An unstable isotope has a half life of 3 days. How many kg of a 820 kg sample will remain after 6 days?

The molecular formula is \(\\C_{7}H_{21} F_{7}\).

The formulas that a compound could have are about three. The empirical formula shows the ratio of the atoms in the compound. The molecular formula shows the actual number of atoms of each element that we have in any particular compound. The structural formula can be used to deduce the arrangement of atoms in the compound.

We can now obtain the empirical formula using;

Mass of carbon = 0.1935 g

Number of moles of carbon =  0.1935 g/ 12 g/mol = 0.016 mole

Mass of hydrogen = 0.0325 g/1 g/mol = 0.0325 moles

Mass of fluorine =  0.2043 g/19 g/mol = 0.011 moles

We now have to divide through by the lowest number of moles ;

C - 0.016 mole/0.011 moles   H - 0.0325 moles/0.011 moles F- 0.011 moles /0.011 moles

C – 1  H – 3 F – 1

The empirical formula is \(\\CH_{3}F\)

Molar mass of the empirical formula = 12 + 3 + 19 = 34 g/mol

Molecular formula = 34n = 240.23

n = 240.23/34

n = 7

The molecular formula is \(\\C_{7}H_{21} F_{7}\).

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The IUPAC name for each of the compounds would be:

A. 2,6-Dimethyl octane

B. Octane

IUPAC naming is a system of naming organic compounds according to the rules set up by the International Union of Pure and Applied Chemistry.

According to these rules:

Applying these rules to the structures in the image, the IUPAC names would be as follows:

A. 2,6-Dimethyl octane

B. Octane

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Answer:

7.50×10¯² lbs = 7.50×10¯² lbs × 453.952 g / 1 lbs

7.50×10¯² lbs = 34.0464 g.

Explanation:

From the question given, the following data were obtained:

1 lbs = 453.952 g

7.50×10¯² lbs =.?

Thus, we can obtain convert 7.50×10¯² lbs to grams as follow:

1 lbs = 453.952 g

Therefore,

7.50×10¯² lbs = 7.50×10¯² lbs × 453.952 g / 1 lbs

7.50×10¯² lbs = 34.0464 g

Therefore, 7.50×10¯² lbs is equivalent to 34.0464 g.

Answer:

molecular weight of C12H22O11 or grams This compound is also known as Lactose or Sucrose or Maltose. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles C12H22O11, or 342.29648 grams.

Explanation:

Answer: unused waste from food processing

Explanation: have you takin the chemical reaction system unit test yet? It’s alternated

Answer:

unused waste from food processing

Explanation:

Answer:

A) –7.8944×10-13 J/atom

Explanation:

Mass defect of a radionuclide (m)

\(=-8.7839\)×\(10^{-30} kg\)

Formula for binding energy

\(E=mc^{2}\)

   \(=(-8.7839x10^{-30} kg)(3x10^{8} m/s)^{2}\)

\(E=-7.8944x10^{-13} J/atom\)

If the mass defect for a radionuclide is – 8.7839 × 10⁻³° kg, the binding energy per atom will be –7.8944 × 10⁻¹³  J/atom. The correct option is A.

A radionuclide is an unstable nuclide because it contains so much charge. The excess energy is used by the gamma radiation by the nucleus, the electron uses energy to move to another orbital.

Radionuclides are particles that are used to scanning or monitor the radioactive chemicals that are in the body due to swallowing or inhaling.

The binding energy per atom can be calculated by the formula

The mass of defect of a radionuclide (m) is – 8.7839 × 10⁻³° kg

E = mc²

E =  – 8.7839 × 10⁻³° x 3 x 10⁸ m/s

E = –7.8944 × 10⁻¹³  J/atom.

Thus, the correct option is A. –7.8944×10-13 J/atom.

