Li atoms in 3.7 moles

Answer:

[tex]Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)[/tex]

Explanation:

Hello.

In this case, for the reaction between aqueous solutions of ammonium chloride and iron (III) hydroxide, we have the following complete molecular reaction:

[tex]3NH_4Cl(aq)+Fe(OH)_3(s)\rightarrow 3NH_4OH+FeCl_3[/tex]

And the full ionic equation, taking into account that the iron (III) hydroxide cannot be dissolved as it is insoluble in water:

[tex]3NH_4^+(aq)+3Cl^-(aq)+Fe(OH)_3(s)\rightarrow 3NH_4^+(aq)+3OH^-(aq)+Fe^{3+}(aq)+3Cl^-(aq)[/tex]

Finally, the net ionic equation, considering that spectator ions are NH₄⁺, Cl⁻ as they are both the left and right side, therefore, the net ionic equation is:

[tex]Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)[/tex]

Best regards.

The net ionic equation for the reaction of aqueous solutions should be Fe(Oh)3 ➡Fe3+(aq) + 3OH^-(aq).

When the reaction lies between the between aqueous solutions of ammonium chloride and iron (III) hydroxide

So, here the total reaction should be

3NH4Cl(aq) + Fe(OH)3(s) ➡ 3NH4OH + FeCl3

So, here net ionic equation, considering that spectator ions are NH₄⁺, Cl⁻ since they are both the left and right sides.

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The correct answer is A. Electron

Explanation:

The model of this atom depicts the nucleus of this in the center of the model, this section of the atom contains sub-particles known as protons and neutrons. Moreover, in the atom, the nucleus is surrounded by three sub-particles that orbit or move around the nucleus. These sub-particles are the electrons; these differ from other sub-particles because they have a negative charge and they are not part of the nucleus. Also, these move around the nucleus is orbits, although they move similarly to waves. According to this, the correct answer is A.

Answer:

I'm on the exact same queston

Answer:

The test variable (independent variable) controls the outcome variable (dependent variable)

Explanation:

its right on study island

Answer:

about 0.13 mol

Explanation:

To find number of mols when given grams you first have to find the molar mass of the compound. This is done by adding up the atomic masses of the element in the compound. So Sr= 88 g/mol S=32 g/mol and O=16 g/mol. Then 88+32+(16x4)=184. Then using this you can convert from grams to mols by dividing the grams by the molar mass. So, 24.23/184 equals about 0.13 mol.

Answer:

6.7

⋅

10

23

atoms of H

Explanation:

Answer:

70.88 mL volume of 1.27 M of HCl is required.

Explanation:

Given data:

Initial volume = ?

Initial  molarity =  1.27 M

Final volume = 197.4 mL

Final molarity = 0.456 M

Solution:

Formula:

M₁V₁ = M₂V₂

Now we will put the values in formula.

1.27 M × V₁ =  0.456 M × 197.4 mL

V₁ = 0.456 M × 197.4 mL/1.27 M

V₁ = 90.014M.mL/1.27 M

V₁ = 70.88 mL

70.88 mL volume of 1.27 M of HCl is required.

Answer:

The empirical formula for C12 H24 O6 is C2 H4 O.

Answer:

We are given the formula of the compound:

C12H24O6

The empirical formula of a molecular formula is the lowest whole number ratio between the number of atoms of each element

The ratio of C to H to O in the given formula is :

12 : 24 : 6

we notice that all 3 of the numbers have 6 in common. Dividing all three of the numbers by 6, we get:

2 : 4 : 1

Hence, the ratio of Carbon to Hydrogen to Oxygen in the empirical formula of the given compound is 2 : 4 : 1 ,

Empirical Formula = C2H4O

Answer:

An element is atoms with the same number of protons.

Explanation:

Protons, electrons, and neutrons.

Answer:

Explanation:

In order to find the original pressure , we use the formula for Boyle's law which is

[tex]P_1V_1 = P_2V_2[/tex]

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the original volume

[tex]V_1 = \frac{P_2V_2}{P_1} \\[/tex]

From the question

P1 = 250020 Pa

P2 = 120870 Pa

V2 = 6 L

We have

[tex]V_1 = \frac{120870 \times 6}{250020} = \frac{725220}{250020} \\ = 2.90064794...[/tex]

We have the final answer as

Hope this helps you

Answer:

41.7 kJ/mol

Explanation:

ln(k) = ln(A) − Eₐ/(RT)

Pick any two points.  I'll choose 100°C and 400°C.

