How many moles of gold are equivalent to 1.204 × 1024 atoms? 0.2 0.5 2 5

Answer:

Two constitutional isomers for the compound C3H8S are shown in the attachment below

Explanation:

Constitutional isomers are isomers that have the same molecular formula but different connectivity. Two constitutional isomers for the compound C3H8S are shown in the attachment below.

As per the concept of structural isomers, the  two constitutional isomers that share the molecular formula C₃H₈S are attached in attachment below.

Structural isomers are defined as the isomers  in which atoms are completely  arranged  in a different order but the molecular formula remains the same.

They are the molecules which have same molecular formula but different  connectivities  of atoms  which depend on the order they are put together. An increase in the number of carbon atoms leads to an increase  in the structural isomers.

Learn more about structural isomers,here:

https://brainly.com/question/33929317

#SPJ6

Answer:

[tex]\huge\boxed{Density}[/tex]

Explanation:

This is because of density. Since Ice is less dense than water, the ice even formed inside the pond then starts floating on the surface of water. Because of this density, Ice floats on water and thus the water freezes at the surface.

Answer:

Density

Explanation:

Water is less dense as a solid than as a liquid, so as ponds freeze, the ice floats at the top and the pond freezes from the top-down.

Answer:

[tex]E=2.06\times 10^{-19}\ J[/tex]

Explanation:

Given that,

The wavelength of electromagnetic radiation is 963.5 nm.

We need to find the energy of a photon with this wavelength.

The formula used to find the energy of a photon is given by :

[tex]E=\dfrac{hc}{\lambda}\\\\E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{963.5\times 10^{-9}}\\\\E=2.06\times 10^{-19}\ J[/tex]

So, the energy of a photon is [tex]2.06\times 10^{-19}\ J[/tex].

Answer:

The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J

Explanation:

Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.

The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:

Q = c * m * ΔT

Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).

In this case:

Replacing:

[tex]Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K[/tex]

Q= 64,121.2 J

The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J

Answer:

406 K.

Explanation:

The following data were obtained from the question:

Number of mole (n) = 0.750 mole

Volume (V) = 10.0 L

Pressure (P) = 2.50 atm

Temperature (T) =.?

Note: Gas constant (R) = 0.0821 atm.L/Kmol

The temperature, T can be obtained by using the ideal gas equation as follow:

PV = nRT

2.5 x 10 = 0.75 x 0.0821 x T

Divide both side by 0.75 x 0.0821

T = (2.5 x 10) /(0.75 x 0.0821 )

T = 406 K.

Therefore, the temperature is 406 K.

Answer: 406 K

Explanation:

We can rewrite the ideal gas law to solve for T:

PV = nRT

T=PV / nR

We are given the following from the problem:

n=0.750 mol P=2.50 atm V=10.0 L

Plugging in our values and using R=0.08206 L⋅atm / K⋅mol we get:

T=(2.50 atm)(10.0 L) / (0.750 mole)(0.08206L ⋅ atm ⋅ mole K) = 406 K

Answer:

See explanation

Explanation:

1)

Mg3N2(s) + 6H2O(l) ------------> 3Mg(OH)2 + 2NH3(g)

Number of moles of magnesium nitride= mass/molar mass= 4.86g/100.9494 g/mol = 0.048 moles

1 mole of magnesium nitride yields 3 moles of magnesium hydroxide

0.048 moles of magnesium nitride yields 0.048 moles × 3= 0.144 moles of magnesium hydroxide

Theoretical yield of magnesium hydroxide = 0.144 moles × 58.3197 g/mol = 8.398 g

Percent yield= actual yield/ theoretical yield × 100

Percent yield= 7.18/8.398 × 100/1 = 85.5%

2)

N2(g) + 3H2(g) -------> 2NH3(g)

Number of moles of hydrogen gas = mass/ molar mass = 6.41g/ 2gmol-1 = 3.205 moles of hydrogen gas.

From the balanced reaction equation;

3 moles of hydrogen gas yields 2 moles of ammonia

3.205 moles of hydrogen gas yields 3.205 × 2/3 = 2.1367 moles of ammonia

Theoretical yield of ammonia = 2.1367 moles × 17 gmol-1 = 36.3 g

Percent yield = actual yield/ theoretical yield ×100

Percent yield = 26.2/36.3 ×100

Percent yield = 72.2%

3)

3Cl2(g) + 3H2O(l) ------> HOCl3(aq) + 5HCl(aq)

Number of moles of water= mass/ molar mass = 3.79g/18 gmol-1 = 0.21 moles

Since

3 mole of water yields 5 mole of HCl

0.21 moles of water yields 0.21 × 5/3 = 0.35 moles of HCl

Theoretical yield of HCl = 0.35 moles × 36.5 gmol-1 = 12.775 g

Percent yield = actual yield/ theoretical yield × 100/1

Percent yield = 8.70/12.775 ×100

Percent yield = 68.1%

Answer:

pH of the solution is 2.0

Explanation:

