Learning ObjectivesIn this activity, you'll practice determining the limiting reagent for a reaction between silver nitrate and copper metal. In part 1 you'll make the measurements you need so you can predict the amount of silver produced. In part 2 you'll be able to measure how much silver was actually produced. Finally you will be asked to consider possible reasons for the discrepancy.Part 1: Describe the ReactionIn this activity, you will be observing a reaction where copper and silver nitrate react to form copper (II) nitrate and silver.1.Write a balanced chemical equation for this reaction.2AgNO3 Cu -> 2Ag Cu(NO3)2Score: 0 / 12.

18 Chromosomes from his mother and 18 Chromosomes from his father.

If a spike has random chromosomes we would expect it to inherit 18 chromosomes from its father and 18 from its mother. All 36 chromosomes are arranged in two sets of chromosomes. Almost every cell in the human body has the same DNA. Most of the DNA is in the cell nucleus.

Must have a dominant allele to be considered a tastant hence Tt. Parents pass on traits such as eye color and blood type to their children through genes. Some health conditions and diseases can also be inherited genetically. Traits can take many forms. Bases are the parts of DNA that store information, giving DNA the ability to encode human phenotypes' visible characteristics. Adenine and guanine are purine bases.

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Answer:

30.8 grams O

Explanation:

In H₂O, there are 2 moles of hydrogen for every 1 moles of oxygen. You can use this to convert between moles of H and moles of O. Then, you can use the molecular weight of oxygen (15.998 g/mol) to convert between moles O and grams. The answer should have 3 sig figs because the given value (3.85) also has 3 sig figs.

3.85 moles H          1 mole O             15.998 g
--------------------  x  ------------------  x  --------------------  = 30.8 grams O
                              2 moles H            1 mole O

The temperature of the countertop changes by 0.032 °C

The quantity of heat transferred to the countertop is given by

Q = CΔT where Q = quantity of heat transferred to the countertop = 5000 J = 5 kJ, C = heat capacity of the countertop = 158.5 kJ/°C and ΔT = temperature change of the countertop.

Since we require the temperature change of the countertop, we make ΔT subject of the formula.

So, ΔT = Q/C

So, substituting the values of the variables into the equation, we have

ΔT = Q/C

ΔT = 5 kJ/158.5 kJ/°C

ΔT = 0.032 °C

So, the temperature of the countertop changes by 0.032 °C

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Metamorphic rock formation involves the transformation of existing rocks under heat and pressure. They form without melting and can appear foliated or non-foliated.

Metamorphic rocks are formed from the transformation of existing rock types in a process called metamorphism, which means 'change in form'. The appropriate phrases that describe metamorphic rock formation are: 'form from existing rocks', 'form without melting', 'appear foliated or non-foliated', and 'require heat and pressure to form'. These rocks are subject to conditions of heat and pressure that cause them to change physically and/or chemically, resulting in a new type of rock. They can either be foliated (layered) or non-foliated. Importantly, metamorphic rock formation does not include a liquid state, meaning they do not 'form from liquid rock' or 'form from deposition'.

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The mass of P-32 remaining after 27 days, if the initial mass is 40g is 10.5 g.

The half-life of a radioactive isotope is the amount of time it takes for one-half of the radioactive isotope to decay.

The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.

The number of half lives in 27 days;

n = 27 days/14 days

n = 1.929

The mass of P-32 remaining after 27 days, if the initial mass is 40g;

mass remaining = 40 g / (2^1.929)

mass remaining = 40 g / 3.808

mass remaining = 10.5 g

Thus, the mass of P-32 remaining after 27 days, if the initial mass is 40g is 10.5 g.

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Answer:

sound waves

Explanation:

hope this helps

The following are categorized into polar or nonpolar molecules:

a) Beryllium chloride (BeCl₂) is a nonpolar molecule. The Be-Cl bond is polar due to the electronegativity difference between beryllium and chlorine, but the molecule is linear with the two polar bonds pointing in opposite directions, resulting in a net dipole moment of zero.

b) Hydrogen sulphide (H₂S) is a polar molecule. The H-S bond is polar due to the electronegativity difference between hydrogen and sulfur, and the molecule has a bent shape, resulting in a net dipole moment that is not zero.

c) Sulphur trioxide (SO₃) is a nonpolar molecule. The S-O bonds are polar due to the electronegativity difference between sulfur and oxygen, but the molecule is trigonal planar with the three polar bonds pointing in different directions, resulting in a net dipole moment of zero.

d) Water (H₂O) is a polar molecule. The H-O bond is polar due to the electronegativity difference between hydrogen and oxygen, and the molecule has a bent shape, resulting in a net dipole moment that is not zero.

e) Trichloromethane (CHCl₃) is a polar molecule. The C-Cl bonds are polar due to the electronegativity difference between carbon and chlorine, and the molecule has a tetrahedral shape, resulting in a net dipole moment that is not zero.

