What type of compound does the formula CuCl2 represent?
A) ionic salt
B) covalent molecule

Answer: Thus volume in liters is [tex]6.14\times 10^3L[/tex]

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of silver oxide}}{\text{Volume of solution in L}}[/tex]     .....(1)

Molarity of silver oxide solution = [tex]7.73\times 10^{-5}mmol/L=7.73\times 10^{-2}\mu mol/L[/tex]      [tex]1mmol=1000\mu mol[/tex]

Moles of silver oxide = [tex]475\mu mol[/tex]      

Volume of solution in L = ?

Putting values in equation 1, we get:

[tex]7.73\times 10^{-2}\mu mol\L=\frac{475\mu mol}{\text{Volume of solution in L}}\\\\\{\text{Volume of solution in L}}=\frac{475\mu mol}{7.73\times 10^{-2}\mu mol\L}=6.14\times 10^3L[/tex]

Thus volume in liters is [tex]6.14\times 10^{3}L[/tex]

Answer:

See explanation for answer

Explanation:

You are missing some important data, in this case, the mass and temperature of the water.

In order to solve for this and explain to you how to do it, I'm going to assume some values for the mass and temperature for water, and then, you only need to replace these values with the values you have to get an accurate answer.

Let's assume the mass of water we have is 0.150 kg of water at 40 °C.

Now, let's remember that the water and ice are exerting heat and the sum of these heats must be zero always:

Q = Qw + Qi  = 0 (1)

In the case of the heat exerted by the ice, let's remember that the ice is passing through several stages. First, it's temperature changes but not it's phase, it remains solid. Second, it's when the ice begins to melt and ends when it's totally melted. And third, the change of temperature of water when the ice melted. So for these three stages, the heat of ice can be calculated with the following expression:

Qi = (mi * Ci * ΔTi) + (mi * Lf) + (mi * Cw * ΔTm)   (2)

The values of Ci, Lf and Cw are tabulated and reported data, and these are the following:

Ci: Specific heat of ice =  2100 J/kg °C

Cw: Specific heat of water = 4190 J/kg °C

Lf: heat of fusion of water =  3.34x10⁵ J/kg

For the case of heat of water it's just the specific heat of water, it's mass and the difference of temperature and the expression is:

Qw = mw * Cw * ΔT  (3)

Now, let's calculate firt the heat of water:

Qw = 0.150 * 4190 * (27.2 - 40)

Qw = -8044.8 J (a)

We have the heat of water, now, let's calculate heat of ice in function of the mass of ice:

Qi = [mi * 2100 * (0 + 10.9)] + (mi * 3.34x10⁵) + [mi * 4190 * (27.2 - 0)]

Qi = 22,890mi + 3.34x10⁵mi + 113,968mi

Qi = 470,858mi (b)

Finally, replace (a) and (b) in equation (1) to solve for mass of ice:

0 = 470,858mi - 8044.8

8044.8 = 470,858mi

mi = 8044.8 / 470,858

mi = 0.0171 kg or 17.1 g of ice

This is the mass required to drop the temperature of the system to 27.2 °C. Now, remember to replace the value of mass of water and temperature in this procedure to get the real and accuraten answer.

(a) The percentage of gold in the jewelry is  100%.

(b) The purity of the gold jewelry in karats is 24 karats.

(a) The percentage of gold in the jewelry is calculated as;

The given parameters include;

Density of gold (ρg) = 19.3 g/cm³

Density of silver (ρs) = 10.5 g/cm³

Volume of the jewelry (Vj) = 0.675 cm³

The total volume of the jewelry is the sum of the volumes of gold and silver:

Vj = Vg + Vs

Vg = Vj (since we are assuming the total volume is the sum of gold and silver volumes)

Vs = Vj - Vg

= 0.675 cm³ - 0.675 cm³

= 0 cm³

Mg = Vg x ρg

= 0.675 cm³ x 19.3 g/cm³

= 13.03 g

Ms = Vs x ρs

= 0 cm³ x 10.5 g/cm³

= 0 g

The percentage of gold is;

Percentage of gold = (Mg / (Mg + Ms)) x 100

= (13.03 g / (13.03 g + 0 g)) x 100

=  100%

(b) Since the jewelry is 100% gold by mass, its purity in karats is 24 karats (the highest purity since it's pure gold).