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Answer:  23.7kPa

Explanation:  Given:

Convert temperatures to Kelvin

\(\\ \sf\longmapsto 50+273=323K\)

\(\\ \sf\longmapsto 30+273=303K\)

Using Charles law

\(\\ \sf\longmapsto \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\)

\(\\ \sf\longmapsto \dfrac{15}{323}=\dfrac{V_2}{303}\)

\(\\ \sf\longmapsto V_2=\dfrac{15(303)}{323}\)

\(\\ \sf\longmapsto V_2=\dfrac{4545}{323}\)

\(\\ \sf\longmapsto V_2=14.07L\)

The equation C2H4(g) + H2(g) + C2H6(g) → Caco3(s) + Cao(s) + CO2(g) shows an increase in entropy due to the formation of a gas as a product. Option A

In this equation, the reactants on the left-hand side consist of gases (C2H4 and H2), while the products on the right-hand side include a solid (Caco3) and a gas (CO2).

When a reaction involves a change from gaseous to solid or liquid states, there is typically a decrease in entropy because the particles become more ordered and constrained in the solid or liquid phase.

Conversely, when a reaction involves the formation of gases, there is generally an increase in entropy because gases have higher degrees of molecular motion and greater freedom of movement compared to solids or liquids.

In the given equation, the reactants include three gaseous compounds (C2H4, H2, and C2H6), and one of the products is a gas (CO2). Therefore, the overall entropy of the system increases during this reaction.

The equation Fe(s) + S(s) → FeS(s) does not show an increase in entropy. Both the reactants (Fe and S) and the product (FeS) are solids. Since solids have lower entropy compared to gases or liquids, the entropy of the system does not increase in this reaction. Option A

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Answer:

No

Explanation:

It is given that :

Amount of the clear aqueous solution = 4 L

Temperature of the clear solution = 22°C

The solution remain clear even after an unknown amount of X compound is dissolved in the solution.

Then the water is evaporated under vacuum and the remaining precipitates weighs 0.56 kg.

Therefore, from the above data, we cannot calculate the solubility of the compound X that is added to the clear aqueous solution at 22°C.

The substance which gets dissolved is known as solute and the substance where the solute is dissolved is known as the solvent.

Answer:

56.67 g of NH₃.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

N₂ + 3H₂ —> 2NH₃

Next, we shall determine the masses of N₂ and H₂ that reacted and the mass of NH₃ produced from the balanced equation. This is illustrated below:

Molar mass of N₂ = 14 × 2 = 28 g/mol

Mass of N₂ from the balanced equation = 1 × 28 = 28 g

Molar mass of H₂ = 1 × 2 = 2 g/mol

Mass of H₂ from the balanced equation = 3 × 2 = 6 g

Molar mass of NH₃ = 14 + (3×1) = 14 + 3 = 17 g/mol

Mass of NH₃ from the balanced equation = 2 × 17 = 34 g

SUMMARY:

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂ to produce 34 g of NH₃

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

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂.

Therefore, 80 g of N₂ will react with = (80 × 6)/28 = 17.14 g of H₂.

We can see from the calculation made above that a higher mass (i.e 17.14 g) of H₂ than what was given (i.e 10 g) is required to react completely with 80 g of N₂. Thus, H₂ is the limiting reactant and N₂ is the excess reactant.

Finally, we shall determine the maximum mass of ammonia (NH₃) produced from the reaction.

To obtain the maximum mass of NH₃, the limiting reactant will be used because all of it is consumed in the reaction.

The limiting reactant is H₂ and the maximum mass of NH₃ can be obtained as follow:

From the balanced equation above,

6 g of H₂ reacted to produce 34 g of NH₃.

Therefore, 10 g of H₂ will react to produce = (10 × 34)/6 = 56.67 g of NH₃.

Therefore, 56.67 g of NH₃ is produced from the reaction.

Classrooms have grown less artificial by deliberately arranging social activities and games that promote healthy interactions. Students can discuss critical subjects with you in a private setting during class time.

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Answer:

Gold

hope this helps :)

Answer: Gold

Explanation: Gold is called a heavy metal because of its high density, which comes from the fact that each of its atoms is individually very heavy. ... In contrast, gold atoms slide past each other relatively easily, which makes the metal soft and malleable.