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Answer:

2.567 mL

Explanation:

For exothermic reactions, ΔH is always negative, energy is considered a product, and energy is released in the reaction.

An exothermic reaction is one in which energy, usually in the form of heat, is released to the environment from the reaction. In other words, the final temperature of an exothermic reaction would always be more than the initial temperature.

This also means that heat energy is a product of exothermic reactions and this heat is released to the environment from the reaction.

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Answer:

a, d, and e is correct on edge of the nuity

Explanation:

have a good day

Answer:

option B

hope it helps! please mark me brainliest

Answer:

15.4%

Explanation:

If Ka = 0.54 mM = 1.51x10⁻⁵

Then;

C₄H₈O₂               -------->            C₄H₇O₂⁻          +           H⁺

I                    0.54x10⁻³                             0                                0

E                   0.54x10⁻³(1-x)                      0.54x10⁻³x                0.54x10⁻³x

Recall that x is the percentage degree of dissociation

From the ICE table;

Ka = [C₄H₇O₂⁻] [ H⁺]/[C₄H₈O₂]

1.51x10⁻⁵=(0.54x10⁻³x) (0.54x10⁻³x)/ 0.54x10⁻³(1-x)  

1.51x10⁻⁵ = 0.54x10⁻³x^2/1-x

1.51x10⁻⁵(1-x) = 0.54x10⁻³x^2

1.51x10⁻⁵ - 1.51x10⁻⁵x = 0.54x10⁻³x^2

Hence;

0.54x10⁻³x^2 + 1.51x10⁻⁵x - 1.51x10⁻⁵=0

x^2 + 0.028x - 0.028 = 0

Solving the quadratic equation here;

x = 0.154 or −0.182

Ignoring the negative result, x = 0.154

Hence, fraction of butanoic acid that is in the dissociated form in this solution = 15.4%

The quantity of NaOH required to reach the third equivalence point is 20mL.

Using the titration formula,

CaVa/CbVb = Na/Nb

Where,

Ca = concentration of citric acid (0.200 M)

Cb = concentration of NaOH (0.750 M)

Va = Volume of citric acid (25.0 mL)

Vb = volume of NaOH (x mL)

Na = number of reacting mole of citric acid (3)

Nb = number of reacting mole of NaOH (1)

Therefore Vb ( x mL) =CaVaNb/CbNa

 = 0.2× 25×3/0.75 ×1

= 15/0.75

                   Vb ( x mL) = 20 mL

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Answer:

x/2

Explanation:

X = 2 molecules of H2

For 2 molecules of H2, there's only 1 molecule of O2. Meaning, there's twice the amount of H2, so O2 = x/2 molecules.

I hope I'm understanding this question right.

The total number of electrons  Al+3. Fe¹3 Mg² Sn¹² Co²Co³ Lit¹ Cr+3,  Rb Pt+2 are 3, 3, 2, 12, 2, 3, 0, 3.0,2 electrons

Electrons are subatomic particles that have a negative charge and are found outside the nucleus of an atom. They orbit the nucleus in shells or energy levels and are involved in chemical bonding and electricity. Electrons are much smaller in size compared to protons and neutrons, which make up the nucleus of an atom.

Al+3 has 10 electrons (13 protons - 3 electrons).

Fe¹3 has 20 electrons (26 protons - 3 electrons).

Mg² has 10 electrons (12 protons - 2 electrons).

Sn¹² has 50 electrons (50 protons - 12 electrons).

Co² has 27 electrons (27 protons - 2 electrons).

Co³ has 26 electrons (27 protons - 3 electrons).

Li¹ has 3 electrons (3 protons - 0 electrons).

Cr+3 has 21 electrons (24 protons - 3 electrons).

Rb¹ has 37 electrons (37 protons - 0 electrons).

Pt+2 has 76 electrons (78 protons - 2 electrons).

Therefore,  The number of electrons in an atom determines its chemical behavior and the way it interacts with other atoms.

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The frequency of this radiation of a quantum of energy 4.14 *10-14 J is calculated to be 2.58 *10-15 Hz.

The Photoelectric Effect occurs when electrons are released from a metal surface when light strikes it. It was once thought that the greater the intensity of incoming light, the greater its energy, independent of hue. The light was viewed and treated as a wave, however, the wave phenomenon could not explain light's photoelectric actions.