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Answer:

The answer to your question is a) pure substance- compound

Explanation:

a) pure substance-compound.  A pure substance can be an element or compound that is not mixed with another substance, glucose is a compound and also a pure substance.

b) mixture-heterogeneous.  This option is wrong because glucose is not mixed with another substance.

c) pure substance-element.  Glucose is a pure substance but it is a compound because it is composed of carbon, hydrogen, and oxygen. This option is incorrect.

d) mixture-homogeneous.  Glucose is not a mixture, it is a pure substance. This option is wrong.

e) none of the above. This option is wrong because the first option is correct.

Answer: A. Pure substance - compound

Explanation: Sugars or sacharrides are complex compounds having Carbon,Hydrogen and Oxygen atoms on its composition. Since a combination of two or more different elements forms a compound and compounds are pure substances.

Answer:

-375.9_KJ/(mol)

Explanation:

H(T2 ) ≈ H(T1)+CPΔT

Specific heat of Carbon is 0.71 J/g K.

At 283.15 the heat capacity is 37.12 J/(mol*K)

Kirchhoff's law

H(T2 ) ≈ H(T1)+CPΔT

Where

H(T1) and H(T2 ) are the heat of formation of CO2 at temperatures T1 and T2

CP is the heat capacity

Thus we have and ΔT is the temperature change

H(T2 ) ≈ -393.51×10^3+CP×(500-25)

= -393.51×10^3+37.12×(500-25)

= -375878 J/(mol)

= -375.9KJ/(mol)

To find the enthalpy of formation of CO2 at 500°C, we can use the equation ΔH = ΔH° + CpΔT, where ΔH° is the enthalpy change at 25°C, Cp is the heat capacity, and ΔT is the change in temperature. Assuming a constant heat capacity over the temperature range, we can calculate the change in enthalpy using the average heat capacity and the given values. The enthalpy of formation at 500°C is approximately -380.76 kJ/mol.

The enthalpy of formation of CO2 at 25°C is -393.51 kJ/mol. To find the enthalpy of formation at 500°C, we need to consider the change in enthalpy with temperature. One approach is to use the equation ΔH = ΔH° + CpΔT, where ΔH is the enthalpy change at the higher temperature, ΔH° is the enthalpy change at 25°C, Cp is the heat capacity, and ΔT is the change in temperature. However, since the question does not provide the heat capacity, an alternative method is to assume that the heat capacity is constant over the temperature range and use the average heat capacity to calculate the change in enthalpy.

For CO2, the average heat capacity over this temperature range is approximately 29.0 J/(mol·K). So, to find the enthalpy of formation at 500°C, we can use the equation ΔH = ΔH° + CpΔT, where ΔH° = -393.51 kJ/mol, Cp = 29.0 J/(mol·K), and ΔT = 500 - 25 = 475 K. Plugging in these values, we get: ΔH = -393.51 kJ/mol + (29.0 J/(mol·K) × 475 K) = -380.76 kJ/mol.

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Answer:

Option (A)

Explanation:

Cosmic radiations are usually defined as a type of radiations that is comprised of high-energy photons and carry harmful ultraviolet radiation from the sun. These are emitted from the sun at a speed that is equivalent to the speed of the light.

When these radiations are incident on earth, it interacts with the upper atmosphere, resulting in the emission of charged particles such as pions, which undergoes decay and releases other smaller particles, commonly known as muons.

Thus, the correct answer is option (A).

Answer: water is a liquid at room temperature because of the hydrogen bonding between its molecule.

10 g of sugar in 100 ml of water is sweeter.

Explanation:

We need to find the concentration of the two solutions from that we can find which one is dilute and which one is concentrated solution.

Concentration is given in terms of molarity (M) = moles/L

Moles can be found by dividing the given mass by the molar mass of the sugar (342.3 g/mol)

Now we can find the molarity as,

1. 100 ml = 0.1 L

moles = 10 g / 342.3 g / mol = 0.029 moles

molarity = 0.029 moles / 0.1 L = 0.29 M

2. molarity = 0.029 / 0.5 L = 0.058 M

So first solution is more concentrated, and also taste sweeter than the second solution.

10 g of sugar in 100 ml of water is more sweeter.