Answer:

As an example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate. The acetate is created by the reaction of acetic acid and the strong base, hydroxide.

When HClO2 is mixed with the appropriate amount of HCl it would create a buffer solution. That is option B.

A buffer solution is the solution that resists a change in pH of a solution when acid or base is added because it is made up of weak acid and the conjugate base or weak base and the conjugate acid.

The methods that can be used to form a buffer solution include:

Although HCl is a strong acid, it can be converted to a weak acid through dilution with water. It is in this context that it can be used to form a buffer solution.

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The mass of 3.40 x 10^35 formula units of KCL is 4.20 * 10^13.

The empirical formula of an ionic or covalent network solid compound that is used as a separate entity for stoichiometric calculations is known as a formula unit in chemistry. It is also the ionic compound with the lowest whole number ratio of ions.

As we know,1 mole = 6.02 x 1023 molecules.

1 = (1 mole)/(6.02x1023 molecules)

So, (3.40x10^35 molecules KCL)*(1 mole KCL)/(6.02x1023 molecules KCL)

5.647 *10^11 moles Na2SO4

Molar mass of KCL = 74.55 g/mol

Mass of KCL will be =  5.647 *10^11  * 74.55

Hence, mass of KCL = 4.20 * 10^13

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Answer:

Ratio = 2 : 3 (Approx.)

Explanation:

Given:

Compound A = 1.278 g

Compound B = 1.904

Find:

Ratio

Computation:

Ratio = Compound A / Compound B

Ratio = 1.278 / 1.904

Ratio = 2 : 3 (Approx.)

Answer:

1. get vitamin c

2. get your bone mineral density tested

3. don't smoke or drink excessively

4. start weight-bearing exercise

Explanation:

helps improve skeletal system

Answer:

The answer is C.

Explanation:

the wall and the ballon will use negative energy to repel eachother

When a charged balloon is brought close to a wall the walls and the balloon repel each other. Therefore, option C is correct.

Non-contact forces, such as electrostatic forces, pull or push on objects without touching them. When certain materials are rubbed together, a phenomenon known as 'charge' is transferred from one surface to the other. Charged objects pull on uncharged objects and can push or pull on charged objects.

We already know that positive and negative charges interact. However, the size of the electrostatic force determines the strength of this interaction.

When the charged balloon is brought close to the wall, it repels some of the wall's negatively charged electrons . As a result, that section of the wall is repelled.

Thus, option C is correct.

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Answer:

0.5 mole

Explanation:

The question isn't even clear

But I'm guessing you want to ask the number of moles

n= Number of molecules/ Avogadros number

n= 1/2

An intramolecular force is : C. A force acting between the atoms within a molecule

In other to have full understanding, we will be breaking the word into three; which are: Intra, Intramolecular and Force.

Intra means within, intramolecular means within a molecule and force means a pull or push.

So combining the words together, we can say that intramolecular force is a force that acts within a molecule, holding the atoms in it together.

We can conclude by saying an intramolecular force is a force acting between the atoms within a molecule

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Answer:

Explanation:

4)6.27x10^20/(6.02x10^23)(u should have this number on your formula sheet)=0.001 mol

5)7.4x6.02x10^23=44.548x10^23atoms

6)molar mass for K is 39.10g/mol

3.27x39.10=127.86g

Im bad at sig figs. Just do it urself(:p

The concentration (molarity) of the final NaCl solution is 175 M.

To find the concentration (molarity) of the final NaCl solution, we can use the equation:

M1V1 = M2V2

Where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.

In this case, we have an initial NaCl solution with a concentration of 2.5 M and a volume of 350 mL (0.350 L). We are diluting this solution to a total volume of 5.0 mL (0.005 L).

Plugging these values into the equation, we have:

(2.5 M) * (0.350 L) = M2 * (0.005 L)

Simplifying the equation:

0.875 = 0.005 * M2

Dividing both sides by 0.005:

M2 = 0.875 / 0.005

M2 = 175M

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Answer:

the molarity of the solution is 0.25 M.

Explanation:

To calculate the molarity of a solution, we need to know the amount of solute in moles and the volume of the solution in liters.

