You have a 0. 7 M solution. Your job is to produce 50 mL of a 0. 4 M solution.


A. How much of the 0. 7 M solution do you need to start with? (Show your work. )


B. How many moles of solute were in the 0. 7 M solution? (Show your work. )


C. How much water do you need to add to the previous amount of the 0. 7 M solution to dilute it to

0. 4 M? (Show your work. )


D. A student decides to put double the amount of water calculated in Part C. Describe what effect this

will have on the overall concentration of the resulting solution. Justify your answer.

Answer: True

Explanation: Pressure is not dependent on mass.

Redox reaction is a type of chemical reaction in which the electrons are transferred between species. The chloride ion is the reducing agent in the reaction.

A reducing agent in a reduction-oxidation reaction is the chemical species that donates or loses electrons to another species and is present in a lower oxidation state.

The reduction-oxidation reaction is shown as,

\(\rm MnO_{2} + 4H^{+} + 2Cl^{-} \rightarrow Mn^{2+} + 2H_{2}O + Cl_{2}\)

In this reaction chloride ions lost two-electron and their oxidation number changed to zero. Hence, the chloride ion is the reducing agent.

Here, manganese is an oxidizing agent as the oxidation number changes from +4 to +2 and gains electrons.

Therefore, option A. chloride ion is the reducing agent.

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Answer:

Answer is A

Explanation:

CI-

The empirical formula of the unknown compound is C₁₂N₁.

To determine the empirical formula of the unknown compound, we need to find the mole ratios of carbon (C) and nitrogen (N) in the compound based on the given mass data.

1. Calculate the moles of N₂O, NO₂, and CO₂ produced:

Moles of N₂O = mass / molar mass = 0.1411 g / 44.013 g/mol ≈ 0.003206 mol

Moles of NO₂ = mass / molar mass = 0.1925 g / 46.0055 g/mol ≈ 0.004187 mol

Moles of CO₂ = (total moles of C - moles of N₂O - 2 x moles of NO₂) / 2

= (0.9500 g / 12.01 g/mol - 0.003206 mol - 2 x 0.004187 mol) / 2

= (0.0791 mol - 0.003206 mol - 2 x 0.004187 mol) / 2 ≈ 0.0322 mol

2. Calculate the moles of C and N:

Moles of C = moles of CO₂ = 0.0322 mol

Moles of N = moles of N₂O + moles of NO₂ = 0.003206 mol + 0.004187 mol ≈ 0.007393 mol

3. Find the mole ratio of C to N:

Mole ratio of C to N = Moles of C / Moles of N = 0.0322 mol / 0.007393 mol ≈ 4.36

4. Adjust the mole ratio to the nearest whole number:

Multiply the mole ratio by a common factor to obtain whole numbers. In this case, multiply by 2:

Mole ratio of C to N = 4.36 x 2 ≈ 8.72

5. Divide the mole ratio by the greatest common divisor to obtain the simplest whole number ratio:

Divide 8.72 by 0.72 (approximately the greatest common divisor) to get 12.11.

Round the ratio to the nearest whole number: 12

Therefore, the empirical formula of the unknown compound is C₁₂N₁.

The correct question is:

An unknown compound contains only C and N. Combustion of 0.9500 g of this unknown compound results in the formation of 0.1411 g of N₂O, 0.1925 g of NO₂ , and some CO₂ , as the only products. Determine the empirical formula of the unknown compound.

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Based on the equation of the reaction and the limiting reactant of the reaction, molarities of the ions are determined from the amount in moles of excess reactants and soluble products formed.

The limiting reactant is he reavtant that is used up in the reaction after which the reaction stops.

The limiting reactant is obtained from the mole ratio of the reactants in the equation of the reaction.