Molarity is defined as number of moles of solute over volume of solution in litres.

So, concentration is given in terms of molarity (M) = moles / L

Number of moles is given mass over molar mass:

Molar mass of sugar = 342.3 g/mol

Number of moles= 10 g / 342.3 g / mol

Number of moles = 0.029 moles

Molarity for two solutions will be:

Molarity = 0.029 moles / 0.1 L

Molarity = 0.29 M

Molarity = 0.029 / 0.5 L

Molarity  = 0.058 M

So, first solution is more concentrated, and also taste sweeter than the second solution.

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Answer : The partial pressure of [tex]NO_2[/tex] is, 12.34  atm

Explanation :

For the given chemical reaction:

[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]

The expression of [tex]K_p[/tex] for above reaction follows:

[tex]K_p=\frac{(P_{NO_2})^2}{P_{N_2O_4}}[/tex]         ........(1)

The equilibrium reaction is:

                     [tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]

Initial             x                    0

At eqm       (x-y)                 2y

Putting values in expression 1, we get:

[tex]70.9=\frac{(2y)^2}{(x-y)}[/tex]           ..............(2)

As we are given that, 25.8 % of [tex]N_2O_4[/tex] remains at equilibrium. That means,

[tex]25.8\% \times x=(x-y)[/tex]

[tex]\frac{25.8}{100}\times x=(x-y)[/tex]

[tex]0.258\times x=(x-y)[/tex]

[tex]0.742x=y[/tex]       ..............(3)

Now put equation 3 in 2, we get the value of 'x'.

[tex]70.9=\frac{(2\times 0.742x)^2}{(x-0.742x)}[/tex]

[tex]x=8.31[/tex]

Now put the value of 'x' in equation 3, we get:

[tex]0.742x=y[/tex]

[tex]0.742\times 8.31=y[/tex]

[tex]y=6.17[/tex]

Now we have to calculate the new partial pressure of [tex]NO_2[/tex] at equilibrium.

Partial pressure of [tex]NO_2[/tex] = (2y) = (2\times 6.17) = 12.34  atm

Hence, the partial pressure of [tex]NO_2[/tex] is, 12.34  atm

To determine the partial pressure of NO2 at equilibrium, use the equilibrium expression and substitute the given values into the equation.

To determine the partial pressure of NO2 at equilibrium, we will use the equilibrium expression for the reaction:

Kp = (P(NO2))^2 / P(N2O4)

Given that the equilibrium constant Kp is 70.9 and the equilibrium concentration of N2O4 is 25.8% of its initial concentration, we can calculate the partial pressure of NO2 at equilibrium:

P(NO2) = sqrt(Kp * P(N2O4))

Substituting the values, we get:

P(NO2) = sqrt(70.9 * (1 - 0.258))

P(NO2) = sqrt(70.9 * 0.742)

Answer:Use the K L M N shell orbit

Explanation: Valence electrons are those electrons on the outermost shell of an atomic nucleus.

The K L M N electron shell is named by a spectroscopist called Charles G. Barkla, who discovered the A X-rays ( high energy rays) and the B X-rays ( low energy rays), he later renamed the A X-rays K X-rays which is the Energy level orbit, and the B X-rays was also renamed to be the L M N and so on.

PLEASE SEE PICTURE ATTACHED

THE PICTURE SHOWS HOW TO DETERMINE VALENCE ELECTRON USING THE As AND B's WHICH ARE THE K L M N SHELL ORBIT, USING SILICON AS AN EXAMPLE.

The maximum number of electron in the K orbit is 2

The maximum number of electron in the L M N shell is 8

Therefore the number of electron found in the outermost shell becomes the number of Valence electron in that element.

Answer:

6.1 h = 6 h and 8 min

Explanation:

First, let's found the rate of disappearing of N2O5. Knowing that it's a first-order reaction, it means that the rate law is:

rate = k*pN2O5

Where k is the rate constant, and pN2O5 is the initial pressure of N2O5 (2.0 atm), so:

rate = 3.4x10⁻⁵*2.0

rate = 6.8x10⁻⁵ atm/s

Thus, at each second, the partial pressure of the reagent decays 6.8x10⁻⁵ atm. The rate is also the variation of the pressure divided by the time. Because it is decreasing, we put a minus signal in the expression.

