Which half reaction correctly describes the oxidation that is taking place? Z n (s) right arrow upper Z n superscript 2 plus (a q) plus e superscript minus. Upper Z n (s) right arrow upper Z n superscript 2 plus (a q) plus 2 e superscript minus. Upper C u superscript 2 plus (a q) plus 2 e superscript minus right arrow upper C u (s). Upper C u superscript 2 plus (a q) plus e superscript minus right arrow upper C u (s).

Answer:

yes it doese

Explanation:

Question:

What is the total combined mass of carbon dioxide and water that is produced?

Answer:

109 kg

Explanation:

When 23 kg of gasoline burns by consuming 86 kg oxygen, they produce carbon dioxide and water. To find the total combined mass of carbon dioxide and water, we will use mass conversation law.

According to mass conversation law, the mass of the product is equal to the mass of reagent.

Mass of reagent = Mass of product

In this reaction,

Gasoline + O2 → CO2 + H2O

23 kg + 86 kg → ?

23 kg + 86 kg =  109 kg

Combined mass of carbon dioxide and water will be 109 kg.

Final answer:

To estimate the CO₂ produced from 40 L of gasoline, we multiply the mass of the gasoline (calculated using the density of 0.75 kg/L) by three, resulting in approximately 90 kg of CO₂, which is comparable to human mass.

Explanation:

Based on the provided combustion reaction 2 C8H18 + 25 O2 → 16 CO₂ + 18 H₂O + energy, we can calculate the mass of CO₂ produced from consuming a 40 L tank of gasoline. First, we determine the mass of gasoline using the given density (0.75 kg/L), which is 40 L × 0.75 kg/L = 30 kg of gasoline. Now, using the factor-of-three ratio of CO₂ mass to input fuel mass, we multiply the gasoline mass by three to estimate the CO₂ mass produced. Hence, 30 kg × 3 = 90 kg of CO₂ are produced after burning 40 L of gasoline. If we compare this to the typical human mass, which is roughly between 50-100 kg, one can see that the mass of CO₂ produced is remarkably similar to or even exceeds the mass of an average human.

Answer:

mass of platinum = 2526.12 g

Explanation:

Given data:

Mass of water = 125 g

Initial temperature of water= 100.0°C

Initial temperature of Pt = 20.0°C

Final temperature = 235°C

Specific heat of Pt = 0.13 j/g°C

Specific heat of water = 4.184 j/g°C

Mass of platinum = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

Q(w) = Q(Pt)

m.c.  (T2 - T1)   =   m.c.   (T2 - T1)

125 g × 4.184 j/g°C ×  (235°C - 100.0°C)  = m × 0.13 j/g°C ×  (235°C - 20°C)

125 g × 4.184 j/g°C × 135°C  = m × 0.13 j/g°C × 215°C

70605 j = m×27.95 j/g

m = 70605 j /27.95 j/g

m = 2526.12 g

To find the mass of platinum dropped into hot water, you would apply the conservation of energy principle and use the specific heat values for both substances. However, the provided temperatures suggest an error in the question, as platinum would not heat to a temperature higher than the water's temperature.

The question involves finding the mass of platinum, which was dropped into hot water, by using the concept of heat transfer between the metal and water. According to the law of conservation of energy, the heat lost by the water is equal to the heat gained by the platinum. To solve for the mass of the platinum, we use the formula q = mcΔT, where 'q' is the heat transfer, 'm' is the mass, 'c' is the specific heat, and 'ΔT' is the change in temperature.

However, the given information seems to contain a mistake, as platinum will not naturally heat up to 235.0°C when dropped into water that is at 100.0°C. There must be an error in the given temperatures. Assuming we had the correct temperature details and using the given specific heats for water (4.184 J/g°C) and platinum (0.13 J/g°C), we could set up the heat transfer equations and solve for the unknown mass of platinum.

Answer:

The Scientific Method.

Final answer:

Each SI unit measures a specific fundamental quantity: kilograms (kg) for mass, kelvin (K) for temperature, seconds (s) for time, and amperes (A) for electric current.

Explanation:

The student has asked to match each SI unit to the quantity it measures among mass, temperature, time, and electric current. Here are the matches:

These four units are part of the metric system, which uses powers of 10 to relate quantities over various ranges of nature. All other physical quantities, such as force and charge, are derived from these fundamental units.