Equation of reaction is given as follows:

Molar mass of AgNO3 = 170 g/mol

Molar mass of Na3P = 100 g/mol

Molar mass of Ag3P = 355 g/mol

Molar mass of NaNO3 = 85 g/mol

Mass of AgNO3 reacting = 83.4 g

Moles of AgNO3 reacting = 83.4/170 = 0.49 moles

Mass of Na3P reacting = 62.9

Moles of Na3P reacting = 62.9/100 = 0.629 moles

Moles ratio of AgNO3 to Na3P = 3 : 1

Mole rational of AgNO3 and NaNO3 1 : 1

Based on the mole ratio;

At the end of the reaction, molarity of the ions are as follows:

Molarity of Na+ = {0.49 + (3 × 0.46)}/1.71

Molarity of P^{3+} = 0.465/1.71

Molarity of NO3^{-1} = 0.49/1.71

Therefore, molarities of the ions are determined from the amount in moles of excess reactants and soluble products formed.

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Answer:

i should be A i hope this helps

Explanation:

Answer:

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The closest value to 1.08 V among the given options is 1.076 V. So, the standard potential (E°) for the given reaction is approximately 1.076 V.

To determine the standard potential (E°) for the given reaction, we need to use the standard reduction potentials from the data tables. The reaction can be broken down into two half-reactions:

1. Oxidation of Sn2+ to Sn4+:
Sn2+(aq) → Sn4+(aq) + 2 e⁻

2. Reduction of O2 with H+ to form H2O:
O2(g) + 4 H+(aq) + 4 e⁻ → 2 H2O(l)

Now, find the standard reduction potentials (E°) for both half-reactions in the data table.

For the oxidation of Sn2+ to Sn4+, E°(Sn4+/Sn2+) is +0.15 V.
For the reduction of O2 to H2O, E°(O2/H2O) is +1.23 V.

Now, we can calculate the standard potential for the overall reaction:
E°(overall) = E°(O2/H2O) - E°(Sn4+/Sn2+)

E°(overall) = 1.23 V - 0.15 V
E°(overall) = 1.08 V

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The heat causes the sugar's atoms to combine with the oxygen in the air, forming new groups of atoms. Energy is released in this chemical reaction in the form of smoke and black soot.         hope this helps

Sugar is made of carbon, hydrogen, and oxygen atoms. When heated over a candle, these elements react with the fire to turn into a liquid. The heat causes the sugar's atoms to combine with the oxygen in the air, forming new groups of atoms. Energy is released in this chemical reaction in the form of smoke and black soot.

Answer:

Astamañana vaya buenas noches

For this problem, we could use the ideal gas equation:

Where P is the pressure, V is the volume, R is a constant and T is the temperature.

Before we start solving, we should remember the units:

P [atm]

V [L]

Answer:

The olive oil will be on top of the vinegar, this is because the density is more

Explanation:

since vinegar has a higher density than oil it will sink to the bottom of the bottle. subsequently oil will raise up. thus oil higher layer vinegar lower layer! :) . concentration in g/L....what u are given is 5.4g and 50ml

so first change the ml into L

50ml = 0.050L

now... 5.4g/0.050L = 108g/L

resolve in significant figures is 1.1 x 10^2 g/L

b. molarity = mol/L

L = 0.050 (calculated from above)

now we only have to convert the grams (5.4g) into mols

there are three elements in sucrose ( C = carbon, H = hydrogen

and O = oxygen)

look at the periodic table and find their molar mass (atomic weight)

Carbon = 12.0g/mol

Hydrogen = 1.00g/mol

Oxygen = 16.0g/mol

Sucrose contains 12 carbons, 22 hydrogens, and 11 oxygens

So the total molar mass of sucrose =

Carbon - 12 x 12.0 = 144

Hydrogen - 22 x 1.00 = 22

Oxygen - 11 x 16.0 = 176

Total = 144 + 22 + 176 = 342g/mol

Now u have 5.4grams of sucrose to change to mols u:

5.4 grams x 1 mol/ 342grams

in other words 5.4 grams / 342 grams leaving u with mols

= 0.0158 mols

now your concentration in molarity will be:

0.0158 mols / 0.050L = 0.32M (two significant figures)

and the unit is M which stands for mol/L

c. you are given 7M (aka mol/L)

you are given 100ml which equals 0.1L

7 mol/L x 0.1L = 0.7mol

all u have left to do is change the 0.7 mol back into grams

to do that we do the opposite of what we did in part b

in part b u took grams and changed it into mols

here we are taking mols and changing it back into grams

from part b we calculated the total molar mass = 342g/mol

so:

0.7 mol x 342 g/mol = 239.4grams

resolve into significant figures = 2 x 10^2 grams

a) Mg is oxidized and O2 is reduced

b) Zn is oxidized while Pb(NO3)2 is reduced.

