A nuclear chemist measures the mass of an artificial element. The mass is . What is the mass in grams

Explanation:

Carbon-14 has a half life of 5730 years, meaning that 5730 years after an organism dies, half of its carbon-14 atoms have decayed to nitrogen atoms. Similarly, 11460 years after an organism dies, only one quarter of its original carbon-14 atoms are still around.

There are numerous subcultures, such as hippies, anti-gun groups, high school jocks, environmental activists, people in the furry community, people in the cosplay community, punks, goths, and many more.

"An identifiable subgroup within a society or group of people, particularly one distinguished by beliefs or interests that differ from those of the larger group."

Subcultures are important in articulating an identity, creating a sense of belonging, and influencing members to think about their relationship to mainstream society; however, subcultures differ from widely recognized identity categories such as ethnicity.

Subcultures exist within a society's dominant culture. Subcultures in America include hippies, punk rockers, beatniks, and hipsters.

Thus, There are numerous subcultures, such as hippies, anti-gun groups, high school jocks, environmental activists, people in the furry community.

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The empirical formula of the given compound is \(CaO\).

A chemical compound's empirical formula in chemistry is the simplest whole number ratio of atoms within a molecule.

Given:

Mass of Ca = 2.514g

Mass of O = 1.004g

Molar mass of Ca = 40.078g/mol

Molar mass of O = 15.999g/mol

To find:

Empirical formula = ?

Formula:

Empirical formula = Moles of Ca : Moles of O

Calculations:

No. of moles of Ca = 2.514 / 40.078

n of Ca = 0.06273 moles

No. of moles of O = 1.004 / 15.999

n of O = 0.06275 moles

Divide both by the lowest no. of moles

n of Ca : n of O = 1 : 1

Empirical formula = \(CaO\)

Result:

\(CaO\) is the empirical formula of the compound.

The complete question is:

In an experiment, a 2.514 g sample of calcium was heated in a pure stream of oxygen, and was found to increase in mass by 1.004 g. Calculate the empirical formula for the given compound.

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In an investigation of the decomposition of water, a student compared the amount of oxygen in one test tube to the amount of hydrogen in the other test tube is There is one - half as much as hydrogen.

In decomposition of water , water can be broken into the hydrogen gas and the oxygen gas. The water decomposition reaction is given as :

2H₂O   ------>     2H₂     +     O₂

The ratio of  hydrogen and oxygen in the decomposition reaction of water is 2 : 1.

Thus, In an investigation of the decomposition of water, a student compared the amount of oxygen in one test tube to the amount of hydrogen in the other test tube is There is one - half as much as hydrogen.

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Answer:

protons and electrons

Explanation:

because two atoms of the same elements have different(mass)isotope so the answer is B

Answer:

there is one atom of oxygen and two atoms of hydrogen

Explanation:

Answer:

If I'm correct i believe that the answer is A.

Answer:

A. Where warm moist air masses traveling north collide with cool dry air masses traveling south

Explanation:

I got it right

Answer:

The rate constant of the reaction at 125˚ is \(0.3115 \ \text{sec}^{-1}\).

Explanation:

The Arrhenius equation is a simple equation that describes the dependent relationship between temperature and the rate constant of a chemical reaction. The Arrhenius equation is written mathematically as

                                                  \(k \ = \ Ae^{\displaystyle\frac{-E_{a}}{RT}}\)

                                               \(\ln k \ = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT}\)

where \(k\) is the rate constant, \(E_{a}\) represents the activation energy of the chemical reaction, \(R\) is the gas constant, \(T\) is the temperature, and \(A\) is the frequency factor.

The frequency factor, \(A\), is a constant that is derived experimentally and numerically that describes the frequency of molecular collisions and their orientation which varies slightly with temperature but this can be assumed to be constant across a small range of temperatures.

Consider that the rate constant be \(k_{1}\) at an initial temperature \(T_{1}\) and the rate constant \(k_{2}\) at a final temperature \(T_{2}\), thus

                         \(\ln k_{2} \ - \ \ln k_{1} = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT_{2}} \ - \ \left(\ln A \ - \ \displaystyle\frac{E_{a}}{RT_{1}}\right) \\ \\ \\ \rule{0.62cm}{0cm} \ln \left(\displaystyle\frac{k_{2}}{k_{1}}\right) \ = \ \displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)\)

                                         \(\rule{1.62cm}{0cm} \displaystyle\frac{k_{2}}{k_{1}} \ = \ e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)} \\ \\ \\ \rule{1.62cm}{0cm} k_{2} \ = \ k_{1}e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)}\)