The Quantum, according to Max Planck, is the smallest unit of energy that is released or absorbed in the form of electromagnetic radiation. The energy released or absorbed is proportional to the frequency of the radiation. It may be expressed mathematically as:

E= hc/λ

or

E= hf        ( because, f = c/λ)

where,

E= Energy of the particle

h= Planck's constant

c= speed of light

λ= wavelength of the particle

f = frequency of the radiation

Thus, E= hf

Given,

E = 4,14*10-14J

E= 1.71017356 * 10^20 eV

h= 6.626 *10-34 m2Kg/s

E=hf

f= E/h

f= (1.71017356 * 10^20 eV) / (6.626 *10-34 m2Kg/s)

f= C

Thus, the frequency of this radiation is 6.626 *10-34 m2Kg/s.

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The methyl orange indicator will change from red to yellow at the titration's endpoint.

An acid-base indicator called methyl orange changes color between the pH ranges of 3.1 and 4.4. In acidic and basic solutions it is red and yellow, respectively

In this instance, an acid, hydrochloric acid, is being used to titrate the barium hydroxide solution. The pH of the solution will fall as we add the acid since it will neutralize the base. The hue of the methyl orange indicator will vary when the pH ranges from 3.1 to 4.4

All of the barium hydroxide will have interacted with the hydrochloric acid by the time the titration is complete leaving a neutral solution. When the methyl orange indicator becomes yellow the solution's pH is in the basic range.

Therefore, the methyl orange indicator will change from red to yellow at the titration's endpoint.

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The name of the compound is thulium (iii) sulfate and it's molecular formula is:

\(Tm_2(SO_4)_3\)

The molecular formula of a compound is the formula showing the ratio in which atoms of elements combine in 1 mole of the compound.

The oxidation state of thulium is +3 and the oxidation state of SO4 is +2.

By exchange of radicals, the molecular formula of the compound is as :

\(Tm_2(SO_4)_3\)

The name of the compound is thulium (iii) sulfate.

Therefore, the compound is thulium (iii) sulfate.

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Answer:  36.1%

Explanation:  The molar mass of the complete molecule, CuSO4*5H2O is 249.6 g/mole.  The molar mass of just the water (5H20) is 90 g/mole.  Assume 1 mole of CuSO4*5H2O, the percent water would be:

90g H2O/249.6g CuSO4*5H2O = 0.361 or 36.1%

The molarity of the solution if 3.2 moles of MgCl, are dissolved in 1.3 liters of water is 2.46 M.

The molarity of a solution is the concentration of a substance in solution, expressed as the number of moles of solute per litre of solution.

The molarity of a solution can be calculated by dividing the number of moles in the substance by its volume as follows:

molarity = no of moles ÷ volume

According to this question, 3.2 moles of MgCl, are dissolved in 1.3 liters of water. The molarity is as follows:

Molarity = 3.2mol ÷ 1.3L

Molarity = 2.46 M

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Answer:

b

Explanation:

The temperature of the gas in Kelvin, after its volume is reduced to 2,450 mL, is approximately 143.27 K.

To determine the temperature of the gas in Kelvin after its volume is reduced, we can use the combined gas law, which relates the initial and final conditions of pressure, volume, and temperature for a given amount of gas.

The combined gas law equation is:

(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂

Where P₁ and P₂ are the initial and final pressures, V₁ and V₂ are the initial and final volumes, T₁ is the initial temperature in Kelvin, and T₂ is the final temperature in Kelvin.

Given that the initial volume V₁ is 3,480 mL, the initial temperature T₁ is -70.0 °C (which needs to be converted to Kelvin), and the final volume V₂ is 2,450 mL, we can substitute these values into the equation.

To convert -70.0 °C to Kelvin, we add 273.15 to it, resulting in T₁ = 203.15 K.

Now we can solve for T₂:

(T₂ * V₁) / T₁ = V₂

T₂ = (V₂ * T₁) / V₁ = (2,450 mL * 203.15 K) / 3,480 mL

Simplifying the equation, we find:

T₂ ≈ 143.27 K

Therefore, the temperature of the gas in Kelvin, after its volume is reduced to 2,450 mL, is approximately 143.27 K.