Oxidation is defined as:

Reduction is defined as:

Looking at the first reaction, the Mg atom gains oxygen to become MgO. This means that Mg is oxidized. The oxidizing agent is O2. At the same time, O2 is being reduced and the reducing agent is Mg.

For the second reaction, the oxidation number of Pb is reduced from +2 to 0. Thus, Pb has been reduced by Zn while Zn itself has been oxidized. The reducing agent here is Zn while the oxidizing agent is Pb(NO3)2.

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The final volume of the reconstituted cefazolin solution is 4 mL.  The volume of the cefazolin powder is 0.6 mL. The final volume of the 100 mg/mL cefazolin solution is 10 mL.

The packaging insert instructs to add 3.4 mL of sterile water without bacteriostat to the 1-g vial of cefazolin. This results in a reconstituted solution with a concentration of 250 mg/mL.

To find the final volume, we can set up the equation:

Concentration of reconstituted solution = Amount of drug / Final volume

Using the given concentration (250 mg/mL) and the amount of drug (1 g = 1000 mg), we can rearrange the equation to find the final volume:

250 mg/mL = 1000 mg / Final volume

Solving for the final volume:

Final volume = 1000 mg / 250 mg/mL = 4 mL

Therefore, the final volume of the reconstituted cefazolin solution is 4 mL.

To find the volume of the cefazolin powder, we need to subtract the volume of sterile water added from the final volume of the reconstituted solution.

Given that 3.4 mL of sterile water is added to the vial, and the final volume of the reconstituted solution is 4 mL, we can calculate the volume of the cefazolin powder as follows:

Volume of cefazolin powder = Final volume - Volume of sterile water added

Volume of cefazolin powder = 4 mL - 3.4 mL = 0.6 mL

Therefore, the volume of the cefazolin powder is 0.6 mL.

To determine the final volume of the 100 mg/mL cefazolin solution, we can use the concentration and the amount of drug.

We are given that the resulting concentration after reconstitution should be 100 mg/mL. Considering the amount of drug is 1 g (1000 mg), we can set up the following equation:

Concentration of reconstituted solution = Amount of drug / Final volume

Using the given concentration (100 mg/mL) and the amount of drug (1000 mg), we can rearrange the equation to find the final volume:

100 mg/mL = 1000 mg / Final volume

Solving for the final volume:

Final volume = 1000 mg / 100 mg/mL

Final volume = 10 mL

Therefore, the final volume of the 100 mg/mL cefazolin solution is 10 mL.

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Answer:

1.31M

Explanation:

Based on the reaction, 1 mole of Zn produce 1 mole of ZnCl₂. As the reaction occurs completely, the moles of Zn added = Moles of ZnCl₂ produced. To find molarity we need the moles of ZnCl₂ and the volume of the solution in liters:

Moles Zn = Moles ZnCl₂ -Molar mass Zn: 65.38g/mol-:

15g * (1mol / 65.38g) = 0.23 moles of ZnCl₂

Volume in Liters:

175mL * (1L / 1000mL) = 0.175L

The molarity is:

0.23moles / 0.175L

The balanced equation for the given reaction is option (C)

3Sn(s) + 2NO3-(aq) + 8H+(aq) → 3Sn2+(aq) + 2NO(g) + 4H2O(l).

