNO3 and the mass of Ag2O produced can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted to produce 232g of Ag2O.

Therefore, 0.2g of AgNO3 will react to produce = (0.2 x 232)/340 = 0.136g of Ag2O.

Therefore, 0.136g of Ag2O was produced from the reaction.

Answer:

pH = 3.39

Explanation:

The equilibrium in water of ascorbic acid (With its conjugate base) is:

H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)

Where the acidic dissociation constant is written as:

Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]

H₂O is not taken in the Ka expression because is a pure liquid.

As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:

[H₂C₆H₆O₆] = 2.5x10⁻³M - X

[HC₆H₆O₆⁻] = X

[H₃O⁺] = X

Replacing in the Ka expression:

7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]

1.975x10⁻⁷ - 7.9x10⁻⁵X = X²

0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷

Solving for X:

X = -0.00048566→  False solution, there is no negative concentrations

X = 0.00040666 → Right solution

As [H₃O⁺] = X, [H₃O⁺] = 0.00040666

pH is defined as -log [H₃O⁺];

pH = -log 0.00040666,

Answer:

A. [OH⁻] = 2.188x10⁻³M

B. pKb = 6.02

Explanation:

When hydrazine is in equilbrium with water, its reaction is:

N₂H₄(aq) + H₂O(l) ⇄ HN₂H₄⁺(aq) + OH⁻(aq)

Where Kb, is defined as the ratio between concentrations in equilibrium of the species, thus:

Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]

A. From pH, you can find [OH⁻], thus:

pH = -log [H⁺]

11.34 = -log [H⁺]

4.57x10⁻¹² = [H⁺]

As 1x10⁻¹⁴ = [OH⁻] [H⁺]

1x10⁻¹⁴ / 4.57x10⁻¹² = [OH⁻]

B. Concentrations in equilibrium of the species are:

[N₂H₄] = 5.0M - X

[HN₂H₄⁺] = X

[OH⁻] = X

Where X is reaction coordinate

As [OH⁻] = 2.188x10⁻³M

X = 2.188x10⁻³M

Replacing:

[N₂H₄] = 5.0M - 2.188x10⁻³M = 4.9978M

[HN₂H₄⁺] = 2.188x10⁻³M

[OH⁻] = 2.188x10⁻³M

Replacing in Kb expression:

Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]

Kb = [2.188x10⁻³M] [2.188x10⁻³M] / [4.9978M]

Kb = 9.577x10⁻⁷

pKb is defined as -log Kb

pKb = -log 9.577x10⁻⁷

Answer:

pH at equivalence point is 8.47

Explanation:

Benzoic acid react with NaOH, thus:

HC₇H₅O₂ + NaOH → C₇H₅O₂⁻ + H₂O + Na⁺

You reach equivalence point when moles of the acid = moles of NaOH.

Moles of benzoic acid are:

0.025L ₓ (0.0988mol / L) = 0.00247 moles

To have 0.00247 moles of NaOH in solution and reach equivalence point you need to add:

0.00247 moles NaOH ₓ (1L / 0.115mol) = 0.0215L of NaOH solution.

Total volume is 0.0465L.

There are produced 0.00247 moles of C₇H₅O₂⁻ and its molarity will be:

0.00247 mol C₇H₅O₂⁻ / 0.0465L = 0.0531M C₇H₅O₂⁻

C₇H₅O₂⁻ is in equilibrium with water, thus:

C₇H₅O₂⁻(aq) + H₂O ⇄ HC₇H₅O₂(aq) + OH⁻(aq)

Where Kb = Kw / Ka = 1x10⁻¹⁴ / 6.5x10⁻⁵ = 1.54x10⁻¹⁰ is:

Kb = 1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]

The concentrations in equilibrium of the species are:

[HC₇H₅O₂] = X

[OH⁻] = X

[C₇H₅O₂⁻] = 0.0531M - X

Where X represents how much C₇H₅O₂⁻ react, X is reaction coordinate

Replacing in Kb expression:

1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]

1.54x10⁻¹⁰ = [X] [X] / [0.0531 - X]

8.169x10⁻¹² - 1.54x10⁻¹⁰X = X²

8.169x10⁻¹² - 1.54x10⁻¹⁰X - X² = 0

Solving for X:

X = -2.858x10⁻⁶M → False solution, there is no negative concentrations

X = 2.858x10⁻⁶M → Right solution

As [OH⁻] = X

[OH⁻] = 2.858x10⁻⁶M

pOH is -log [OH⁻]

pOH = 5.54

pH = 14 - pOH

pH = 8.46

Answer:

The correct answer is 0.00080 gram per liter.