When T = 100°C = 373 K, k = 1.10×10⁻⁹ L/mol s:

ln(1.10×10⁻⁹) = ln(A) − Eₐ/(R × 373)

When T = 400°C = 673 K, k = 4.40×10⁻⁷ L/mol s:

ln(4.40×10⁻⁷) = ln(A) − Eₐ/(R × 673)

Subtract the two equations and solve:

ln(4.40×10⁻⁷) −  ln(1.10×10⁻⁹) = -Eₐ/(R × 673) + Eₐ/(R × 373)

5.991 = 0.00120 Eₐ/R

Eₐ/R = 5013.4

Eₐ = 41700 J/mol

Eₐ = 41.7 kJ/mol

Answer:

[tex]T_f=25.0\°C[/tex]

Explanation:

Hello.

In this case, considering that the sample of hot copper is submerged into the water and the container is isolated, the heat lost by the copper is gained by the water so we can write:

[tex]Q_{Cu}=-Q_w[/tex]

In terms of mass, specific heat and temperature we write:

[tex]m_{Cu}C_{Cu}(T_f-T_{Cu})=-m_wC_w(T_f-T_w)[/tex]

Whereas the final temperature is the same for both copper and water because they are in contact until thermal equilibrium is reached. In such a way, the required maximum temperature no more than the equilibrium temperature and is computed as shown below:

[tex]T_f=\frac{m_{Cu}C_{Cu}T_{Cu}+m_wC_wT_w}{m_{Cu}C_{Cu}+m_wC_w}[/tex]

Thus, plugging the given data in the formula, we obtain:

[tex]T_f=\frac{10.3g*0.385\frac{J}{g\°C}*100\°C +47.0g*4.184\frac{J}{g\°C}*23.5\°C }{10.3g*0.385\frac{J}{g\°C}+47.0g*4.184\frac{J}{g\°C}}\\\\T_f=25.0\°C[/tex]

Which is a small change considering the initial one, because the mass of water is greater than the mass of copper as well as for the specific heats.

Best regards!

The maximum temperature of the water in the insulated container after the copper metal is added is 25 °C

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

Mass of water (Mᵥᵥ) = 47 g

Temperature of water (Tᵥᵥ) = 23.5°C

Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Mass of copper (M꜀) = 10.3 g

Temperature of copper (M꜀) = 100 °C

Specific heat capacity of copper (C꜀) = 0.385 J/gºC

The equilibrium temperature of the mixture can be obtained as follow:

Heat loss by copper = Heat gained by water

Q꜀ = Qᵥᵥ

M꜀C꜀(M꜀ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ– Mᵥᵥ)

10.3 × 0.385 (100 – Tₑ) = 47 × 4.184 (Tₑ – 23.5)

3.9655 (100 – Tₑ) = 196.648 (Tₑ – 23.5)

Clear bracket

396.55 – 3.9655Tₑ = 196.648Tₑ – 4621.228

Collect like terms

396.55 + 4621.228 = 196.648Tₑ + 3.9655Tₑ

5017.778 = 200.6135Tₑ

Divide both side by 200.6135

Tₑ = 5017.778 / 200.613

Thus, the equilibrium temperature of the mixture is 25 °C. Therefore, the maximum temperature of the water in the insulated container is 25 °C

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Answer: 0 I think

Explanation:

pretty sure its zero because  i learned it last year but im in middle school so you might want to look it up.

Answer:

In this example, a 3 kilogram mass, at a height of 5 meters, while acted on by Earth's gravity would have 147.15 Joules of potential energy, PE = 3kg * 9.81 m/s 2 * 5m = 147.15 J. 9.81 meters per second squared (or more accurately 9.80665 m/s 2 ) is widely accepted among scientists as a working average value for Earth's gravitational pull.

Explanation:

Answer:

29.8

Explanation:

The formula for volume is mass/ density, so 59/1.98. 29.8 is the answer.

Answer:

175

Explanation:

Answer:

The Group 2 metals become more reactive towards the water as you go down the Group.

Explanation:

These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.

Please mark me brainliest! hope this helped!

God bless!

Answer:

C

Explanation:

Because sodium is basically always soluble with any compound, it is between a and c. a is part of the reactant so it cant be A. So C.

The statement, that describes the compounds would be the precipitate that forms in the reaction is "Ba3(PO4)2"

Precipitate is a solid generated by a change in a solution, usually due to a chemical reaction or a change in temperature that reduces a solid's solubility.

The combination of more than one element will be identified ad compound.

When cations and anions in aqueous solution combine to create an insoluble ionic solid called a precipitate, double displacement reactions occur, resulting in the formation of a solid form residue. Except for salts of Group 1 metals and ammonium, salts of phosphates and carbonates ions are insoluble, according to the solubility flow chart. The production of a solid white precipitate is used to demonstrate the insoluble nature of barium phosphate.

Hence the correct option is c.

