The appropriate mixing ratio for a solution to kill bloodborne pathogens is 1 part chlorine to 10 parts water. If you are mixing a solution in a spray bottle with 15 ounces of water, how much chlorine do you need to add for the correct ratio

The amount of energy need to melt 25.4 grams of iodine, I₂ is 1.57×10³ Joules

How to calculate the amount of energy needed to to melt the the iodine, I₂?

The following useful information were given from the question:

The amount of energy need to melt 25.4 grams of iodine, I₂ can be calculated as shown below:

Heat energy needed (Q) = Mass (m) × Latent heat of fusion (Hf)

= 25.4 grams × 61.7 J/g

= 1.57×10³ Joules

Thus, the amount of heat energy needed to melt the iodine,I₂ is 1.57×10³ Joules

Pure chemistry involves the study of the basic principles and theories of chemistry while applied chemistry uses these principles to solve real-world problems or develop new products. They are interconnected in the sense that advancements in pure chemistry can lead to new opportunities in applied chemistry and vice versa.

The relationship between pure chemistry and applied chemistry is similar to that of theory and practice. Pure chemistry, also known as basic chemistry, is the study of the basic principles and theories of chemistry for the sake of knowledge itself. It seeks to understand the fundamental principles that explain how matter behaves and interacts. Examples include studying chemical reactions, the properties of elements, and how atoms form compounds.

On the other hand, applied chemistry utilizes these principles to solve real-world problems or develop new products. For instance, an applied chemist might work on creating new material, developing new drugs, improving environmental sustainability, or advancing food science.

In essence, applied chemistry is an application of pure chemistry. Both fields are interconnected; advancements in pure chemistry can lead to new opportunities in applied chemistry and vice versa.

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To measure and add 50 mL of a stock solution, a volumetric pipette or graduated cylinder should be used for accuracy and precision in chemical laboratory work, as volumetric flasks are not suitable for measuring out volumes other than their calibration capacity.

If you need to add 50 mL of a stock solution to a reaction vessel, the most appropriate and precise piece of glassware to use would be ab this case. When dealing with stock solutions, it's important to use glassware that ensures precision to ensure the final concentrations of your solutions are accurate.

When mixing solutions or reagents, accuracy and precision are crucial, especially in a chemical laboratory or medical environment. For instance, preparing an IV solution with a specific concentration involves accurate measurements of the stock solution and addition of the solvent up to a desired volume. Misjudgements in the volume of stock solution added could alter the final concentration of the solution, which could have significant implications, for example, in medical applications.

Actually, the ionic equation for this is a reversible equation since codeine is a weak base. Any weak base or weak acids do not completely dissociate which makes them a reversible process. The ionic equation for this case is:

C18H21O3N  +  H3O+  <=> C18H21O3NH+ +  H2O 

Ions with six d electrons in the outermost d subshell can be found in transition metal elements such as Cr³+(d³), Co²+(d²), Ni²+(d³), Cu²+(d³), and Zn²+(d¹⁰)

The ions with six d electrons in the outermost d subshell are typically found in the transition metal elements.

For example, the ion with six d electrons in the outermost subshell is Cr³+(d³).

Other ions with six d electrons in the outermost subshell include Co²+(d²), Ni²+(d³), Cu²+(d³), and Zn²+(d¹⁰).

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The formula for energy or enthalpy is:

E = m Cp (T2 – T1)

where E is energy = 63 J, m is mass = 8 g, Cp is the specific heat, T is temperature

63 J = 8 g * Cp * (340 K – 314 K)

Cp = 0.3 J / g K

Answer : The specific heat of substance is 0.30 J/g.K

Explanation :

Formula used :

[tex]Q=m\times c\times \Delta T[/tex]

or,

[tex]Q=m\times c\times (T_2-T_1)[/tex]

where,

Q = heat = 63 J

m = mass of substance = 8.0 g

[tex]C_w[/tex] = specific heat of substance = ?

[tex]T_1[/tex] = initial temperature  = 314 K

[tex]T_2[/tex] = final temperature  = 340 K

Now put all the given value in the above formula, we get:

[tex]63J=8.0g\times c\times (340-314)K[/tex]

[tex]c=0.30J/g.K[/tex]

Therefore, the specific heat of substance is 0.30 J/g.K

In the reaction AgNO3(aq)+NaCl(aq)------>AgCl(s)+NaNO3(aq), the chemicals dissolved in water are AgNO3, NaCl, and NaNO3.