Explanation:

Based on the given information, the solubility of water is 0.00380 gram per liter, the temperature mentioned is 25 degree C, the partial pressure of oxygen gas is 760 torr, and the mole fraction of oxygen is 0.210. There is a need to determine the solubility of oxygen in water.  

Based on Henry's law,  

Solubility of oxygen gas = Henry's constant × partial pressure of oxygen gas

Henry's constant, K = solubility of oxygen gas / partial pressure of oxygen gas

= 0.00380 g/L × 1 mol/32 grams / 760 torr × 1 atm/760 torr

= 0.00012 mol/L/atm

= 0.00012 M/atm

Now the partial pressure of the oxygen gas = mole fraction of oxygen × atmospheric pressure

= 0.210 × 1 atm

= 0.210 atm

Now putting the values in Henry's law equation we get,  

Solubility of oxygen gas = 0.00012 mol/L/atm × 0.210 at,

= 0.000025 mol/L × 32 gram/mol

= 0.00080 gram per liter

Answer: That most of an atom's mass was packed in a central nucleus

Answer:

That most of an atom's mass was packed in the central nucleus.

Explanation:

Rutherford disproved Thomson's model in 1911 with his gold foil experiment, in which he demonstrated that the atom has a high-mass nucleus.

Answer:

B. Valence Electron Pairs

Explanation:

Valence-shell electron-pair repulsion, or VSEPR, describes the shape of molecules by determining the repulsion of valence electrons. Therefore, our answer is B.

Answer:

0.48 dm3  (or 480 cm3)

Explanation:

First find the original no. of moles existing in the sulphuric acid:

no. of moles = volume (in dm3) x concentration

                     = 120/1000 x 10

                     = 1.2 mol

Then let the total volume of the diluted acid be v dm3.

Since

Concentration = no. of moles / volume,

so by substituting the given information,

2 = 1.2 / v

v = 0.6 dm3

Hence, the volume of water required

= 0.6 - 120/1000

= 0.48 dm3  (or 480 cm3)

Considering the definition of dilution, 600 cm³ of water is required to dilute 120 cm³ of 10 [tex]\frac{mol}{dm^{3} }[/tex] sulphuric acid to a concentration of 2 [tex]\frac{mol}{dm^{3} }[/tex].

First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.

Dilution is the procedure followed to prepare a less concentrated solution from a more concentrated one and consists of reducing the amount of solute per unit volume of solution. This is accomplished simply by adding more solvent to the solution in the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

In this case, you know:

Replacing in the definition of dilution:

10[tex]\frac{mol}{dm^{3} }[/tex]× 120 cm³= 2 [tex]\frac{mol}{dm^{3} }[/tex]× Vf

Solving:

Vf= (10[tex]\frac{mol}{dm^{3} }[/tex]× 120 cm³) ÷2 [tex]\frac{mol}{dm^{3} }[/tex]

Vf= 600 cm³

In summary, 600 cm³ of water is required to dilute 120 cm³ of 10 [tex]\frac{mol}{dm^{3} }[/tex] sulphuric acid to a concentration of 2 [tex]\frac{mol}{dm^{3} }[/tex].

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Answer:

This reaction isn't yet at an equilibrium. It must shift in the direction of the reactant (namely [tex]\rm NOBr\; (g)[/tex]) in order to reach an equilibrium.

For this mixture, the reaction quotient is [tex]Q_c = 0.0126[/tex].

Explanation:

A reversible reaction is at equilibrium if and only if its reaction quotient [tex]Q_c[/tex] is equal to the equilibrium constant [tex]K_c[/tex].

Start by calculating the equilibrium quotient [tex]Q_c[/tex] of this reaction. Given the reaction:

[tex]\rm 2\; NOBr\; (g) \rightleftharpoons 2\; NO\; (g) + Br_2\; (g)[/tex].

Let [tex][\mathrm{NOBr\; (g)}][/tex], [tex][\mathrm{NO\; (g)}][/tex], and [tex][\mathrm{Br_2\; (g)}][/tex] denote the concentration of the three species. The formula for the reaction quotient of this system will be:

[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2}[/tex].

(Note, that in this formula, both [tex][\mathrm{NO\; (g)}][/tex] and [tex][\mathrm{NOBr\; (g)}][/tex] are raised to a power of two. That corresponds to the coefficients in the balanced reaction.)