Given that \(E_{a} \ = \ 26.5 \ \ \text{kJ/mol}\), \(R \ = \ 8.3145 \ \ \text{J mol}^{-1} \ \text{K}^{-1}\), \(T_{1} \ = \ \left(40 \ + \ 273\right) \ K\), \(T_{2} \ = \ \left(125 \ + \ 273\right) \ K\), and \(k_{1} \ = \ 0.0354 \ \ \text{sec}^{-1}\), therefore,

           \(k_{2} \ = \ \left(0.0354 \ \ \text{sec}^{-1}\right)e^{\displaystyle\frac{26500 \ \text{J mol}^{-1}}{8.3145 \ \text{J mol}^{-1} \ \text{K}^{-1}}\left(\displaystyle\frac{1}{313 \ \text{K}} \ - \ \displaystyle\frac{1}{398 \ \text{K}} \right)} \\ \\ \\ k_{2} \ = \ 0.3115 \ \ \text{sec}^{-1}\)                      

Answer:

Volume = 3.4 L

Explanation:

In order to calculate the volume of CO₂ produced when 15.2 g of CaCO₃ is heated, we need to first write out the balanced equation of the thermal decomposition of CaCO₃:

CaCO₃ (s) + [Heat] ⇒ CaO (s) + CO₂ (g)

Now, let's calculate the number of moles in 15.2 g CaCO₃:

mole no. = \(\mathrm{\frac{mass}{molar \ mass}}\)

                 = \(\frac{15.2}{40.1 + 12 + (16 \times 3)}\)

                 = 0.1518 moles

From the balanced equation above, we can see that the stoichiometric molar ratios of CaCO₃ and CO₂ are equal. Therefore, the number of moles of CO₂ produced is also 0.1518 moles.

Hence, from the formula for the number of moles of a gas, we can calculate the volume of  CO₂:

mole no. = \(\mathrm{\frac{Volume \ in \ L}{22.4}}\)

⇒ \(0.1518 = \mathrm{\frac{Volume}{22.4}}\)

⇒ Volume = 0.1518 × 22.4

                 = 3.4 L

Therefore, if 15.2 g of CaCO₃ is heated, 3.4 L of CO₂ is produced at STP.

Cuso4 + NaoH -》cu(oH)2 +Na2So4

Cuso4 + 2NaoH -》cu(oH)2 +Na2So4

Explanation:

this is balanced equation

Statement a is true, as the melting points of saturated fatty acids do increase with increasing chain length. Statement b is false, as saturated fatty acids do not contain double bonds. Statement c is false, as △9, 12-all cis, 18:3 represents alpha-linolenic acid, not linoleic acid. Statement d is true, as glycerol is indeed a by-product of the hydrolysis of fats.

a. True. The melting points of saturated fatty acids increase with increasing chain length. This is because longer fatty acid chains have stronger intermolecular forces, such as van der Waals forces, which require more energy to break and result in higher melting points.

b. False. Saturated fatty acids do not have double bonds. They are composed of only single carbon-carbon bonds. Double bonds are found in unsaturated fatty acids.

c. False. △9, 12-all cis, 18:3 is not linoleic acid. It represents the structure of alpha-linolenic acid. Linoleic acid is △9, 12-18:2, which means it has two double bonds located at the 9th and 12th carbon positions.

d. True. A by-product of the hydrolysis of fats is glycerol. When fats undergo hydrolysis, they are broken down into their constituent fatty acids and glycerol. Glycerol is a three-carbon molecule that is a component of triglycerides (fats).

During hydrolysis, the ester bonds between the fatty acids and glycerol are cleaved, resulting in the release of fatty acids and glycerol.

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Answer:

25 mL of NaOH

Explanation:

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

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

Mole ratio of the acid, HCl (nₐ) = 1

Mole ratio of the base, NaOH (n₆) = 1

Finally, we shall determine the volume of NaOH required for the reaction. This can be obtained as follow:

Molarity of the base, NaOH (M₆) = 0.2 M

Molarity of the acid, HCl (Mₐ) = 0.1 M

Volume of the acid, HCl (Vₐ) = 50 mL

Mole ratio of the acid, HCl (nₐ) = 1

Mole ratio of the base, NaOH (n₆) = 1

Volume of the base, NaOH (V₆) =?