The first step is to calculate the amount of KOH in moles using its molar mass. The molar mass of KOH is the sum of the atomic weights of potassium (39.1 g/mol), oxygen (16.0 g/mol), and hydrogen (1.0 g/mol):

Molar mass of KOH = 39.1 g/mol + 16.0 g/mol + 1.0 g/mol = 56.1 g/mol

The amount of KOH in moles is:

moles of KOH = mass of KOH / molar mass of KOH

moles of KOH = 42.1 g / 56.1 g/mol

moles of KOH = 0.750 moles

Now that we know the amount of KOH in moles, we can calculate the molarity of the solution:

Molarity = moles of solute / liters of solution

Molarity = 0.750 moles / 3.0 L

Molarity = 0.25 M

Therefore, the molarity of the solution is 0.25 M.

Answer: 9.23 moles of solute

Explanation: It's a hard explanation so you can just have the answer.

Answer:

True

Explanation:

Salt is used to keep ice from forming but it only helps with certain tempuratures is still able to freeze

Answer:

yes it is true have a great time

A reaction that looks to follow second-order kinetics but is not actually a second-order reaction is described by the pseudo-second order (PSO) model.

A chemical reaction that looks to follow second-order kinetics but is not actually a second-order reaction is referred to as pseudo second order (PSO) in a kinetic model.

It is frequently seen in reactions where one reactant's concentration is substantially higher than that of the other reactant, resulting in an abundance of the abundant reactant that remains essentially constant throughout the reaction.

The rate equation in PSO kinetics has the following structure:

1/t = k * [A] * [B]

where [A] and [B] stand for the reactant concentrations, k for the rate constant, and t for the passage of time. According to the equation, the rate of the reaction is inversely proportional to the product of the reactant concentrations.

The consumption of the limiting reactant, which determines the total reaction rate, is what causes the apparent second-order behaviour. The rate of the reaction falls together with the concentration of the limiting reactant with time.

Adsorption reactions, surface reactions, or situations where one reactant is present in excess of the other are frequently described by the PSO model. The PSO model is an approximation and does not imply a real second-order reaction mechanism, it is vital to remember this.

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The desired reaction is 2Al(s) + 3CO2 from Al2O3(s) + 3CO(g) (g) The reactions include 2 Al(s), 3/2 O2(g), and Al2O3(s), with H = 1675.7kJ. ————————— (1) CO(g) = CO2 + 1/2 O2(g) (g).

As a result, the enthalpies of a reactants and products are added together, and the result is used to compute the enthalpy of a reaction. This endothermic process generates and absorbs environmental heat if H is positive. This reaction is exothermic so emits heat into the environment if H is negative.

A negative H indicates that heat is transferred from the a system towards its surroundings, whereas a positive H indicates that heat is transferred from the surroundings into the system. An enthalpy of reaction (Hrxn) for a chemical reaction is the difference of enthalpy between the products and reactants; Hrxn is measured in kilojoules per mole.

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0.498 moles of copper(II) phthalocyanine would be produced by

the complete cyclotetramerization of 255 grams of phthalonitrile in the

presence of excess copper(ll) chloride.

Copper(ll) phthalocyanine (Cu(C₃₂H₁₆N₈)) is produced by the cyclotetramerization of phthalonitrile (C₈H₄N₂) according to the following reaction: 4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g) How many moles of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 255 grams of phthalonitrile in the presence of excess copper(II) chloride?

Let's consider the following balanced equation.

4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)

The molar mass of C₈H₄N₂ is 128.13 g/mol. The moles corresponding to 255 g of C₈H₄N₂ are:

\(255 g \times \frac{1mol}{128.13 g} = 1.99 mol\)

The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 1.99 moles of C₈H₄N₂ are:

\(1.99 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.498 mol Cu(C_{32}H_{16}N_8)\)

0.498 moles of copper(II) phthalocyanine would be produced by

the complete cyclotetramerization of 255 grams of phthalonitrile in the

presence of excess copper(ll) chloride.

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