The given reaction involves the following species:

Sn(s) | Sn2+(aq) || NO(g) | NO3-(aq), H+(aq) | Pt(s)

The left-hand side (LHS) of the reaction involves Sn(s) and Sn2+(aq) which are oxidized, while NO(g) and NO3-(aq) are reduced. Hence, the reaction can be written as:

Sn(s) → Sn2+(aq) + 2e- ...(1)NO3-(aq) + 4H+(aq) + 3e- → NO(g) + 2H2O(l) ...(2)

Balancing equations (1) and (2) gives:

3Sn(s) → 3Sn2+(aq) + 6e- ...(3)

2NO3-(aq) + 8H+(aq) + 6e- → 2NO(g) + 4H2O(l) ...(4)

Multiplying equation (3) by 2 and adding it to equation (4) gives the balanced equation for the reaction:

3Sn(s) + 2NO3-(aq) + 8H+(aq) → 3Sn2+(aq) + 2NO(g) + 4H2O(l)

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Answer:

0.6 Ω

Explanation:

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

Voltage (V) = 12 V

Current (I) = 20 A

Resistance (R) =?

From Ohm's law,

V = IR

Where:

V => is the voltage

I => is the current

R => R is the resistance

With the above formula, we can obtain the resistance as follow:

Voltage (V) = 12 V

Current (I) = 20 A

Resistance (R) =?

V = IR

12 = 20 × R

Divide both side by 20

R = 12 / 20

R = 0.6 Ω

Thus the resistance is 0.6 Ω

2.87 g of silver chloride can be made by reacting 200.0 mL of calcium chloride with 170.0 mL of silver nitrate, both of which have a 0.23 M silver nitrate concentration. Chemically, silver chloride is a substance.

AgCl is the chemical formula for this. It is generally known that this white crystalline substance is only weakly soluble in water. Silver chloride changes from grey to black or purplish in hue upon illumination or heating, indicating the presence of silver. As the mineral chlorargyrite, AgCl can be found in nature. Chemical processes take place everywhere around us, from the food our bodies metabolize to how the sunlight we receive is produced.

silver chloride mass: m (AgCl)=0.02*143,32g/mole=2.87 g, where n (AgCl) = (0.02*2)/2=0.02 moles.

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Answer:

1) conduction

2) Radiation

Explanation:

Conduction is a mode of heat transfer by which heat energy is transferred through a material, the average position of the particles of the material remaining the same.

Radiation is a process of heat transfer by which heat is transferred from a hotter to a cooler point without any intervening medium.

The pan used to fry the egg is a conductor of heat hence heat can pass through it by conduction thereby enabling the eggs to cook.

Heat can travel without an intervening medium hence Zoe feels hot near the stove. This ability of the heat to travel without an intervening medium is called radiation.

Answer:

Conduction first then radiation.

Explanation:

Answer:

C. Neither accurate nor precise

Explanation:

Thus the results are neither accurate nor precise.

The hospital's experimentation with peroral endoscopic myotomy (POEM) filling is part of the tertiary care position.

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Answer: T=6.90 K

Explanation:

For this problem, you would use Ideal Gas Law.

\(PV=nRT\)

P=pressure

V=volume

n=moles

R=ideal gas constant

T=temperature (in Kelvin)

We can fill in these values with what we are given.

P=1.41 atm

V=1.00 L

n=2.49 mol

R=0.08206 Latm/Kmol

T=?K

Since we are looking for T, we can get T alone.

\(T=\frac{PV}{nR}\)

\(T=\frac{(1.41 atm)(1.00L)}{(2.49mol)(0.08206Latm/Kmol)}\)

\(T=6.9006 K\)

With significant figures, we get \(T=6.90 K\).

Answer:

It is most of the time cheaper and more fresh than regular store bought food.

Explanation:

Answer:

I beileve it is NORTH KOREA!! XD actually D.

Explanation:

When alkaline hydrolysis was first invented, people were hired for various roles related to the process and implementation of this technology. Some of the jobs that emerged include Chemical engineers, Technicians and operators, Waste management specialists, Scientists and researchers.

Chemical engineers: These professionals played a crucial role in developing and optimizing the alkaline hydrolysis process. They were responsible for designing the equipment, developing the necessary chemical reactions, and ensuring the efficient operation of the system.

Technicians and operators: Skilled technicians and operators were hired to operate and maintain the alkaline hydrolysis equipment. They were trained to monitor the process parameters, handle the chemicals involved, and ensure the proper functioning of the system.