MₐVₐ / M₆V₆ = nₐ / n₆

0.1 × 50 / 0.2 × V₆ = 1

5 / 0.2 × V₆ = 1

Cross multiply

5 = 0.2 × V₆

Divide both side by 0.2

V₆ = 5/0.2

V₆ = 25 mL

Thus, 25 mL of NaOH is needed for the reaction

Answer:

EL Niño  had hurricanes and so many damages in nature. :)

Explanation:

it creates warm weather, there is low pressure towards canada

Numerous health issues, such as disease outbreaks, malnutrition, heat stress, and respiratory illnesses are being brought on by the severe drought and associated food insecurity, flooding, precipitation, and temperature increases brought on by El Nino.

The exceptional warming of surface waters in the eastern tropical Pacific Ocean is referred to as El Nino, a climate pattern. El Nino is a bigger phenomena known as the El Nino-Southern oscillation, and it is its warm phase.

Climate change brought on by El Nino can be severe and extensive. Increased precipitation results from convection over warmer surface waters. In Ecuador and northern Peru, rainfall rises dramatically, causing coastal erosion and flooding.

Homes, businesses, schools, and hospitals may be destroyed by heavy rains and flooding. They hinder transportation and obliterate crops. As reservoirs dry up and rivers carry less water, the droughts pose a threat to the area's water resources. Agriculture is in danger since it depends on water for irrigation.

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The starting mixture contained approximately 23.62% SiO2.

Silicon dioxide (SiO2) has many important uses in various fields:

Glassmaking: SiO2 is a primary component of most types of glass. It is added to glass to improve its hardness, clarity, and resistance to heat and chemicals.

Ceramics: SiO2 is used in the production of ceramics and pottery as it gives the material added strength and durability.

Electronics: SiO2 is used as a dielectric material in electronic devices like transistors, integrated circuits, and microchips. It is an important component of the insulation layers that protect the electrical components and prevent them from overheating.

Construction: SiO2 is used as an important component in construction materials like concrete, bricks, and roofing tiles. Its hardness and durability make it ideal for building materials.

Cosmetics: SiO2 is used in many cosmetic products like face powders, sunscreens, and lotions. It is used as an absorbent or bulking agent that helps to give products a silky texture.

To determine the percentage of SiO2 in the starting mixture, we need to calculate the total mass of the starting mixture and the mass of SiO2 in it.

The total mass of the starting mixture is the sum of the masses of NaCl, SiO2, and CaCO3:

total mass = 1.3012 g + 0.5410 g + 0.4503 g = 2.2925 g

The mass of SiO2 in the starting mixture is given as 0.5410 g.

To calculate the percentage of SiO2 in the starting mixture, we divide the mass of SiO2 by the total mass of the mixture and multiply by 100:

% SiO2 = (mass of SiO2 / total mass) x 100

% SiO2 = (0.5410 g / 2.2925 g) x 100

% SiO2 = 23.62%

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Explanation:

MnO4−+ SO32− → Mn2++ SO42−

Splitting into half equations;

MnO4−  → Mn2+

SO32− → SO42−

Balancing the electrons;

2 MnO4−  + 10 e- → 2Mn2+

5SO32− → 5SO42− + 10 e-

In an acidic medium, it becomes;

2 MnO4−  +  8 H+ → 2 Mn2+ + 4 H2O

5 SO32−  + H2O → 5 SO42−  +  2 H+

Final equation is;

2 MnO4- + 5 SO32- + 6 H+ → 2 (Mn)2+ + 5 SO42- + 3 H2O

Coefficient of H+ = 6

Coefficient of H2O = 3

Coefficient of MnO4- = 2

Coefficient of SO32- = 5

Coefficient of (Mn)2+- = 2

Coefficient of SO42- = 5

Answer:

\(5SO_3^{2-}+2MnO_4^{-}+6H^+ \rightarrow 5SO_4^{2-}+ 2Mn^{2+}+3H_2O\)

Explanation:

Hello,

In this case, given the reaction:

\(MnO_4^{-}+ SO_3^{2-} \rightarrow Mn^{2+}+ SO_4^{2-}\)

We first identify the oxidation state of both manganese and sulfur at each side:

\(Mn^{7+}O_4^{-}+ S^{4+}O_3^{2-} \rightarrow Mn^{2+}+ S^{6+}O_4^{2-}\)

So we have the oxidation and reduction half-reactions below, including the addition of water and hydronium as it is in acidic media:

\(S^{4+}O_3^{2-}+H_2O \rightarrow S^{6+}O_4^{2-}+2H^++2e^-\)

\(Mn^{7+}O_4^{-}+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\)