Calculate the reaction quotient given the concentration of each species:

[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2} \approx 1.26\times 10^{-2} = 0.0126[/tex].

(Note that the unit is ignored.)

Apparently, [tex]Q_c > K_c[/tex]. Since [tex]Q_c[/tex] and [tex]K_c[/tex] are not equal, this reaction is not at an equilibrium. If external factors like temperature stays the same,

Keep in mind that [tex]Q_c[/tex] denotes a quotient. To reduce the value of a quotient, one may:

In [tex]Q_c[/tex], that means reducing the concentration of the products while increasing the concentration of the reactants. In other words, the system needs to shift in the direction of the reactants before it could reach an equilibrium.

Answer:

550 m/s

Explanation:

The average molecular speed (v) is the speed associated with a group of molecules on average. We can calculate it using the following expression.

[tex]v = \sqrt{\frac{3 \times R \times T}{M} }[/tex]

where,

We can use the info of argon to calculate the temperature for both samples.

[tex]T = \frac{v^{2} \times M}{3 \times R} = \frac{(391m/s)^{2} \times 39.95g/mol}{3 \times 8.314J/k.mol} = 2.45 \times 10^{5} K[/tex]

Now, we can use the same expression to find the average molecular speed in a sample of Ne gas.

[tex]v = \sqrt{\frac{3 \times R \times T}{M} } = \sqrt{\frac{3 \times (8.314J/k.mol) \times 2.45 \times 10^{5}K }{20.18g/mol} } = 550 m/s[/tex]

Answer:

7p

Explanation:

principal quantum number is 7

n=7( principle shell)

angular momentum quantum number gives sub shell

l = 1 means it is p orbital

so answer is 7p orbital

Answer:

B. Reactant + Reactant -> Product + Product

Explanation:

Reactants are substances that- as the name suggests- reacts with other substances at the beginning of a reaction

Products are substances that are produced as a result of the reaction

Typically, when writing a chemical reaction, an arrow is used to show the direction the reaction is moving.  In this case, the arrows in options B and C suggest that the reaction only moves in one direction- forwards

And as mentioned above, reactants are the substances at the start of the reaction, they're what mixes together to form a new product.  

To keep things simple:

Products can't be at the beginning of a reaction since they weren't formed yet.

Similarly, reactants can't be part of the products since they already existed and didn't need to be made. In a lot cases, the reactants would be completely used up to make the products

As such, only one possible chemical reaction would follow that reasoning:

    Reactant + Reactant ->  Product + Product

Reactant + Reactant → Product + Product is the correctly written chemical reaction. Hence, option B is correct.

A chemical equation is a mathematical expression of the chemical reaction which represents the product formation from the reactants.

In an equation, the reactants are written on the left-hand side and the products are written on the right-hand side demonstrated by one-headed or two-headed arrows.

Hence, option B is correct.

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Answer:

The element is strontium and the number of neutrons it have is 51.

Explanation:

Based on the given information, the ionic compound is,  

XCl₂ ⇔ X₂⁺ + 2Cl⁻

X2+ is the ion of the mentioned element

As mentioned in the given question, the number of electrons of the element X is 36 and as seen from the reaction the charge present on the ion is +2. Now the atomic number will be,  

No. of electrons = atomic number - charge

36 = atomic number - 2

Atomic number = 38

Based on the periodic table, the atomic number 38 is for strontium element, and the sign of strontium is Sr. Hence, the element X is Sr.  

Now based on the given information, the mass number of the element is 89. Now the no. of neutrons will be,  

No. of neutrons = mass number - atomic number

= 89 - 38

= 51 neutrons.  

Answer:

You can take 1mL of your stock solution in a 100mL volumetric flask and complete to volume.

Explanation:

You need to create a 0.00200M solution of Fe(NO₃)₃. First, you have to obtain the concentration of the first solution you made. That is:

0.8g Fe(NO₃)₃.9H₂O × (1mol / 403.9972g) =

0.0020 moles of Fe(NO₃)₃.9H₂O = Moles of Fe(NO₃)₃

In 10mL = 0.010L:

0.0020 moles of Fe(NO₃)₃ / 0.010L = 0.20M Fe(NO₃)₃

This is the concentration of your stock solution, as you want to obtain a 0.0020M solution, you dilution factor must be:

0.20M / 0.0020M = 100

That means you need to dilute your stock solution 100 times.

You can make this dilution, for example,

Answer:

Option B

Explanation:

Scientific question are answered through experimentation, through testing the theory about the physical world.

Answer: its A

through observing and measuring the physical world

Explanation: