What volume, in milliliters, of 2.0 calcium chloride stock solution would you use to make 500 ml of 0.300 m of calcium chloride cacl2 solution?

Answer:

                                     Molarity  =  1.66 mol.L⁻¹

Explanation:

Answer:

Lead, Pb

Explanation:

Electron configuration of an element depicts the arrangement or the distribution of electrons in different energy levels (orbitals) within the atom. This arrangement is based on the Aufbau principle which states that the electrons are added such the orbital with the lowest energy gets filled up first.

The given electron configuration is:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p²

The total number of electrons = 82

For a neutral atom; number of electrons = atomic number(Z)

Based on the periodic table, the element with atomic number = 82 is lead (Pb).

The correct answer is false.

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Final answer:

The characteristics of Wassily Kandinsky's Improvisation 31 show his innovative use of color and line for their own sake, reflecting his move towards pure abstraction and exploring the emotional and spiritual qualities of art.

Explanation:

A characteristic of Wassily Kandinsky's Improvisation 31 is the use of color and line for their own sake, not to represent something specific. Kandinsky aimed to create a sense of rhythm and staccato in the painting, infusing it with musicality through abstract elements.

Improvisation 31, like many of Kandinsky's works, reflects his move towards pure abstraction and his interest in the spiritual and emotional aspects of art. The painting is a representation of the Expressionist movement in early 20th-century art, seeking to convey inner emotions rather than outward impressions.

Impressionism 31 showcases Kandinsky's innovative approach to art, breaking away from traditional representation and embracing the freedom of abstract expression, marking a significant shift in the art world. Through his use of color, line, and form, Kandinsky creates a visually dynamic and emotionally evocative piece that invites viewers to explore the depths of their subconscious.

[tex]\boxed{\text{4.79 g}}[/tex] of KBr is required to make 350 mL of 0.115 M KBr solution.

Further Explanation:

Concentration is defined as the amount or quantity of solute present in specific amount of solution. In order to evaluate concentration of various solutions, different concentration terms are used. Some of these are mentioned below.

1. Molarity (M)

2. Molality (m)  

3. Mole fraction (X)  

4. Parts per million (ppm)  

5. Mass percent ((w/w) %)  

6. Volume percent ((v/v) %)  

Molarity is defined as the number of moles of solute that can be dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate molarity of KBr solution is as follows:

  [tex]\text{Molarity of KBr solution}=\dfrac{\text{Moles of KBr}}{\text{Volume (L) of KBr solution}}[/tex]  ...... (1)

Rearrange equation (1) to calculate moles of KBr.

[tex]\begin{aligned}{\text{Moles of KBr}&=\left[{(\text{Molarity of KBr solution})\\&\text{ }\text{ }\text{ }(\text{Volume (L) of KBr solution})]\right]\end{aligned}}[/tex]        ...... (2)

Substitute 0.115 M for molarity of KBr solution and 350 mL for volume of KBr  solution in equation (2).

[tex]\begin{aligned}\text{Moles of KBr}&=(\text{0.115 M})(\text{350 mL)}\left(\dfrac{\text{10}^{-3}\text{ L}}{\text{1 mL}}\right)\\&=\text{0.04025 mol}\end{aligned}[/tex]

The formula to calculate moles of KBr is as follows:

[tex]\text{Moles of KBr}=\dfrac{\text{Mass of KBr}}{\text{Molar mass of KBr}}[/tex]                                       ...... (3)

Rearrange equation (3) for mass of KBr.

[tex]\text{Mass of KBr}=(\text{Moles of KBr})(\text{Molar mass of KBr})[/tex]               ...... (4)

Substitute 0.04025 mol for moles of KBr and 119.002 g/mol for molar mass of KBr in equation (4).

[tex]\begin{aligned}{\text{Mass of KBr}&=(\text{0.004025 mol})(\text{119.002 g/mol})\\&=\text{4.79 g}}\end{aligned}[/tex]

Therefore 4.79 g of KBr is required to make 350 mL of a 0.115 M KBr solution.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity, KBr, 350 mL, 0.115 M, 4.79 g, mass, molar mass, concentration terms, concentration, solute, moles of solute, volume.

Answer is: There is only one reactant.

Synthesis reaction is a type of reaction in which multiple reactants combine to form a single product.

For example, balanced chemical reaction: Ba + F₂ → BaF₂.

In barium fluoride, barium has oxidation number +2 and fluorine has oxidation number -1, so compound has neutral charge.

Second chemical reaction: CaO(s) + H₂O(l) → Ca(OH)₂(aq).

One molecule is produced from two molecules.

A synthesis reaction typically involves substrates bonding to form larger molecules, but the process is not always linear, may involve multiple steps and intermediates, and can include unexpected outcomes like isomerization or complex cyclization patterns.

One aspect that is not always true about a synthesis reaction is that they do not necessarily occur in a linear fashion. A synthesis reaction, in chemistry, usually involves two or more substrates reacting to form a larger molecule, commonly expressed as A + B → AB. This process generally requires energy to form new bonds. However, it is not always a straightforward single step; multiple-step synthesis may be involved, where intermediate products are formed before the final molecule is synthesized.

Moreover, certain synthesis reactions, like those leading to the formation of allenes and alkynes, may competewith isomerization processes, leading to a mix of products rather than a single, pure compound. Also, not all synthesis reactions strictly follow predictable rules like Markovikov's rule, which is often used to predict the outcome of electrophilic addition reactions. In cases where the molecular geometry affects reactivity, the result of a synthesis reaction might diverge from expected patterns.

Finally, the complexity of the molecular structure being synthesized can affect the nature of the synthesis reaction. For example, schemes involving cyclization can lead to the formation of multiple new rings in a compound, demonstrating that synthesis can be a versatile tool for constructing complex molecules.

Answer: 127 g

Explanation:

Final answer:

To make a 0.50 molal solution of NaCl with 750.0 g of water, 21.92 grams of sodium chloride must be dissolved in the water.

Explanation:

To calculate how many grams of sodium chloride (NaCl) are needed to make a 0.50 molal (m) solution with 750.0 g of water, we should understand the definition of molality. Molality is the number of moles of solute per kilogram of solvent. Therefore, a 0.50 molal solution of NaCl requires 0.50 moles of NaCl dissolved in each kilogram (1000 grams) of water. Since we have 750.0 g or 0.750 kg of water, we will adjust this proportionally.

First, calculate the moles of NaCl needed:

Next, convert the moles of NaCl to grams. The molar mass of NaCl is approximately 58.44 g/mol:

Therefore, 21.92 grams of sodium chloride must dissolve in 750.0 g of water to make a 0.50 molal solution.

Answer:

C JULY 23RD

Explanation

The approximate date of the next full moon can be estimated to be approximately 22.5 days after July 2.

The approximate date of the next full moon can be estimated by calculating the time it takes for the Moon to go through its phases. According to the information provided, it takes about 29.5 days for the Moon to complete a full cycle of phases. Since the third quarter moon occurs about a week after the full moon, we can estimate that the next full moon would be approximately 22.5 days after July 2. So, if the third-quarter moon was observed on July 2, the approximate date for the next full moon would be roughly July 9 or 10, depending on the exact time of the third-quarter phase.

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Answer:

Net ionic reaction: 2PO₄³⁻(aq) + 3Ca²⁺(aq) → Ca₃(PO₄)₂(s).

The chlorine atom released from the ultraviolet decomposition of CFCs acts as a catalyst for the decomposition of ozone in the atmosphere.

The product of the ultraviolet decomposition of chlorofluorocarbons (CFCs) that acts as a catalyst for ozone decomposition is the highly reactive chlorine atom (Cl). When CFCs like CCl₂F₂ are exposed to UV radiation, a chlorine atom is released due to the photolytic cleavage of the CFC molecule. This chlorine atom can then catalyze the destruction of ozone by participating in a series of reactions where it converts ozone (O3) into oxygen gas (O2), regenerating the chlorine atom to continue the cycle of ozone depletion.

A simplified mechanism of this process is as follows:

These reactions demonstrate how chlorine liberated from CFCs can lead to the breakdown of a large number of ozone molecules, thus significantly contributing to ozone layer depletion.

Answer:

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To find the number of moles of potassium bromide in the beaker after adding additional water, calculate the new concentration and multiply it by the new volume.

To find the number of moles of potassium bromide in the beaker, we can use the formula:

Moles = Concentration x Volume

Initially, the beaker contains 0.50 mol of potassium bromide in 600 ml of water. When an additional 600 ml of water is added, the total volume becomes 1200 ml. To find the new concentration, we divide the number of moles (0.50 mol) by the new volume (1200 ml):

Concentration = Moles / Volume = 0.50 mol / 1200 ml = 0.00042 mol/ml

The new number of moles of potassium bromide in the beaker is:

New Moles = Concentration x Volume = 0.00042 mol/ml x 1200 ml = 0.50 mol

Answer:

See explanation below

Explanation:

As the problem states, this is an acid base titration, and both titrations are already saying that they were both neutralized. When an acid base titration is neutralized, means that it reach it's equivalence point. In this point, we can say that the moles of the acid are the same moles of the base. In other words the following:

n₁ = n₂   (1)

1 is the acid and 2 is the base.

You should note that the above expression is real when the mole ratio is 1:1. When it's not, we need to see the mole ratio and then, adjust the expression to that.

the moles can also be expressed as:

n = M * V

Replacing in the first expression we have:

M₁V₁ = M₂V₂  (2)

With this expression we can calculate either the volume or concentration of the compounds given. Let's do this by parts:

Titration 1:

In this case we have KOH and H₂SO₄, so the balanced reaction would be:

2KOH + H₂SO₄ -------> K₂SO₄ + 2H₂O

As you can see, we have 2 moles of KOH and 1 mole of the acid, so the mole ratio is 2:1, therefore, expression (2) becomes:

M₁V₁ = 2M₂V₂

From here, we solve for concentration of the acid (M₁)

M₁ = 2M₂V₂ / V₁

Replacing the given values we have:

M₁ = 2 * 25 * 0.15 / 15

Now, how can we fill the chart? Is easy, we just put the obtained values:

For the acid it would be:

Solution: H₂SO₄;   Molar ratio: 1;    Volume: 15 mL;  Concentration: 0.5 M

For the base:

Solution: KOH;   Molar ratio: 2;    Volume: 25 mL;    Concentration: 0.15 M

Titration 2:

In this case we do the same thing as before but with different data. First the equation:

HBr + NaOH --------> NaBr + H₂O

The equation is already balanced and we can see a mole ratio of 1:1, so we can use expression (2) and solve for concentration of the base instead:

M₁V₁ = M₂V₂

M₂ = M₁V₁ / V₂

M₂ = 30 * 0.250 / 20

The chart can be filled the same way as in titration 1:

For the acid it would be:

Solution: HBr;   Molar ratio: 1;    Volume: 30 mL;  Concentration: 0.25 M

For the base:

Solution: NaOH;   Molar ratio: 1;    Volume: 20 mL;    Concentration: 0.375 M

Answer:

For 3: The total mass change of the reaction is [tex]4.255\times 10^{3}kg[/tex]

For 4: The mass defect is [tex]0.911\times 10^{-27}kg[/tex] and energy equivalent to this mass is [tex]8.199\times 10^{-14}kJ[/tex]

For 5: The equivalent mass of the reaction is [tex]1.5755\times 10^{-11}kg[/tex]

Explanation:

To calculate the mass change of the reaction for given energy released, we use Einstein's equation:

[tex]E=\Delta mc^2[/tex]

E = Energy released = [tex]3.83\times 10^{-12}J[/tex]

[tex]\Delta m[/tex] = mass change = ?

c = speed of light = [tex]3\times 10^8m/s[/tex]

Putting values in above equation, we get:

[tex]3.83\times 10^{-12}Kgm^2/s^2=\Delta m\times (3\times 10^8m/s)^2\\\\\Delta m=4.255\times 10^3kg[/tex]

Hence, the total mass change of the reaction is [tex]4.255\times 10^{3}kg[/tex]

For the given isotopic representation:  [tex]_{27}^{60}\textrm{Co}[/tex]

Atomic number = Number of protons = 27

Mass number = 60

Number of neutrons = Mass number - Atomic number = 60 - 27 = 33

To calculate the mass defect of the nucleus, we use the equation:

[tex]\Delta m=[(n_p\times m_p)+(n_n\times m_n)+]-M[/tex]

where,

[tex]n_p[/tex] = number of protons  = 27

[tex]m_p[/tex] = mass of one proton  = 1.00728 amu

[tex]n_n[/tex] = number of neutrons  = 33

[tex]m_n[/tex] = mass of one neutron = 1.00867 amu

M = Nuclear mass number = 59.9338 amu

Putting values in above equation, we get:

[tex]\Delta m=[(27\times 1.00728)+(33\times 1.00867)]-[59.9338]\\\\\Delta m=0.54887amu[/tex]

Converting the value of amu into kilograms, we use the conversion factor:

[tex]1amu=1.66\times 10^{-27}kg[/tex]

So, [tex]0.54887amu=0.54887\times 1.66\times 10^{-27}kg=0.911\times 10^{-27}kg[/tex]

To calculate the equivalent energy, we use the equation:

[tex]E=\Delta mc^2[/tex]

E = Energy released = ?