Next, we exchange the transferred electrons:

\(5*(S^{4+}O_3^{2-}+H_2O \rightarrow S^{6+}O_4^{2-}+2H^++2e^-)\\2*(Mn^{7+}O_4^{-}+8H^++5e^- \rightarrow Mn^{2+}+4H_2O)\\\\5S^{4+}O_3^{2-}+5H_2O \rightarrow 5S^{6+}O_4^{2-}+10H^++10e^-\\2Mn^{7+}O_4^{-}+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\)

Then we add the resulting half-reactions and simplify the transferred electrons:

\(5S^{4+}O_3^{2-}+5H_2O+2Mn^{7+}O_4^{-}+16H^+ \rightarrow 5S^{6+}O_4^{2-}+10H^++ 2Mn^{2+}+8H_2O\)

We rearrange the terms in order to simplify water and hydronium molecules:

\(5S^{4+}O_3^{2-}+2Mn^{7+}O_4^{-}+16H^+-10H^+ \rightarrow 5S^{6+}O_4^{2-}+ 2Mn^{2+}+8H_2O-5H_2O\\\\5S^{4+}O_3^{2-}+2Mn^{7+}O_4^{-}+6H^+ \rightarrow 5S^{6+}O_4^{2-}+ 2Mn^{2+}+3H_2O\)

Finally we write the balanced reaction in acidic media:

\(5SO_3^{2-}+2MnO_4^{-}+6H^+ \rightarrow 5SO_4^{2-}+ 2Mn^{2+}+3H_2O\)

Best regards.

Let's start by balancing the reaction:

As we can see, C appears only on two comopunds, CO and CO₂, and since both have 1 C each, their coefficients have to be the same for C to be balanced. However, CO has 1 O and CO₂ has 2, so there is a difference of 1 O betwee them.

The other source of O is Fe₂O₃, that has 3 O. So, we must choose a coefficient for CO and CO₂ such that the difference between the numbers of O is a multiple of 3, that way we can fix this difference with the O from Fe₂O₃. So, we can put coefficients of 3 on both of them:

That way, we maintained C balanced (3 on each side) and now we have 3 + 3 O on the left side and 6 O on the right side, so the same amount.

Now, we just have to calance Fe, but it is easy since we have it alone in Fe. Since we have 2 on the left side, it is enough to put a coefficient of 2 on Fe to get the balanced reaction:

Now, to convert from mass to number of moles, we need the molar masses of the reactants, which we can calculate from the atomic weights of the elemnts in each of them:

Now, we can convert their masses to number of moles:

Now, to determine the limiting reactant, we need to divide both the number of mole by their coefficients on the balanced reaction, so we can see how many we need per reaction of each:

Now, the limiting reactant is the one we have less number of moles per reaction. We can see that we have less CO than Fe₂O₃, so the limiting reactant is CO.

Answer:

water molecule...glucose..chloroplast...cell...leaf

The structure of the order from smallest to the largest should be shown below:

The order of the series are as follows:

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One time explanation of one observation is a hypothesis.

Whenever, an assumption is made juts by observation and without proper testing then that is referred as an hypothesis.

For example- If my friend is late, so I might assume that he or she must have been stuck in traffic. So, this assumption is not tested yet and is made based on my observation. So, this is a hypothesis.

Whenever, a conclusion is given after several experiments and testing after observing something in science is referred as scientific theory.

For example- Atomic theory given by Dalton was based on several experiments. However, theories can also be proved wrong as Dalton's theory was proved wrong later on my JJ Thomson.

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The final volume of the ladybug's prison is approximately 0.4 liters.

To determine the final volume of the ladybug's prison, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature. The equation for Boyle's Law is:

P1 * V1 = P2 * V2

Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume, respectively.

In this scenario, the initial volume (V1) is given as 0.5 L, and the initial pressure (P1) is 1 atm. The final pressure (P2) is 1.25 atm. We need to find the final volume (V2).

Plugging the given values into the equation, we have:

1 atm * 0.5 L = 1.25 atm * V2

Simplifying the equation, we find:

0.5 L = 1.25 atm * V2

Dividing both sides of the equation by 1.25 atm, we get:

0.5 L / 1.25 atm = V2

V2 ≈ 0.4 L

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Answer:

10 neutrons

Explanation:

The mass of is made up of the total protons and neutrons in an atom.

Each particle has a mass of 1 amu (atomic mass unit).

All fluorine atoms have 9 protons.

If each proton is 1 amu, the protons must contribute 9 amu to the total mass of the atom (9 x 1 amu = 9 amu).