1 atm = 760 torr, so 380torr/760 = 0.5 atm

rate = -Δp/t

6.8x10⁻⁵ = -(0.5 - 2.0)/t

t = 1.5/6.8x10⁻⁵

t = 22,058 s (÷60)

t = 368 min (÷60)

t = 6.1 h = 6 h and 8 min

To determine the time it takes for a sample of N2O5 to reach a certain partial pressure, we can use the first-order rate equation and solve for time.

The decomposition of N2O5(g) into NO2(g) and O2(g) at 25°C follows first-order kinetics. The rate constant, k, is given as 3.4×10−5 s−1.

To determine how long it will take for a sample initially containing 2.0 atm of N2O5 to reach a partial pressure of 380 torr, we can use the first-order rate equation:

ln([N2O5]t/[N2O5]0) = -kt

Rearranging the equation and substituting the given values, we get:

ln(380/760) = -(3.4×10−5 t)

Solving for t, we find that it will take approximately 2,370 seconds or about 39.5 minutes.

Explanation:

The given reaction is as follows.

         [tex]SO_{2} + O_{2} \rightarrow SO_{3}[/tex]

Now, balancing the given equation by putting appropriate coefficients.

              [tex]2SO_{2} + O_{2} \rightarrow 2SO_{3}[/tex]

It is given that equimolar [tex]SO_{2}[/tex] and [tex]O_{2}[/tex]. Hence,

Therefore, during completion of the reaction,

          [tex]SO_{2} + O_{2} \rightarrow SO_{3} + \frac{1}{2}O_{2}[/tex]    

Temperature (T) = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K

Pressure (P) = 1.75 atm

R (gas constant) = 0.0820 L atm/mol K

As on completion of reaction there is [tex]O_{2}[/tex] and [tex]SO_{3}[/tex] remains in the mixture. Therefore, molar mass of the mixture is equal to the sum of molar mass of

Total molar mass = [tex]O_{2} + SO_{3}[/tex]

                             = (32 + 80) g/mol

                             = 112 g/mol

Hence, according to the formula we will calculate the density as follows.

       Density = [tex]\frac{P \times \text{molar mass}}{R \times T}[/tex]

                     = [tex]\frac{1.25 atm \times 112 g/mol}{0.0820 Latm/mol K \times 298 K}[/tex]

                     = 5.73 g/L

Thus, we can conclude that density of the product gas mixture is 5.73 g/L.

Answer:

d = 3.27g/L

Explanation:

[tex]2SO_{2} + O_2 => 2SO_3[/tex]

ICF table

                     2SO_{2} + O_2   => 2SO_3            

initial               1 mol      1 mol         0 mol

change          -1 mol     0.5 mol       1 mol

final                0 mol     0.5 mol       1 mol

calculate the mole fractions

[tex]X_O_2 = \frac{0.5 mol}{1.5 mol} = \frac{1}{3}\\\\X_S_O_3 = \frac{1 mol}{1.5 mol} = \frac{2}{3}\\[/tex]  

[tex]\frac{n}{V} = \frac{P}{RT} = 0.0511 mol / L\\\\d = \frac{n}{RT} * X_O_2 * MM_O_2 + X_S_O_3 * MM_S_O_3\\\\d = (0.0511 mol/L) * (\frac{1}{3} mol * 32.0 g/mol + \frac{2}{3} * 80.06 g/mol)\\\\d = 3.27 g/L[/tex]

Explanation:

When we are working in a laboratory then it is necessary that certain appropriate measures have to be taken in order to avoid any king of accident or mistake.

So, when we are mixing liquids in a separatory funnel then following safety measures should be followed.

1. Everything should be done in your hood

2. You should wear gloves

3. Your hood sash should be between your face and the separatory funnel

4. Don't aim the separatory funnel at anyone

5. You should have a firm grip on the stopper and vent the funnel frequently by using the stopcock

Therefore, we can conclude that all the given statements are true about mixing liquids in a separatory funnel.

Answer:

Options (A), (B) and (C)

Explanation:

In an ecosystem, the transfer of nutrients takes place from one place o another. This input and output of nutrients continuously takes place in the ecosystem.

Thus, the correct answers are options (A), (B) and (C).