The lactic acid is a weak acid that is in equilibrium with water as follows:

Lactic acid + H2O ⇄ Lactate + H₃O⁺

And Ka for lactic acid: 1.38x10⁻⁴

Ka = 1.38x10⁻⁴ = [Lactate] [H₃O⁺] / [Lactic acid]

Initial concentration of lactic acid is (MW: 112.06g/mol):

80g * (1mol / 112.06g) / 1L = 0.714M

The equilibrium concentration of the species in the equilibrium are:

[Lactate] = X

[H₃O⁺] = X

[Lactic acid] = 0.714-X

Replacing in Ka expression:

1.38x10⁻⁴ = [X] [X] / [0.714-X]

9.8532x10⁻⁵ -  1.38x10⁻⁴X = X²

9.8532x10⁻⁵ -  1.38x10⁻⁴X - X² = 0

Solving for X:

X = -1.0x10⁻². False solution, there is no negative concentrations

X = 9.86x10⁻³M. Right solution.

As [H₃O⁺] = X

[H₃O⁺] = 9.86x10⁻³M

and pH = -log [H₃O⁺] = -log 9.86x10⁻³M

pH = 2.0

Answer:

Explanation:

The equation for the decay is given as;

¹⁴₆C  --> X + ⁰₋₁e

For conservation of matter, the mass number and atomic number has to be the same in both the reactant and product side f he equation;

Mass number;

14 = x + 0

x = 14

Atomic Number;

6 = x + (-1)

x = 6 + 1 =7

¹⁴₆C  --> ¹⁴₇N + ⁰₋₁e

The name of the product nuclide with atomic number of 7 is Nitrogen. The symbol is; ¹⁴₇N

Answer:

Temperature of the water

Explanation:

In every study, there must be independent and dependent variables. An independent variable is the variable that is changed in order to obtain a response. In this case, the temperature of the water is being changed, the response in this experiment is the respiration rate of the goldfish.

Thus the respiration rate of the goldfish is the dependent variable because it is controlled by the temperature of the water and changes accordingly.

Summarily, the independent variable is the temperature of the water while the dependent variable is the respiration rate of the goldfish.

Answer: It's the anode broski (B)

Explanation: I'm taking the Chem summer course too broski, this was the correct answer. Cheers broski

Answer:

C) SO3

Explanation:

Lewis formula shows the bonding between atoms of a molecule and expresses the lone pair present in the atoms.

SO3 or Sulfur trioxide cannot be adequately described by a single Lewis formula because it has majorly 3 resonance structures because Sulfur does not follow the octet rule and can expand electrons in its outer shell.

Hence, the correct answer is C) SO3

Answer:

AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.

Explanation:

If a student mixes 1.0 mL of aqueous silver nitrate AgNO3 (aq)  with 1.0 mL of aqueous sodium chloride, NaCl (aq), in a clean test tube.

The sodium chloride is being acidified with dilute trioxonitrate (V) acid. Then a few drops of  silver trioxonitrate(V) is added afterwards. A  white precipitate of silver chloride, which dissolves readily in aqueous ammonia indicates the presence of sodium chloride.

The reaction proceeds as follows:

[tex]\mathtt{AgNO_{3(aq)} + NaCl _{(aq)} \to AgCl _{(s)} + NaNO_3_{(aq)}}[/tex]

From the reaction between AgNO3 (aq) and NaCl (aq), AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.

Answer:

In basic conditions that is ( pH > 7 ), the equilibrium shifts towards right and produces a lot of (-ve) negatively charged fatty acids which are polar, In water, since they have long hydrophobic hydrocarbon part, this form micelles where the hydrocarbon part  remain inside the  sphere and -coo- group remain outside the sphere due to H-bonding interaction with water.

At ( pH < 5 ) I.e acidic conditions, the equilibrium shift to the left giving neutral molecules which can not have stronger H-bonding interaction .

So micelles cant form as they become insoluble.

Answer:

concentration   in    mmol/L = 8.97 × 10⁻¹ mmol/L

Explanation:

Given that:

the number of moles of  silver(I) nitrate(AgNO3) the chemist used in preparing a solution = 269 mmol = 269 × 10⁻³ mmol

The volume of the volumetric flask = 300 mL  = 300 × 10⁻³ L

In order to calculate the concentration in mmol/L of the chemist's silver(I) nitrate (AgNO3) solution , we used the formula which can be expressed as;

[tex]concentration \ in \ mmol/L = \dfrac{ number \ of \ mmol}{vol. \ of \ the \ solution}[/tex]

[tex]concentration \ in \ mmol/L = \dfrac{ 269 * 10^{-3 } \ mmol }{300 * 10^{-3} \ L }[/tex]

concentration   in    mmol/L = 0.8966   mmol/L

concentration   in    mmol/L = 8.97 × 10⁻¹ mmol/L

The two layers are part of the thermosphere are  exosphere and ionosphere.