Answer:

Molality of solution = 0.973 m

Explanation:

Molality : It is defined as the moles of the solute per Kg mass of solvent.It is not temperature dependent.

Solute = Substance which is present in less quantity in the solution is called the solute. Here , Sucrose is the solute.

Solvent = Substance which is present in more quantity is the solvent. Here water is solvent.

[tex]Molality=\frac{moles\ of\ solute}{mass\ of\ solvent}[/tex]

Density = It is defined as the mass per unit volume.

[tex]Density=\frac{mass}{Volume}[/tex]

Mass of Solute = 137.9 g

[tex]Moles=\frac{mass}{Molar\ mass}[/tex]

Molar mass of sucrose =

[tex]C_{12}H_{22}O_{11}[/tex]= 12(mass of C)+22(mass of H)+11(mass of O)

= 12(12)+22(1)+11(16)

= 144+22+176

= 342 g/mol

[tex]Moles=\frac{mass}{Molar\ mass}[/tex]

[tex]Moles=\frac{137.9}{342}[/tex]

[tex]Moles=0.403[/tex]

Moles = 0.403 moles

Mass of Solvent = 414.1 g  (water)

[tex]Molality=\frac{moles\ of\ solute}{mass\ of\ solvent(g)}(1000)[/tex]

[tex]Molality =\frac{0.403}{414.1}\times 1000[/tex]

Molality = 0.973 m

To test how mass of a reactant affects speed of a reaction, set up an experiment using a consistent reactant, like hydrochloric acid, and alter the mass of another reactant, like sodium bicarbonate. By keeping all other variables constant, you can measure how the varying mass affects the speed of the reaction indicated by when bubbling ceases.

To test how the mass of a reactant affects the speed of a reaction, an experiment could be set up in the following way: Obtain a substance that reacts with a certain reactant. This could be an acid-base reaction or a redox reaction. Let's say the reaction is between hydrochloric acid (HCl) and sodium bicarbonate (NaHCO3) which produces carbon dioxide gas.

Maintain controlled conditions: all other variables such as temperature, pressure, and volume of HCl should be kept constant. The only changing factor would be the mass of the sodium bicarbonate.

Measure the time it takes for the reaction to complete for various masses of NaHCO3. You do this by observing when the bubbling (indicative of CO2 production) stops. You would likely see that increasing the mass of the reactant (NaHCO3) increases the speed of the reaction.

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The correct option is D.

The best experimental design for testing how the mass of a reactant affects the speed of reaction would be 'The mass of one reactant at a time is varied, and the time it takes the reaction to finish is measured', while maintaining other factors like temperature and concentration of other reactants constant. This method is known as the method of initial rates.

The best example of a controlled experiment to test how the mass of a reactant affects the speed of a chemical reaction would be option D: The mass of one reactant at a time is varied, and the time it takes the reaction to finish is measured. This method is known as the method of initial rates, often employed in chemistry to measure reaction rates using different initial reactant concentrations. It is crucial to vary only one aspect while keeping others constant (temperature, concentration of other reactants etc.) to accurately determine the effect one factor has on the reaction speed.

The temperature of the reactants and concentration of the reactants also significantly impacts the rate of a chemical reaction. Higher the temperature, or the concentration, faster the reactions typically occur. However, these other factors need to be controlled in this experiment to singularly test the effect of mass of one reactant.

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The complete question is given below:

You want to test how the mass of a reactant affects the speed of a reaction.

Which of the following is an example of a controlled experiment to test this?

A. The mass of one reactant and the temperature of the reaction mixture are increased until the reaction is finished.

B. The mass of all the reactants is varied, and the time it takes the reaction to finish is measured.

C. The mass of all of the reactants is kept the same, and the mixtures are allowed to react for different lengths of time.

D. The mass of one reactant at a time is varied, and the time it takes the reaction to finish is measured.

True because the biosphere contains

life such as animals and plants Hint the word “bio”

Answer:

its to hard

Explanation:

Answer:

its to hard

Explanation:

Hardness, lustre, and colour.

These Rocks shown in photographs are made of different types of minerals which have properties as follows:

The color of the rock is grey, brown and yellow after it is ground into a powder its color is streak.