Examples of nutrient input into a terrestrial ecosystem include nutrients dissolved in rain, nitrogen fixation by bacteria, and chemical weathering of rocks by organic acids released by plant roots.

An example of nutrient input into a terrestrial ecosystem includes:

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Answer:

The answer to your question is Empirical formula  InCl₃

Explanation:

Data

InCl = 0.5 g

Cl = 0.2404 g

Empirical formula = ?

Process

1.- Calculate the mass of Indium

    Total mass = mass of Indium + mass of Chlorine

     0.50 = mass of Indium + 0.2404

     mass of Indium = 0.50 - 0.2404

     mass of Indium = 0.2596 g

2.- Calculate the moles of Indium and Chloride

Atomic mass Indium = 115 g

Atomic mass Chlorine = 35.5 g

                       115 g of In ------------------ 1 mol

                     0.2596 g of In ------------- x

                        x = (0.2596 x 1) / 115

                        x = 0.0023 moles of Indium

                        35.5 g of Cl ------------- 1 mol

                         0.2404 g    --------------- x

                          x = (0.2404 x 1) / 35.5

                          x = 0.0068 moles of Cl

3.- Divide by the lowest number of moles

Indium     0.0023 / 0.0023 = 1

Chlorine  0.0068 / 0.0023 = 3

4.- Write the empirical formula

                                       InCl₃

The empirical formula of the indium compound is [tex]InCl_3[/tex]

Mass of indium = 0.500-0.2404

= 0.2596g

And,  

Mass of chlorine = 0.2404g

Now  

Divide each mass by atomic mass

[tex]In = 0.2596\div 114.818 = 0.002261\\\\Cl = 0.2404\div 35.453 = 0.00678[/tex]

Divide by smaller:

In = 1

Cl = 3

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Answer: ATP Adenosine triphosphate

Explanation: During light reaction plants produce glucose and as well as ATP and NADPH as another products. Now during the Dark or light independent reaction there is no presence of light energy to fuel the process so ATP is used as an alternative to fix CO2 molecules.

Answer:

The answer to your question is 432 g of CO₂

Explanation:

Data

CaCO₃  = 983 g

CaO = 551 g

CO₂ = ?

Balanced reaction

                               CaCO₃ (s)   ⇒   CaO (s)   +  CO₂ (g)

This reaction is balanced, to solve this problem just remember the Lavoisier Law of conservation of mass that states that the mass of the reactants is equal to the mass of the products.

                    Mass of reactants = Mass of products

                    Mass of CaCO₃   = Mass of CaO + Mass of CO₂

Solve for CO₂

                    Mass of CO₂  = Mass of CaCO₃ - Mass of CaO                    

                     Mass of CO₂ = 983 g - 551 g

Simplification

                     Mass of CO₂ = 432 g                        

To achieve the greatest accuracy using the ideal gas law, the scientist should ensure low pressure and high temperature, accurate measurement instruments, and where possible, simple gas molecules. Applying corrections for non-ideal gas behavior can further enhance accuracy.

To ensure the greatest accuracy when calculating the number of moles using the ideal gas law, a scientist should make sure the conditions under which the gas measurements are taken approximate those of an ideal gas as closely as possible. The scientist should ensure that the gas is at a low pressure and a high temperature to minimize the interactions between gas molecules and the divergence from ideal gas behavior. Real gases deviate from ideal behavior at high pressures and low temperatures due to the volume occupied by the gas molecules and intermolecular forces.

It is essential to use a highly accurate pressure gauge, thermometer, and volume measurement method to reduce experimental error. Additionally, since deviations from ideal behavior increase with the complexity of the gas molecules, using a simple gas, such as a noble gas or a diatomic molecule like N2 or O2, can help produce more accurate results.

Finally, applying corrections for non-ideal conditions using the van der Waals equation or another real gas model can improve accuracy if the experimental conditions cannot be made to closely mimic those of an ideal gas.

Answer:

In 7.80 g of styrene, we have 3.60×10²³ atoms of H

Explanation:

Empirical formula of styrene is CH

Molecular formula of styrene is C₈H₈

So, 1 mol of styrene has 8 moles of C and 8 moles of H and 1 mol weighs 104.14 grams. Let's make a rule of three:

104.14 g (1 mol of C₈H₈) have 8 moles of H

Then 7.80 g would have ( 7.80  .8) / 104.14 = 0.599 moles

As we know, 1 mol of anything has NA particles (Avogadro's Number,  6.02×10²³), so 0.599 moles will have (mol . NA) particles

0.599 mol . 6.02×10²³ atoms / 1 mol = 3.60×10²³ atoms

Answer:

6.25 gallons solution should be drained and replaced 6.25 gallons with bleach to increase the bleach content to the recommended level.