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Answer:

Salt domes storage has advantages in cost, security, environmental risk, and maintenance. Salt formations offer the lowest cost, most environmentally secure way to store crude oil for long periods of time. Stockpiling oil in artificially-created caverns deep within the rock-hard salt costs historically about $3.50 per barrel in capital costs. Storing oil in above ground tanks, by comparison, can cost $15 to $18 per barrel - or at least five times the expense. Also, because the salt caverns are 2,000-4,000 feet below the surface, geologic pressures will sea; any crack that develops in the salt formation, assuring that no crude oil leaks from the cavern. An added benefit is the natural temperature differential between the top of the caverns and the bottom - a distance of around 2,000 feet; the temperature differential keeps the crude oil continuously circulating in the caverns, giving the oil a consistent quality.

Answer:

Boron and Aluminum

Explanation:

If you write the electron configuration for boron and aluminum, you get:

[tex]1s^22s^22p^1[/tex] for boron and [tex]1s^22s^22p^63s^23p^1[/tex] for aluminum. Both have 3 valance electrons and has 2 electrons in a s-orbital and 1 in a p-orbital. These valance electron similarities are based on the column/group the elements are. Therefore, Boron and Aluminum have similar chemical behaviours and similar arrangement of outer/valance electrons.

Answer:

it is outside the nucleus  F

it has a positive charge  T

it has no mass  F

it has a negative charge  F

it is inside the nucleus  ...it is part OF the nucleus.

it is the same as the atomic number  T

it is the same as the number of neutrons  F

75% of the isotopes have a mass ima just guess cuz i dunno about this one...i think it matters on the atom element.

Explanation:

Answer:

acceleration = force/mass

                 = (68.6+mg)/7

                 = 19.6 m/s²

Explanation:

9.8 m/s² is the acceleration acting on 7 kg mass if force of 68.6 N  is used to move it towards earth.

Force is defined as a cause which is capable of changing the motion of an object. It can cause an object which has mass to change it's velocity. It is also simply a push or a pull . It has both magnitude as well as direction.Hence, it is a vector quantity.

It has SI units of Newton and is represented by'F'.Newton's second law states that force which acts on an object is equal to momentum which changes with time. If mass of object is constant, acceleration is directly proportional to net force acting on an object.

The concepts which related to force are thrust and torque .Thrust increases the velocity of an object and torque produces change in rotational speed of an object.

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Answer:

ΔG° = -1.45 × 10⁵ J

Explanation:

Step 1: Given data

Step 2: Calculate the standard Gibbs free energy change (ΔG°)

We will use the following expression.

ΔG° = -n × F × E°cell

ΔG° = -1 mol e⁻ × 96,485 C/mol e⁻ × 1.50 V

ΔG° = -1.45 × 10⁵ J

By apply Gibbs's free energy, the value of delta G is equal to -144727.5 Joules.

Given the following data:

To determine the value of delta G, we would apply Gibbs's free energy:

Mathematically, Gibbs's free energy is given by the formula:

[tex]\Delta G^\circ = -nFE^{ \circ}_{cell}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\Delta G^\circ = -1 \times 96485 \times 1.50[/tex]

[tex]\Delta G^\circ = -144727.5[/tex]

Delta G = -144727.5 Joules

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Answer:

I am sure they are why won't they

Answer:

6.02 × 10²³ atoms Cl₂

Explanation:

Avagadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Step 1: Define

1.00 mol Cl₂ (g)

Step 2: Use Dimensional Analysis

[tex]1.00 \hspace{3} mol \hspace{3} Cl_2(\frac{6.02(10)^23 \hspace{3} atoms \hspace{3} Cl_2}{1 \hspace{3} mol \hspace{3} Cl_2} )[/tex] = 6.02 × 10²³ atoms Cl₂

Answer:

From fastest to slowest its: (4)A to B, (1)E to F, (3)C to D, (2)D to E

Explanation:

The steeper the line is the faster she went. D to E she didn't make any progress because the line is straight. Sry I'm terrible at explaining things.

Complete Question

The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).

(a) before addition of any HCl (b) after addition of 25.0 mL of HCl

Answer:

a The value  is  [tex]pH =12.81[/tex]

b [tex]pH  = 11.9[/tex]

Explanation:

From the question we are told that

  The first pKb value  for B is [tex]pK_b_1  =  2.10[/tex]

   The second pKb value  for B is [tex]pK_b_2  =  7.54[/tex]

     The volume is  [tex]V =   50.0 mL   =[/tex]

     The  concentration  of  B is  [tex][B]  =  0.60 M[/tex]

     The concentration of [tex]C_A =  0.60 M[/tex]

Generally the reaction equation showing the first dissociation of B is  

[tex]\ce{B_{(aq) } + H_2O _{(l)} <=> BH^+ _{(aq)}  +  OH^- _{(aq)} }[/tex]

Here the ionic  constant for B is mathematically represented as

      [tex]K_i  =  \frac{[BH^+] [OH^-]}{[B]}[/tex]