The lustre of the rock tells how shiny the rocks are.

Other properties include hardness, texture, shape, and size.

Final answer:

The area of a rectangular field measuring 6.0 m by 8.0 m is 480,000 cm², which is 4.8 × 10⁵ cm² in scientific notation. So the correct option is B.

Explanation:

To calculate the area of a rectangular field in square centimeters, you would use the formula Area = length × width. First, we must ensure that both dimensions are in the same units, so we convert the dimensions from meters to centimeters. There are 100 centimeters in a meter, so:

Next, we calculate the area using the converted measurements:

Area = 600 cm × 800 cm = 480,000 cm²

Therefore, the area of the field in square centimeters is 480,000 cm², which can be expressed in scientific notation as 4.8 × 105 cm2, making option B the correct answer.

Answer:                  34.0147 g/mol

Explanation:

Answer:34.02g/mol

Explanation:H2O2

2x1.01 + 2x16 grams/mols

2.02 + 32 =34 .2g/mol

Answer:

1. b

2. a

3. d

4. d

Explanation:

Answer:

                       4.57 × 10⁶  nanometers  

Explanation:

                    In this problem we are asked to convert between two units i.e. a meter into nanometer. In sciences, the different units are used for a same quantity (achieved by multiplying conversion factors) to get rid of very small values and get a readable and  intelligible values.

For Example:

In given statement the value of small distance is 4.5 × 10⁻³ meter the real number form of this number is 0.00457. Hence, this number can be converted to a very large number by multiplying it with 10⁹ or 1000000000. Hence,

                         4.5 × 10⁻³  × 1.0 × 10⁹  =  4.57 × 10⁶

Or,

                 4.5 × 10⁻³  meters  =  4.57 × 10⁶  nanometers    

Although the quantity is the same for same units but the number has changed.          

Answer: Solution attached.

Each equation is now balanced.

Explanation:

Answer:

3.5 mol·L⁻¹  

Explanation:

1. Set up an ICE table.

[tex]\begin{array}{cccccc}\text{2NO} & + & \text{I}_{2} &\, \rightleftharpoons \, & \text{2NOI} & & \\ 2.0 & & 4.0 & & 1.0 & & \\ -2x & & -x & & +2x & & \\ 2.0-2x & & 4.0-x & & 1.0+2x & & \\\end{array}[/tex]

2. Solve for x

The equilibrium concentration of NO is 1.0 mol·L⁻¹, so

       1.0 = 2.0 - 2x

2x + 1.0 = 2.0

        2x =  1.0

          x = 0.5

3. Calculate the equilibrium concentration of I₂

[I₂] = 4.0 - x = 4.0 - 0.5 = 3.5 mol·L⁻¹

The concentration of I₂ at equilibrium is calculated to be 3.5 M by using the initial concentration of NO to determine the stoichiometric change in I₂ concentration based on the reaction 2NO(g) + I₂(g) → 2NOI(g).

To find the concentration of I₂ at equilibrium for the reaction 2NO(g) + I₂(g) → 2NOI(g), we use the initial and equilibrium concentrations of NO to determine the change in concentration of I₂. Given the stoichiometry of the reaction, for every 1 mole decrease in NO, there is a 0.5 mole decrease in I₂. The initial concentration of NO is 2.0 M, and at equilibrium, it is 1.0 M, which means there has been a 1.0 M decrease (2.0 M - 1.0 M). The I₂ concentration at equilibrium can be found by subtracting half of this change from the initial I₂ concentration. Since initially the concentration of I₂ is 4.0 M, the equilibrium concentration is calculated as 4.0 M - (1.0 M / 2) = 3.5 M.

Answer:

[tex]\large\boxed{\text{The volume of the gas at a temperature of 300K is 575 liters}}[/tex]

Explanation:

You already know the relation between the volume (V) of the gas and its temperature (T):

           [tex]V/T=1.75[/tex]

The units of V is liters and of T is Kelvin (K).

Thus, the units of the constant 1.75 is liters/K.

Hence, to find the volume (V) of the gas at a temperature (T) of 300 K, you just must solve for V and substitute the temperature to compute V:

         [tex]V=1.75T[/tex]

          [tex]V=1.75liters/K\times 300K=575liters[/tex]