Explanation:

Volume of the solution in the tank = 150 gal

Percentage of bleach in solution = 4%

Volume of bleach present = (4% of 150)gal

Let the volume of solution removed = x

Volume of bleach removed = (4% of x )gal

Desired percentage of bleach solution = 8%

Volume of bleach in 8% solution = 8% of 150 gal

(4% of 150)gal +x - (4% of x )gal = 8% of 150 gal

(4% of 150) +x - (4% of x ) = 8% of 150

[tex]6+x-0.04x=12[/tex]

[tex]0.96x=6[/tex]

x = 6.25

6.25 gallons solution should be drained and replaced 6.25 gallons with bleach to increase the bleach content to the recommended level.

To increase the bleach solution from 4% to 8% in a 150-gallon tank, 6.25 gallons of the existing solution should be drained and replaced with pure bleach.

To adjust the concentration of bleach in the sterilization tank from 4% to 8%, we need to find out how much of the current solution should be removed and replaced with pure bleach. Initially, there are 150 gallons of a 4% bleach solution. If x gallons of this solution are drained and replaced with pure bleach, the amount of bleach in the tank remains the same because pure bleach is being added to compensate for the amount drained.

The initial amount of bleach in the solution is 4% of 150 gallons, which is 0.04 × 150 = 6 gallons of bleach. We need to end up with 8% bleach in a 150-gallon tank, so the final amount of bleach needed is 0.08 × 150 = 12 gallons. Since the initial and final amount of bleach has to be the same to ensure complete sterilization, we can set up the following equation:

Initial Bleach + Replaced Bleach - Removed Bleach = Final Bleach

6 + x - (0.04 × x) = 12

This simplifies to:

6 + 0.96x = 12

Solving for x gives:

0.96x = 6

x = 6 / 0.96

x = 6.25 gallons

Therefore, to achieve an 8% bleach concentration, 6.25 gallons of the 4% bleach solution should be drained from the tank and replaced with pure bleach.

Answer: Hydrogen bonding

Explanation:

Answer:

The answer to your question is 2 ml

Explanation:

Data

Initial volume = ?

Initial concentration = 220 mg/ml

Final volume = 10 ml

Final concentration = 43 mg/ml

Equation

    Initial volume x Initial concentration = Final volume x Final concentration

Solve for Initial volume

 Initial volume = (Final volume x Final concentration) / Initial concentration

Substitution

 Initial volume = (10 x 43) / 220

Simplification

 Initial volume = 430 / 220

Result

 Initial volume = 1.95 ml ≈ 2.0 ml

Answer:

The answer to your question is  21 g of CO₂  

Explanation:

Balanced Reaction

                               C₃H₈  +  6O₂    ⇒    3CO₂   +   4H₂O

Data

mass of C₃H₈ = 7 g

mass of O₂  = 98 g

Determine the limiting reactant

Molecular mass of C₃H₈ = (12 x3) + (1 x 8) = 36 + 8 = 44 g

Molecular mass of Oxygen = 16 x 12 = 192 g

Theoretical proportion  C₃H₈ / O₂ = 44 / 192 = 0.23

Experimental proportion  C₃H₈ / O₂ = 7 / 98 = 0.07

As the proportion diminishes in the experiment, the excess reagent is the oxygen and the limiting reactant is propane.

- Calculate the mass of CO₂

                     44 g of C₃H₈   ---------------- 3(44) of CO₂

                       7 g of C₃H₈    ---------------   x

                       x = (7 x 3(44)) / 44

                       x = 924 / 44

                       x = 21 g of CO₂                        

Answer:

The non-cyclic photophosphorylation which is the light-requiring part of photosynthesis in some higher plants, in which an electron donor is required, and oxygen is also produced as a waste product. this consists of two photoreactions, resulting in the synthesis of ATP and NADPH 2.