In the chemical reaction AgNO3(aq)+NaCl(aq)------>AgCl(s)+NaNO3(aq), the chemicals that are dissolved in water, also known as aquatic solutions, are AgNO3 (Silver Nitrate) and NaCl (Sodium Chloride) on the reactant side, and NaNO3 (Sodium Nitrate) on the product side. These are shown with the (aq) notation which indicates they are aquatic, or dissolved in water.

The product AgCl (Silver Chloride), shown with the (s) notation, demonstrates that it precipitates, or forms a solid in the solution instead of dissolving in the water.

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Since it was specifically stated that heat is released, therefore heat is located on the right side of the reaction equation. Also stating the states of each compound, the complete balanced reaction would be:

C2H5OH (l) + 3 02 (g) -----> 2 CO2 (g) + 3 H2O (g) + 1235 kJ

The balanced combustion reaction will be

[tex]\rm \bold {C_2H_5OH_ (_l_) + 3 0_2 _(_g_) \rightarrow 2 CO_2 _(_g_) + 3 H_2O (g) + 1235 kJ}[/tex]

To write a balanced equation

Hence, balanced reaction will be

[tex]\rm \bold {C_2H_5OH_ (_l_) + 3 0_2 _(_g_) \rightarrow 2 CO_2 _(_g_) + 3 H_2O (g) + 1235 kJ}[/tex]

To know more about chemical equation, refer to the link:

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Actually, the uncertainty point is when we have to estimate one place smaller than what the graduated cylinder can accurately read. So in this case, the smallest that we can accurately read is given by the minor marks of 0.2 mL. So if it is in between two minor marks we can only estimate the number, therefore the uncertainty is +/- 0.1ml.

Answer:

The estimated uncertainty is ±0.1 ml

Explanation:

Measurements are usually associated with uncertainty which depends on the instrument used or the procedure employed.

It is given that the graduated cylinder of volume 25 ml has a major scale marked every 2 ml and minor scale with 0.2 ml. In general measurements made with graduated cylinders can be estimated one decimal place beyond the smallest division. In this case, the volume of the liquid can be read to the nearest 0.1 division. Hence, the uncertainty would be ± 0.1 ml.

To find the number of moles in a given mass, convert the mass to grams (if necessary) and divide by the molar mass. Hence, in 5.00 mg of aspartame, there are approximately 1.70 x 10^-5 moles.

To calculate the number of moles of aspartame in a given mass, we would use the molar mass of aspartame. First, we need to convert the mass from milligrams to grams because molar mass is usually given in g/mol. As 1 gram (g) is equal to 1000 milligrams (mg), 5.00 mg is equal to 0.005 g.

The molar mass of aspartame (C14H18N2O5) is approximately 294.31 g/mol. After converting the mass to grams, we use the formula: number of moles = mass (g) / molar mass (g/mol).

Therefore, the number of moles of aspartame in 5.00 mg is 0.005g / 294.31 g/mol = 1.70 x 10^-5 mol. So, there are roughly 1.70 x 10^-5 moles of aspartame in 5.00 mg.

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Answer: The volume of liquid is 16.22 mL.

Explanation:

Density is defined as the ratio of mass of the object and volume of the object.

Mathematically,

[tex]\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}[/tex]

We are given:

Density of liquid = [tex]0.789g/mL[/tex]

Mass of liquid = 12.8 g

Putting values in above equation, we get:

[tex]0.789g/mL=\frac{12.8g}{\text{Volume of liquid}}\\\\\text{Volume of liquid}=16.22mL[/tex]

Hence, the volume of liquid is 16.22 mL.

Solving this problem only requires the knowledge of stoichiometry. We are given the compound propane C3H8. From the chemical formula, we can see that for every 3 moles of C, there are 8 moles of H. Therefore:

moles of H = 0.20 moles C * (8 moles H / 3 moles C)

moles of H = 0.53 moles H

Answer: In the given sample, 0.533 moles of hydrogen will be present.

Explanation:

We are given a chemical compound having formula [tex]C_3H_8[/tex]

We are provided with 0.20 moles of carbon atoms.

As, for every 3 moles of carbon atoms, 8 moles of hydrogen atoms are present.

So, for 0.20 moles of carbon atoms, [tex]\frac{8}{3}\times 0.20=0.533moles[/tex] of hydrogen atoms will be present.