[tex]\Delta m[/tex] = mass change = [tex]0.911\times 10^{-27}kg[/tex]

c = speed of light = [tex]3\times 10^8m/s[/tex]

Putting values in above equation, we get:

[tex]E=(0.911\times 10^{-27}kg)\times (3\times 10^8m/s)^2\\\\E=8.199\times 10^{-11}J[/tex]

Converting this into kilojoules, we use the conversion factor:

1 kJ = 1000 J

So, [tex]8.199\times 10^{-11}J=8.199\times 10^{-14}kJ[/tex]

Hence, the mass defect is [tex]0.911\times 10^{-27}kg[/tex] and energy equivalent to this mass is [tex]8.199\times 10^{-14}kJ[/tex]

For the given chemical reaction:

[tex]C_2H_5OH(l)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(l);\Delta H=-1418kJ/mol[/tex]

To calculate the equivalent mass of the reaction for given energy released, we use Einstein's equation:

[tex]E=\Delta mc^2[/tex]

E = Energy released = [tex]1418kJ=1418\times 10^3J[/tex]

[tex]\Delta m[/tex] = mass change = ?

c = speed of light = [tex]3\times 10^8m/s[/tex]

Putting values in above equation, we get:

[tex]1418\times 10^{3}Kgm^2/s^2=\Delta m\times (3\times 10^8m/s)^2\\\\\Delta m=1.5755\times 10^{-11}kg[/tex]

Hence, the equivalent mass of the reaction is [tex]1.5755\times 10^{-11}kg[/tex]

Water is often referred as a universal solvent because it is capable dissolving much more solutes as compared to any other solvent.  This is because, water is a high polar molecule. In water, H has partial positive charge while O has partial negative charge. 

Due to this, water favors dissociation of molecules into positively and negatively charged ions. Positively charge ions gets attracted  towards oxygen i.e. negatively charges, while negatively charged ions get attracted towards positive end of water molecule. 

However, it is worth nothing that, despite water being referred as universal solvent, many compounds are insoluble or partially soluble in water. For instance, most of the hydroxide displays poor solubility in water.

Answer:

Water is polar with a positive and negative side.

I hope this helps! Good luck on your test!

Answer:

The nucleus of an atom of U-235 splits, resulting in two smaller fragments and the release of a large amount of energy

Explanation:

In the fission reaction U-235 absorbs one neutron and breaks into two new atoms (also known as  fission fragments) and three neutrons with the release of massive amount of energy.  

The nuclear fission reaction is as follows:

¹₀n + ²³⁵₉₂U → ¹⁴¹₅₆Ba + ⁹²₃₉Kr + 3 ¹₀n

U-235 has a half life of 703.8 million years. In the fission reaction massive amount of energy 202.5 MeV is released.  

Answer: iron (III) chloride is the excess reactant in the reaction.

Explanation:

i just did the assignment

Option D : [tex]Cu^{2+}(aq)+2 e^{-}\rightarrow Cu(s)[/tex]

Reduction take place when oxidation state of atom of an element decrease. Here, addition of electron/s takes place. Opposite to that in oxidation, oxidation state increases and here, loss of electron/s take place.

The balanced chemical equation is as follows:

[tex]2Al(s)+3Cu^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Cu(s)[/tex]

Here, oxidation state of Al changes from zero to +3 thus, it undergoes oxidation and oxidation state of Cu changes from +2 to zero thus, it undergoes reduction.

The half reactions will be:

Oxidation: [tex]Al(s)\rightarrow Al^{3+}(aq)+3e^{-}[/tex]

Reduction: [tex]Cu^{2+}(aq)+2 e^{-}\rightarrow Cu(s)[/tex]

Therefore, option D is correct.

The ionic equation represents the electrolytes in the dissociated ion form in the aqueous solution. Half-reaction which represents the reduction is the increased number of electrons in the copper atom.

Reduction in the chemical reaction is the increase in the number of electrons in the atoms, while oxidation is the decrease in the number of electrons from an atom.

The balanced chemical reaction is shown as:

[tex]\rm 2Al + 3Cu^{2+} \rightarrow 2Al^{3+} + 3Cu[/tex]

Reduction occurs when the oxidation number of the atom decreases and electrons are added to it. In the reaction the oxidation number of copper changes from +2 to 0 hence, undergoes reduction.

The half-reaction of the equation is written as,

Oxidation: [tex]\rm Al \rightarrow Al^{3+} + 3e^{-}[/tex]

Reduction: [tex]\rm Cu^{2+} + 2e^{-} \rightarrow Cu[/tex]

Therefore, reduction takes place at option D. [tex]\rm Cu^{2+} + 2e^{-} \rightarrow Cu.[/tex]

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