Explanation:

   A. When photosystem II absorbs light, an electron excited to a higher energy level in the reaction center chlorophyll (P680) is captured by the primary electron acceptor.  The oxidized chlorophyll is now a very strong oxidizing agent; its electron “hole” must be filled.

  B.  An enzyme extracts electrons from water and supplies them to P680, replacing the electrons that the chlorophyll molecule lost when it absorbed light energy.  This reaction splits a water molecule into two hydrogen ions and an oxygen atom, which immediately combines with another oxygen atom to form O2.  This splitting of water is responsible for the release of O2 into the air.

  C.  Each photoexcited electron (energized by light) passes from the primary electron acceptor in photosystem II to photosystem I via an electron transport chain.  This electron transport chain is very similar to the one in cellular respiration; however, the carrier proteins in the chloroplast ETC are different from those in the mitochondrial ETC.

  D.  As electron move down the chain, their exergonic “fall”to a lower energy level is harnessed by the thylakoid membrane to produce ATP (by chemiosmosis).  The production of ATP in the chloroplast is called photophosphorylation because the energy harnessed in the process originally came from light.  This process of ATP production is called non-cyclic photophosphorylation.  The ATP generated in this process will provide the energy for the synthesis of glucose during the Calvin cycle (light independent reactions).

  E.  When an electron reaches the “bottom” of the electron transport chain, it fills an electron “hole” in the chlorophyll a molecule in the reaction center of photosystem I (P700).  The hole was created when light energy drives an electron from P700 to the primary electron acceptor of photosystem I.

   F. The primary electron acceptor of photosystem I passes the excited electrons to a second electron transport chain which transmits them to an iron-containing protein.  An enzyme reaction transfers the electrons from the protein to NADP+ that forms NADPH (which has high chemical energy due to the energy of the electrons).  NADPH is the reducing agent needed for the synthesis of glucose in the Calvin cycle.

Molar mass of C_3H_6 = 36 g mol^-1 +6g mol^-1 =42g mol^-1

Empirical formula of Sample = (CH_3)

Explanation:

Mass sample having only C and H = 42g

Mass of C = 12g mol^-1

Molar mass of C = 12g mol^-1

Moles of C = 36 g / 12g mol^-1

= 3 mol

Mass of H in sample = 6g

Molar mass of H = 1 g mol^-1

Moles of H = 6g / 1g mol^-1

= 6 mol

Therefore in 42g sample 3 mol of C + 6 mol of H present, that is C_3H_6

Molar mass of C_3H_6 = 36 g mol^-1 +6g mol^-1 =42g mol^-1

Empirical formula of Sample = (CH_3)

Answer:

We can solve the question 'What is the empirical formula of the compound?' The answer is CH2

Explanation:

A 42.0g sample of compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6.0g of H. Which of the following questions about the compound can be answered using the results of the analysis?

A) What was the volume of the sample?

B) What is the molar mass of the compound?

C) What is the chemical stability of the compound?

D) What is the empirical formula of the compound?

Step 1: Data given

Mass of the compound = 42.0 grams

⇒ 36.0 grams = carbon

⇒ 6.0 grams = hydrogen

Molar mass of carbon = 12.01 g/mol

Molar mass of hydrogen = 1.01 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles Carbon = 36.0 grams / 12.01 g/mol

Moles carbon = 3.00 moles

Moles hydrogen = 6.0 grams / 1.01 g/mol

Moles hydrogen = 5.95 moles

Step 3: Calculate mol ratio

We divide by the smallest number of moles

Carbon: 3.00 / 3.00 = 1

Hydrogen: 5.95 / 3.00 = 2

The empirical formula is CH2

Answer:

In a neutral solution the concentration of hydrogen ions is equal to the concentration of hydroxide ions

Explanation:

When:

[H⁺] > [OH⁻]  the pH is acid, so the solution is practically acid.

When

[H⁺] < [OH⁻], the pH is basic, so the solution is practically basic

In neutral solutions:

[H⁺] = [OH⁻], so the pH is neutral (7)

pH > 7 → BASIC SOLUTIONS

pH < 7 → ACID SOLUTION

In a neutral solution,d) the concentration of hydrogen ions is equal to the concentration of hydroxide ions. This balance comes from water's autoionization process, with discrepancy coming from the addition of acids or bases.