This means the neutrons must contribute a total mass of 10 amu (18.998 amu - 9 amu = ~10 amu).

If each neutron has a mass of 1 amu, there must be 10 neutrons in a fluorine atom (10 amu / 1 amu = 10 amu).

The bulk density at 20 percent moisture, given that the soil dries to 20 percent by weight is 1.4 g/cm³

First, we shall obtain the initial volume of the soil. Details below:

Volume = mass / density

Initial volume of soil = 100 / 1.3

Initial volume of soil = 76.93 cm³

Next, we shall obtain the volume after the soil dries to 20 percent by weight. Details below:

We were told that the soil shrinks by an amount equal to the water loss, thus we have

Initial mass of water = 40% = 40 g

Water lost = 20% = 20 g

New mass of water = 40 - 20 = 20 g

Thus,

Volume = Equal value of mass of water lost = 20 cm³

Finally, we shall obtain the bulk density at 20 percent moisture. Details below:

Bulk density at 20 percent moisture = Total mass / new volume

Bulk density at 20 percent moisture = 80 /  5.93

Bulk density at 20 percent moisture = 1.4 g/cm³

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Answer:

11,1

Explanation:

-log(1.3x10^-11)

The pH of a solution with a hydrogen ion concentration of 1.3X10^-11 is -11.1. Therefore, option A is correct.

The potential of hydrogen ion concentration is referred to as pH. a way to gauge how basic or acidic a material is. The pH scale represents from 0 to 14.

A pH value of 7 is considered neutral on this scale, which implies it is neither acidic nor basic. It is more than acidic if the pH value is below seven, and more basic if the pH value is above seven.

The pH can control the presence of nutrients, biological processes, microbial activity, and chemical behaviour.

Given:

Hydrogen ion concentration = 1.3X10^-11

pH = -log [H+]

pH = -log(1.3x10^-11)

pH = -11.1

Thus,-11.1 is the pH of a solution with a hydrogen ion concentration of 1.3X10^-11, option A is correct.

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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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When burning magnesium is lowered in a gas jar full of carbon dioxide, the burning process continues because the carbon dioxide acts as an oxidizing agent.

A combustion reaction is a type of chemical reaction that occurs when a fuel, such as a hydrocarbon, reacts with an oxidizing agent, such as oxygen, to produce heat and light in the form of a flame. The basic equation for a combustion reaction is:

Fuel + Oxidant → Heat + Light + Products

In the case of burning magnesium, the fuel is magnesium and the oxidant is carbon dioxide. The products of the reaction are magnesium oxide and carbon dioxide. Combustion reactions are exothermic, meaning they release energy in the form of heat.

This means that the carbon dioxide provides the oxygen necessary for the combustion reaction to continue, allowing the magnesium to continue burning. Additionally, the carbon dioxide also helps to extinguish the flame by removing the heat and the oxygen from the combustion zone.

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Answer:

heated the water

Explanation:

The correct answer is A.

The nuclear equation that correctly characterizes a nuclear reaction is one where the sum of the atomic numbers and the sum of the mass numbers on both sides of the equation are equal.

\(_{92}^{235}\textrm{U}+_{0}^{1}\textrm{n}\rightarrow _{54}^{140}\textrm{Xe}+_{38}^{94}\textrm{Sr}+3_{0}^{1}\textrm{n}\)

         \(_{92}^{235}\textrm{U}+_{0}^{1}\textrm{n}\rightarrow _{54}^{140}\textrm{Xe}+_{38}^{94}\textrm{Sr}+3_{0}^{1}\textrm{n}\)

The correct answer is A.

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Answer:

water

Explanation:

Answer:

God

Explanation:

1.15 g of Cu are there in a sample of Cu that contains\(1.09*10^{24}\)atoms

The number of grams of Cu in a sample containing\(1.09*10^{24}\)atoms can be determined using Avogadro's number, which is \(6.02*10^{23}\)atoms/mol.

First, we need to calculate the number of moles of Cu in the sample. To do this, we divide the number of atoms by Avogadro's number to get the number of moles:

\(\frac{1.09*10^{24} atoms }{6.02*10^{23} atoms/mol }= 1.81*10^{-1 }mol\)

The sample's mass needs to be determined next. Since copper has an atomic mass of 63.55 g/mol, we can use this to determine the sample's mass:

\(1.81*10^{-1} mol * 63.55 g/mol = 1.15 g\)

Therefore, the sample contains 1.15 g of Cu.

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