Let denot the concentration of  [BH^+]  as  z  and  since [tex][BH^+] =  [OH^-][/tex] then [tex][OH^-][/tex] is also  z

So  [B] =  0.60  -  z  

Here [tex]K_i[/tex] is ionic constant for the first reaction of a dibasic base B and the value is

   [tex]K_i  =  7.94 *10^{-3}[/tex]

So

      [tex] 7.94 *10^{-3}=  \frac{z^2}{ 0.60 - z}[/tex]

=>   [tex]z^ 2 + 0.00794 z - 0.00476[/tex]

using quadratic formula to solve this equation

     [tex]z = 0.0651[/tex]

Hence the concentration of  [tex]OH^{-}[/tex] is   [tex][OH^-] =0.0651[/tex]

Generally  [tex]pOH =  -log [OH^-][/tex]

=>    [tex]pOH =  -log (0.065)[/tex]

=>    [tex]pOH = 1.187 [/tex]

Generally the pH is mathematically represented as

    [tex]pH = 14 - 1.187[/tex]

      [tex]pH =12.81[/tex]

Generally the volume of [tex]HCl[/tex] at the second dissociation of the base B is   [tex] 50 mL [/tex]

The volume of the [tex]HCl[/tex] half way to the first dissociation of the base is 25mL

Now the pOH at half way to the first dissociation of the base is  

     [tex]pOH  =  -log(K_i)[/tex]

=>   [tex]pOH  =  -log(0.00794)[/tex]

=>   [tex]pOH  =  2.100[/tex]

Generally the pH after addition of 25.0 mL of HCl is  

    [tex]pH  =  14 -  2.100[/tex]\

=>   [tex]pH  = 11.9[/tex]

The first dissociation's equation is as follows:

[tex]B(aq) + H_2O(l) \leftrightharpoons BH^{+} (aq) + OH^{-}(aq) \\\\[/tex]

Constant of base ionization

[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 7.94\times 10^{-3} = \frac{x\times x}{(0.95- x)} \\\\\to 7.94\times 10^{-3} = \frac{x^2}{(0.95- x)} \\\\\to x^2=7.94\times 10^{-3} (0.95-x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x = 0.0830\ M\\\\[/tex]

So,

[tex]\to [OH^{-}] = 0.0830\ M\\\\[/tex]

The second dissociation of the base equation is

[tex]BH^{+}\ (aq) + H_20\ (l) \leftrightharpoons BH_2^{2+}\ (aq) + OH^{-}\ (aq) \\\\[/tex]

Constant of base ionization

[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 3.2 \times 10^{-8} =\frac{y \times (0.0830+y)}{(0.0830- y)}\\\\[/tex]

[tex]\to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y = 3.2\times 10^{-8}[/tex]

So,

[tex]\to [OH^{-}] = 0.0830\ M \\\\\to pOH = 1.08 \\\\\to pH = 14.00 - pOH = 12.92\\\\[/tex]

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Answer:

Explanation:

The percentage error of a certain measurement can be found by using the formula

[tex]P(\%) = \frac{error}{actual \: \: number} \times 100\% \\ [/tex]

From the question

actual mass = 200 g

error = 200 - 196.59 = 3.41

We have

[tex]p(\%) = \frac{3.41}{200} \times 100 \\ = 1.705[/tex]

We have the final answer as

Hope this helps you

Answer:

266Lr

Thirteen isotopes of lawrencium are currently known; the most stable is 266Lr with a half-life of 11 hours, but the shorter-lived 260Lr (half-life 2.7 minutes) is most commonly used in chemistry because it can be produced on a larger scale.

Explanation:

hopefully that helps you

Answer:

3.55 L.

Explanation:

We'll begin by calculating the number of mole in 12 g of MnO2. This can be obtained as follow:

Molar mass of MnO2 = 55 + (16×2)

= 55 + 32

= 87 g/mol

Mass of MnO2 = 12 g

Mole of MnO2 =...?

Mole = mass /Molar mass

Mole of MnO2 = 12 / 87

Mole of MnO2 = 0.138 mole

Next, we shall determine the number of mole Cl2 produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)

From the balanced equation above,

1 mole of MnO2 reacted to produce 1 mole of Cl2.

Therefore, 0.138 mole of MnO2 will also produce 0.138 mole of Cl2.

Finally, we shall determine the volume of Cl2 gas obtained from the reaction. This can be obtained as shown below:

Temperature (T) = 25 °C = 25 °C + 273 = 298 K

Pressure (P) = 0.950 atm

Number of mole (n) = 0.138 mole

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =.?

PV = nRT

0.950 × V = 0.138 × 0.0821 × 298

Divide both side by 0.950

V = (0.138 × 0.0821 × 298) / 0.950

V = 3.55 L

Therefore, 3.55 L of chlorine gas were obtained from reaction.