In a neutral solution, d) the concentration of hydrogen ions (H+) is equal to the concentration of hydroxide ions (OH-).

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Final answer:

The concentration after 17.0 minutes will be approximately 0.067 M.

Explanation:

The rate constant for a reaction is a constant that relates the rate of the reaction to the concentration of the reactants. The rate law for a reaction is given by the equation: rate = k[A]^[x][B]^[y], where k is the rate constant, [A] and [B] are the concentrations of the reactants, and x and y are the reaction orders concerning A and B, respectively.

In this case, since the rate constant (k) is given as 5.40 × 10^(-3) s^(-1) and the initial reactant concentration is 0.100 M, we can use the integrated rate law for a first-order reaction to find the concentration after a certain time:

[A] = [A]_0 * e^(-kt)

Substituting the given values, we have:

[A] = (0.100 M) * e^(-5.40 × 10^(-3) s^(-1) * (17.0 * 60 s))

Simplifying this equation gives:

[A] ≈ 0.067 M

The concentration of the reactant after 17.0 minutes is approximately 0.000405 M.

This was determined using the formula for a first-order reaction. Initial concentration, rate constant, and time were given and plugged into the formula for accuracy.

To solve this problem, we need to use the formula for a first-order reaction:

[A] = [A]0 e-kt

Where:

Given:

Plug these values into the formula:

Simplify the exponent:

Using a calculator to find e-5.508:

The concentration of the reactant after 17.0 minutes is approximately 0.000405 M.

Compounds have the lowest percentage of gold content by weight is Aui3

M(Au) =197g/mol

M(O) =16g/mol

M(H)=1g/mol

M(Cl)=35g/mol

M(I)=53 g/mol

1) M=214g/mol =197+1+16

197/214= 0.921

2) M=248g/mol =[tex]197 +3 \times ( 1+16)[/tex]

197/248=0.794

3) M=302g/mol = [tex]197+3 \times 35[/tex]

197/302=0.652

4) M=356g/mol =[tex]197+3 \times 53[/tex]

197/356=0.5533

Explanation:

Iodine is the heaviest out of I, Cl and OH. The items on your list hold less and less gold by mass as you go down. The gold dissolution percentage in a hypochlorite-iodide mix-up is completely conditioned upon the solution pH. So the iodine has the lowest gold content percentage.

Answer: AuI3

Explanation:

Answer:

dH = 1.06 x 10⁻¹⁰ m

Explanation:

Scientific notation is a way of minimizing large figures in smaller decimal form. For example, 2300000 is a large figure so, its size can be minimized by converting it into scientific notation form, 2300000 can also be written as 2.3 x 106.

There are number of formats to write dH=1.06×10−10m in scientific notation by shifting the decimal in right or left direction. For example, if we shift the decimal in right direction, the addition of +1 will occur in the power of 10 i.e. 1.06 x 10-10 will become 10.6 x 10-11 or 106 x 10-12 (these formats of scientific notation are also correct).

In some softwares and programming languages, scientific notation are written as 1.06E-10 so, avoid using these kinds of notations since, E indicates as variable in mathematics.

Variables are alphabetical value in an equation. For example, in equation 2a + 3b, a and b are variables while 2 and 3 are constants.

To express the diameter of a hydrogen atom in scientific notation, it would be 1.06×10⁻¹⁰ meters. Scientific notation simplifies the representation of very large or small numbers and is particularly useful for expressing measurements in physics, like the mass of a hydrogen atom, which is 1.67×10⁻¹⁼ kilograms.

The diameter of a hydrogen atom in its ground state is given as dH=1.06×10⁻¹⁰ meters. To enter this number in scientific notation, you place the decimal point such that there is only one non-zero digit to the left of the decimal point. In this case, the diameter would be 1.06×10⁻¹⁰ m.

Scientific notation is a convenient way to express large or small numbers. For instance, the mass of a hydrogen atom can be expressed as 1.67×10⁻¹⁼ kg. This system lets us write numbers as a product of a number between 1 and 10 and a power of 10.

An example of applying scientific notation is a scale model of a hydrogen atom: if one were building a scale model where the atom's diameter is 1.00 m, finding the proportional size of the nucleus would require understanding the actual size relationship, which in scientific notation is a much simpler task than with standard numeral representation.