The Exosphere is the topmost layer of the Earth's atmosphere.

and its gradually disappear into the vacuum of space.

It consist two parts that are:

exosphere and ionosphere.

Thus, option "A" is correct, the rest of the option is not a part of thermosphere.

To learn more about  atmospheric layers click here:

https://brainly.com/question/25881294

#SPJ2

Answer:

hi hope your doing great the answer is A

Explanation:

its on Edge 2020

hope i helped :)

Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 [tex]\frac{J}{g*C}[/tex] and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 [tex]\frac{J}{g*C}[/tex] *20 C and solving: Q=420 J

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g*[tex]6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}[/tex]= 3.333 kJ= 3,333 J (being kJ=1,000 J)

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 [tex]\frac{J}{g*C}[/tex] and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 [tex]\frac{J}{g*C}[/tex] *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

The amount of heat to absorb is 6,261 J.

Heat required to raise the temperature of ice from -20 °C to 0 °C.

The formula for specific heat is used to calculate the amount of heat

Q = c * m * ΔT

Where,

Q =heat exchanged by a body,

m= mass of the body

c= specific heat

ΔT= change in temperature

Given:

m= 10 g,

specific heat of the ice= 2.1

ΔT=0 C - (-20 C)= 20 C

On substituting the values:

Q= 10 g*2.1  *20 C

Q=420 J

Heat required to convert 0 °C ice to 0 °C water.

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

Heat required to raise the temperature of water from 0 °C to 60 °C

m= 10 g,

Specific heat of the water= 4.18  

ΔT=60 C - (0 C)= 60 C

On substituting:

Q= 10 g*4.18  *60 C

Q=2,508 J

Thus, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

Find more information about Specific heat here:

brainly.com/question/13439286

Answer:

mol LiCl = 4.83 m

Explanation:

GIven:

Solution of LiCl in water XLiCl = 0.0800

Mol of water in kg = 55.55 mole

Find:

Molality

Computation:

mole fraction = mol LiCl / (mol water + mol LiCl)

0.0800 = mol LiCl / (55.55 mol + mol LiCl)

0.0800 mol LiCl + 4.444 mol = mol LiCl

mol LiCl - 0.0800 mol LiCl = 4.444 mol

0.92 mol LiCl = 4.444 mol

mol LiCl = 4.83 m

Answer:

Depends on the chemicals.

Explanation:

The creation of something depends on the chemical we have. If I have chemicals such as acid and base so I can produce salt and water by mixing acid and base. If I have ethanol, vegetable oil and sodium hydroxide which is  a catalyst, I can produce biodiesel which can be used in vehicles instead of fossil fuel. If I have sodium element and chlorine gas, I can produce sodium chloride also known as table salt.

Answer:

Atomic mass = 127.198 amu

Explanation:

The average atomic mass is obtained by summing the masses of the isotopes each multiplied by its abundance.

Atomic mass = (97.62 * 0.0825) + (109.3 * 0.2671) + (138.3 * 0.6504)

Atomic mass = 8.05365 +  29.19403 + 89.95032

Atomic mass = 127.198 amu

Answer:Cobb

Explanation:What y'all

Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.

hydro = water

Answer:

Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!

Explanation:

The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)

Answer:

0.05 mole of Br2.

Explanation:

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

C2H4 + Br2 —› C2H4Br2

From the balanced equation above,

1 mole of C2H4 reacted with 1 mole of Br2 to produce 1 mole of C2H4 Br2.

Finally, we shall determine the number of moles bromine that will react with 0.05 mole of C2H2.

The number of mole of Br2 needed for the reaction can be obtained as follow:

From the balanced equation above,

1 mole of C2H4 reacted with 1 mole of Br2.

Therefore, 0.05 mole of C2H4 will also react with 0.05 mole of Br2.

Therefore, 0.05 mole of Br2 is needed for the reaction.