Waste management specialists: With the introduction of alkaline hydrolysis as a method for disposal of organic waste, specialized professionals in waste management were employed to oversee the proper handling and treatment of the waste materials. They were responsible for implementing safety protocols, managing waste streams, and complying with environmental regulations.

Scientists and researchers: Alkaline hydrolysis required scientific expertise for continuous improvement and innovation. Scientists and researchers were hired to study the process, analyze the results, and explore potential applications in various fields such as biofuel production and chemical synthesis.

Overall, the introduction of alkaline hydrolysis created employment opportunities for professionals in engineering, chemistry, waste management, and research, among others, as this technology gained recognition and adoption.

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The balanced net ionic equation is:

Ag⁺(aq) + ClO₄⁻(aq) + Al³⁺(aq) + 3Cl⁻(aq) → AgCl(s) + AlCl₃(aq)

The balanced net ionic equation for the reaction between silver perchlorate and aluminum chloride in aqueous solution is shown above. In this reaction, the silver ion (Ag⁺) and perchlorate ion (ClO₄⁻) from silver perchlorate (AgClO₄) react with the aluminum ion (Al³⁺) and chloride ion (Cl⁻) from aluminum chloride (AlCl₃) to form solid silver chloride (AgCl) and aluminum chloride in aqueous solution (AlCl₃).

The net ionic equation shows only the species that participate in the reaction and are involved in the formation of the product (AgCl), while the spectator ions (Cl⁻ and Al³⁺) are omitted.

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Answer:

We can use the ideal gas law to solve for the number of moles of gas:

PV = nRT

where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin.

Plugging in the given values:

(1.35 atm)(25 L) = n(0.0821 L·atm/mol·K)(300 K)

n = (1.35 atm)(25 L) / (0.0821 L·atm/mol·K)(300 K)

n = 1.29 mol

Therefore, there are 1.29 moles of gas in the container.

At room temperature, the reaction with the fastest reaction rate would likely be option 3) N2(g) + 2O2(g) --> 2NO2. This is because this reaction involves the breaking of a relatively weak triple bond between the two nitrogen atoms, allowing them to form bonds with oxygen atoms to create NO2 molecules.

In contrast, options 1) and 2) involve the formation of relatively strong covalent bonds, which would require more energy to break and therefore have slower reaction rates. Option 4) involves the decomposition of KClO3, which requires a high activation energy to break the bond between the K and ClO3 ions, and so would also have a slower reaction rate than option 3).

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Approximately 53.55 grams of water will form when 10.54 grams of \(H_2\)reacts with 95.10 grams of \(O_2\).

To determine the grams of water formed in the reaction between hydrogen (\(H_2\)) and oxygen (\(O_2\)), we need to calculate the limiting reagent and use the stoichiometry of the balanced chemical equation.

First, let's write the balanced equation for the reaction:

2\(H_2\) + \(O_2\)→ 2\(H_2O\)

The molar mass of \(H_2\)is 2.016 g/mol, and the molar mass of \(O_2\)is 31.998 g/mol. We can use these values to convert the given masses of \(H_2\)and O2 into moles.

Moles of \(H_2\)= 10.54 g / 2.016 g/mol ≈ 5.221 mol

Moles of \(O_2\)= 95.10 g / 31.998 g/mol ≈ 2.972 mol

According to the balanced equation, the ratio of \(H_2\)to \(O_2\)is 2:1. Therefore, we can determine that \(O_2\)is the limiting reagent since there is less \(O_2\)available compared to the stoichiometric ratio.

To find the moles of water formed, we use the stoichiometry of the balanced equation. From the equation, we see that for every 2 moles of , 2 moles of water are formed.

Moles of water formed = (2 mol \(H_2O\)/ 2 mol \(H_2\)) * 2.972 mol \(H_2\)≈ 2.972 mol \(H_2O\)

Now, we can calculate the mass of water formed using the molar mass of water, which is 18.015 g/mol.

Mass of water formed = 2.972 mol \(H_2O\)* 18.015 g/mol ≈ 53.55 g

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