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Answer: first you have to calculate the amount ionized. We will say it is x mol / L

then % ionization = (amount ionized / initial concentration) * 100%

x can be calculated using an ice chart

HC3H5O2           -----> H+    + C3H5O2-

Initial HC3H5O2 = 0.250  

change              = -x

equilibrium         = 0.250 - x

initial H+            = 0

change              = +x

equilibrium         = x

C3H5O2- initial   = 0

change              = +x

equilibrium         = x

Ka         = [H=][C3H5O2-] / HC3H5O2]

 1.3 * 10 ^ -5    = [x][x] / (0.250 - x)  

So 1.3 * 10 ^ -5 * (0.250 - x)   = x ^ 2  

     3.25 * 10^ -6 - (1.3 * 10^-5)x = x^2   now this is a quadratic equation and you have to rearrange it and solve for x

x^2 + 1.3 * 10^-5)x - 3.25 * 10^ -6 = 0

use the equation x = {-b (+ or -)[b^2 - 4.a.c] ^ 1/2} / 2a

you should get x = 1.80 * 10 ^ -3 or x = -1.80* 10^-3

but x can not be negative..

so x = 1.80 * 10 ^ -3

so percent ionization = (1.80 * 10 ^ -3 / 0.250) * 100%

                               =0.72 %

the other way which is more easier is  

assuming that x is very small and therefore 0.250 - x is approximately equals to 0.250  

then 1.3 * 10^-5 = x^2 / 0.250

so x^2    = 1.3 * 10^-5 * 0.250

        x   = 1.80 * 10 ^-3

then percent ionization is = (1.80 * 10 ^ -3 / 0.250) * 100%

                               =0.72 %

if the percent ionization is > 5 % you can not do that approximation. in such a case you have to solve the quadratic equation. that is why I showed both methods.

now you can do the parts b and c

b answer : percent ionization = 1.27 %  

c answer : 2.54%

good luck

The identity of a cation solution that produces a mix of red and yellow colors in a flame test, but appears red when viewed through a cobalt filter, can be attributed to the presence of the strontium (Sr) cation.

During a flame test, different metal cations emit characteristic colors due to the excitation of electrons and their subsequent emission of light. Strontium, in particular, is known to produce a vibrant red color in flame tests.

The presence of both red and yellow colors indicates the possibility of multiple metal cations in the solution. While the specific metal responsible for the yellow color is uncertain, it could potentially be sodium or another metal that emits a yellow flame.

When the flame is viewed through a cobalt filter, which absorbs yellow wavelengths of light, the yellow color is filtered out, resulting in only the red color being observed. Since strontium is known for its distinctive red flame color and its emission is not affected by the cobalt filter, it is likely the metal cation responsible for the observed red color. Therefore, based on these characteristics, the identity of the cation solution is most likely strontium (Sr).

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Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

Some types of solids that are insoluble in water are:

Many solids are insoluble in water, meaning they do not dissolve in water to a significant extent. Here are some examples of common solids that are generally insoluble in water:

Metals: Most metals, such as gold, silver, platinum, and copper, are insoluble in water.

Non-Metallic Elements: Many non-metallic elements, such as carbon (in the form of graphite or diamond), sulfur, phosphorus, and iodine, are insoluble in water.

Metal Oxides: Some metal oxides, particularly those of less reactive metals, are insoluble in water. Examples include aluminum oxide (Al2O3), iron(III) oxide (Fe2O3), and lead(II) oxide (PbO).

Metal Carbonates: Most metal carbonates are insoluble in water. Examples include calcium carbonate (CaCO3), lead(II) carbonate (PbCO3), and copper(II) carbonate (CuCO3).

Metal Sulfides: Many metal sulfides are insoluble in water. Examples include lead(II) sulfide (PbS), silver sulfide (Ag2S), and mercury(II) sulfide (HgS).

Insoluble Salts: Certain salts have limited solubility in water. Examples include silver chloride (AgCl), lead(II) iodide (PbI2), and calcium sulfate (CaSO4).

It's important to note that while these solids are generally insoluble in water, they may exhibit some solubility to a small extent. The solubility of a solid in water can vary depending on factors such as temperature, pressure, and the presence of other solutes.

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Answer:

Explanation:

From the given information:

The concentration of metal ions are:

\([Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}\)

\([Ca^{2+}]=0.007118 \ M\)

\([Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}\)

\(=0.001994 \ M\)

Mass of Ca²⁺ in 2.00 L urine sample is:

\(= 2.00 L \times 0.001994 \dfrac{mol}{L} \times \dfrac{40.08 \ g}{1 \ mol}\)

= 0.1598 g

Mass of Ca²⁺ = 159.0 mg

Mass of Mg²⁺ in 2.00 L urine sample is:

\(= 2.00 L \times 0.007118 \dfrac{mol}{L} \times \dfrac{24.31 \ g}{1 \ mol}\)

= 0.3461 g

Mass of Mg²⁺ = 346.1 mg

Answer:

Erosion can happen very quickly or very slowly.

e.g splash erosion occurs slowly while gulley erosion occurs very fast.