Answer:

[tex]T=469.1K\\\\T=195.9\°C[/tex]

Explanation:

Hello,

In this case, for the given decomposition reaction, we can compute the enthalpy of reaction considering the enthalpy of formation of each involved species (products minus reactants):

[tex]\Delta _rH=2\Delta _fH_{Ag}+\frac{1}{2} \Delta _fH_{O_2}-\Delta _fH_{Ag_2O}\\\\\Delta _rH=2*0.00+\frac{1}{2} *0.00-(-31.1)=31.1kJ/mol[/tex]

Next, the entropy of reaction considering the standard entropy for each involved species (products minus reactants):

[tex]\Delta _rS=2S_{Ag}+\frac{1}{2} S_{O_2}-S_{Ag_2O}\\\\\Delta _rS=2(42.55)+\frac{1}{2} (205.0)-(121.3)=66.3J/(mol*K)[/tex]

Next, since the Gibbs free energy of reaction is computed in terms of both the enthalpy and entropy of reaction at the unknown temperature, for such Gibbs energy equaling 0, the temperature (in K and °C) turns out:

[tex]\Delta _rG=\Delta _rH-T\Delta _rS\\\\0=31.1kJ/mol-T(66.3\frac{J}{mol*K}*\frac{1kJ}{1000J} )\\\\T=\frac{31.1kJ/mol}{0.0663kJ/(mol*K)} =469.1K\\\\T=195.9\°C[/tex]

Which is within the given rank.

Best regards.

Answer: 1.2 millimoles will be left after 2250 years

Explanation:

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]

[tex]k=\frac{0.693}{5730}=0.00012years^{-1}[/tex]

b) Amount left after 2250 years

[tex]2250=\frac{2.303}{k}\log\frac{1.6}{a-x}[/tex]

[tex]2250=\frac{2.303}{0.00012}\log\frac{1.6}{a-x}[/tex]

[tex]\log\frac{1.6}{a-x}=0.117[/tex]

[tex]\frac{1.6}{a-x}=1.31[/tex]

[tex]{a-x}=\frac{1.6}{1.31}=1.2[/tex]

Thus 1.2 millimoles will be left after 2250 years

Answer:

[OH-] = 1.0 x 10-10 M

Explanation:

The acidity of a solution can be determined directly from the concentration of the hydrogen ions and indirectly from the concentrations of the hydroxide ions.

Generally, for a neutral solution we have;

[H3O+] = [OH-] = 1.0 x 10-7 M

For an acidic solution;

[H3O+] > 1.0 x 10-7 M

[OH-] < 1.0 x 10-7 M

Comparing the options the correct option is;

[OH-] = 1.0 x 10-10 M

Answer:

It will react with water

Explanation:

I know this is an older question, but I just wanted to provide the correct answer.

Since we are dealing with a weak base, and the acid is somewhat stronger, it will react with the water molecules to produce hydronium, which will affect the pH of the solution.

The conjugate base of a weak acid affect pH because it will react with hydronium ion.

A weak acid is an acid that does not dissociate completely in water. On the other hand, a strong acid achieves almost 100% dissociation in water.

Acids dissociate in water to yield the hydronium ion and a conjugate base. For instance, the weak acid, acetic acid is dissociated as follows;

CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+(aq)

We can see that the conjugate base( CH3COO-) could react with the hydronium ions thereby moving the equilibrium position to the left hand side and affecting the pH by decreasing the hydronium ion concentration.

Learn more: https://brainly.com/question/8592296

Missing parts;

Why will the conjugate base of a weak acid affect pH? Select the correct answer below: O it will react with hydroxide O it will react with water O it will react with hydronium O none of the above

Answer:

the mass does not change

Explanation:

Answer:

1. 36

2. Two

Explanation:

The Lewis structure shows the valence electrons present in a compound. Usually the valence electrons are shown as dot structures around the symbol of the elements involved in the compound.

For a compound AB4 where B is more electronegative than A and A has 8 electrons in its valence shell, there will be thirty six valence electrons on the outermost shell of the molecule.

There are six electron pair domains present in the molecule, four bond pairs and two lone pairs. The molecule is in a square planar geometry.

Answer: a- 36 valence electrons

b- 14 lone pairs

Explanation:

Valence is equal to A + 4B = 8 + 4(7)

With 4 bonds between A and the 4 B, that is 36 valence minus 8 electrons in those pairs = 28. 28 is 14 lone pairs.

Answer:

anilinium ion > ammonium ion > amide > aniline > secondary amine

Explanation:

Acidity of amine derivatives can derived from their pKa values.

The rule of thumb for acidity with relation to pKa values is that:

As the pKa decreases the acid strength increases and the conjugate base decreases. Similarly, as the pKa increases, the acid strength decreases and the conjugate base increase.

Hence the stronger the acid , the  lower pKa value  and the weaker the acid , the stronger the pKa value.

So the pKa value for anilinium ion = 4.6

ammonium ion = 9.4

Amide = 15

Similarly, for aniline and secondary amine, in order to determine the derivative with the higher acidity, we will consider the electron withdrawing substituent group.

The more difficult the electron are being withdraw from the electron withdrawing substituent , the more acidic the compound.

In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.

Thus, from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).

the amine derivatives ranking is as follows:

anilinium ion > ammonium ion > amide > aniline > secondary amine