Hence, in the given sample, 0.533 moles of hydrogen will be present.

The correct way to write the title of a poem is by enclosing it in quotation marks, making option (a) "Afternoon on a Hill" the correct formatting for a handwritten poem title.

When formatting the title of a poem in writing, it should either be enclosed in quotation marks or italicized if typed, depending on the style guide being used. The three options given for the poem title 'Afternoon on a Hill' vary in formatting. Based on standard English conventions for titling creative works such as poems, the handwritten version of the poem title would be formatted using either quotation marks or italicization.

Therefore, the correct answer is option (a) "Afternoon on a Hill" as it follows the guideline of using quotation marks for the title of shorter works such as poems. Option (b) and option (c) are missing the necessary punctuation or formatting emphases that set apart the titles of creative works from the rest of the text.

After one half-life (10 seconds), 8 grams of krypton-91 will remain. After 50 seconds (or 5 half-lives), only 0.5 grams will be left.

The question pertains to the concept of half-life in radioactive decay. Half-life is the time taken for half of the radioactive element to decay. With a half-life of 10 seconds, krypton-91 follows an exponential decay pattern. For example, if you start with 16 grams of krypton-91, after one half-life (10 seconds), half of it will decay, leaving you with 8 grams of krypton-91.

After 50 seconds or five half-lives, the remaining krypton-91 would be 16g * (1/2)^5 = 0.5 grams. It is important to note that the radioactive atoms don't disappear but are replaced by their decay products or daughter elements.

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601kg as integers can be written as 601000

In mathematics an integer is a number that can be written without a fractional or decimal component. For example: 21, 4, and −2048 are integers; 9.75, 5½, and √2 are not integers. If zero comes in the middle of the integer it is taken into account but if it comes in the start then it is discarded.  Similarly we have scientific notations and we know that kg is equal to 1000 so we can write 601 kg as 601000 which fulfills both the definition of integer and scientific notation.

601 kg expressed as an integer remains 601 kg, since it is already a whole number and does not require any conversion.

To express 601 kg as an integer, we do not need to apply any conversions, as 601 kg is already an integer value. The term 'integer' refers to a whole number that is not a fraction or a decimal. Therefore, 601 kg is an integer as it stands. It's important to note that when measuring mass or weight, it's beneficial to use units that keep the number manageable and avoid unnecessary complexity, as described with the examples of different weight measurements and their ease based on the unit used.

Final answer:

Nitric acid and lithium hydroxide are classified as strong electrolytes, meaning they fully dissociate into ions in solution. Hydrofluoric acid and ammonia are weak electrolytes, partially dissociating into ions. Ethanol and glucose are nonelectrolytes, dissolving without forming ions.

Explanation:

When classifying solutes as strong electrolytes, weak electrolytes, or nonelectrolytes, we examine how they dissociate in water. A strong electrolyte dissociates completely into ions, a weak electrolyte only partially dissociates, and a nonelectrolyte does not dissociate at all in solution.

We take as a basis for solving this item, 100 g of sample. If 15.77% of the sample is carbon, we solve for the amount of carbon in the sample,

    Carbon = (100 g)(0.1577)  = 15.77

Then, we solve for the amount sulphur in the sample by,

    Sulfur = (100 g)(1 – 0.1577) = 84.23 g

 Next, we solve for the number of moles of the elements by dividing the calculated masses by the molar mass.

 Carbon = (15.77g)(1 mol/12 g) = 1.3 moles

 Sulphur = (84.23 g)(1 mol/32 g) = 2.63 moles

 From the ratio of the calculated number of moles, we can say that each mol of carbon will have to be paired up with two other sulphur atoms. Thus, the shape of the molecule is LINEAR. 

There are 10,080 minutes in one week. To calculate the number of minutes in one week, multiply the number of minutes in an hour (60) by the number of hours in a day (24), and then multiply that result by the number of days in a week (7).

To calculate the number of minutes in one week using unit conversion, you first need to know how many minutes are in an hour, how many hours are in a day, and how many days are in a week. There are 60 minutes in an hour. In each day, there are 24 hours. Therefore, in one day, there are  60 x 24 = 1440 minutes. A week consists of 7 days, so to find out the number of minutes in a week, you would do 1440 x 7 = 10,080. So, there are 10,080 minutes in one week.

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