what process is demonstrated by an acid reacting with a base

Total pressure is defined as the sum of all partial pressures in a container, therefore in our question we have:

Ptotal = 5.76 atm

PO2 = 1.31 atm

PN = 2.22 atm

PAr = 0.77 atm

PCO2 = ?

Since the total pressure is the sum of all partial pressures, if we subtract the partial pressures that we know from the total pressure we will have the value of partial pressure for CO2

5.76 - 1.31 = 4.45

4.45 - 2.22 = 2.23

2.23 - 0.77 = 1.46 atm, this is the Partial Pressure for CO2

Explanation:

Copper has an atomic number of 29, so it contains 29 protons and 29 electrons. ... The atomic weight of copper is 64; it has 29 protons and 35 neutrons.

5. To meet production in August, 71,000 pounds of raw materials are needed.

Since each unit of finished goods requires 5 pounds of raw materials, we can calculate the number of units needed in August as 71,000 pounds / 5 pounds per unit = 14,200 units.

The ending To determine the answers to the questions, we need to calculate the required information based on the provided data. inventory in July is 10% of the following month's raw materials production needs, which is 10% of 14,200 units = 1,420 units. Since each unit requires 5 pounds of raw materials, the total pounds of raw materials to be purchased in July is 1,420 units * 5 pounds per unit = 7,100 pounds. 6. To calculate the estimated cost of raw materials purchases for July, we need to determine the cost per pound of raw materials. Given that the raw materials cost $2.00 per pound, the estimated cost of raw materials purchases for July is 7,100 pounds * $2.00 per pound = $14,200. 7. The total estimated cash disbursements for raw materials purchases in July can be calculated by considering the payment terms. Twenty percent of raw materials purchases are paid for in the month of purchase, and 80% are paid in the following month. Given that the cost of raw material purchases in June is $93,040, the payment made in July is 20% * $93,040 = $18,608. The payment made in August would be 80% * $93,040 = $74,432. Therefore, the total estimated cash disbursements for raw materials purchases in July would be $18,608. 8. To calculate the estimated accounts payable balance at the end of July, we need to consider the payment terms. Since 80% of raw materials purchases are paid in the following month, 80% * $14,200 = $11,360 will be paid in August. Therefore, the estimated accounts payable balance at the end of July would be $14,200 - $11,360 = $2,840. 9. The estimated raw materials inventory balance at the end of July can be calculated by considering the ending raw materials inventory in July. Given that the ending raw materials inventory equals 10% of the following month's raw materials production needs, which is 10% of 14,200 units = 1,420 units. Since each unit requires 5 pounds of raw materials, the estimated raw materials inventory balance at the end of July would be 1,420 units * 5 pounds per unit = 7,100 pounds. 10. To calculate the total estimated direct labor cost for July, we need to consider the number of units produced and the direct labor wage rate. The budgeted unit sales for July are 12,000 units. Each unit of finished goods requires 2 direct labor-hours, so the total direct labor-hours for July would be 12,000 units * 2 direct labor-hours per unit = 24,000 direct labor-hours. Given that the direct labor wage rate is $13 per hour, the total estimated direct labor cost for July would be 24,000 direct labor-hours * $13 per hour = $312,000.

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5) 14,200 units of finished goods will be produced in August.

6) The estimated cost of raw materials purchases for July is  $2,800.

7) The total estimated cash disbursements for raw materials purchases in July is $18,608 + $74,432 = $93,040.

8) The estimated accounts payable balance at the end of July is  $59,546.40.

9)  Raw materials inventory balance is 700 pounds.

10) Total estimated direct labor cost for July is $312,000.

Let's solve  in detail:

5. To meet production in August, 71,000 pounds of raw materials are needed. Since each unit of finished goods requires 5 pounds of raw materials, the total number of units to be produced is 71,000 pounds / 5 pounds per unit = 14,200 units. Therefore, 14,200 units of finished goods will be produced in August.

6. If 7,000 pounds of raw materials are needed to meet production in August, and each unit of finished goods requires 5 pounds of raw materials, the total number of units to be produced is 7,000 pounds / 5 pounds per unit = 1,400 units.

The estimated cost of raw materials purchases for July is 1,400 units * $2.00 per pound = $2,800.

7. To calculate the total estimated cash disbursements for raw materials purchases in July, we need to consider the payment terms.

In July, 20% of raw materials purchases are paid for in the month of purchase, and 80% is paid in the following month.

The raw materials purchased in June amount to $93,040. Assuming 20% is paid in June, the cash disbursement for June is $93,040 * 20% = $18,608.

For the remaining 80% of the June purchases, which will be paid in July, we need to calculate the amount. This is given by 80% of the June purchases - 20% of the July purchases.

Amount to be paid in July = ($93,040 * 80%) - (Total purchases for July * 20%)

Solving for the Total purchases for July:

($93,040 * 80%) - (Total purchases for July * 20%) = Total purchases for July * 80%

Rearranging the equation:

($93,040 * 80%) = (Total purchases for July * 80% + Total purchases for July * 20%)

($93,040 * 80%) = Total purchases for July * 100%

Total purchases for July = ($93,040 * 80%) / 100%

Substituting the value of June purchases, we get:

Total purchases for July = ($93,040 * 80%) / 100% = $74,432

Therefore, the total estimated cash disbursements for raw materials purchases in July is $18,608 + $74,432 = $93,040.

8. If 7,000 pounds of raw materials are needed to meet production in August, and 80% of raw materials purchases are paid in the following month, the estimated accounts payable balance at the end of July would be 80% of the purchases for July.

Accounts payable balance = Total purchases for July * 80% = $74,432 * 80% = $59,546.40.

9. If 7,000 pounds of raw materials are needed to meet production in August, and the ending raw materials inventory equals 10% of the following month's raw materials production needs, the estimated raw materials inventory balance at the end of July would be 10% of the raw materials needed for August.

Raw materials inventory balance = 10% of the raw materials needed for August = 10% of 7,000 pounds = 700 pounds.

10. The total estimated direct labor cost for July can be calculated by multiplying the direct labor wage rate per hour by the total direct labor hours required for production in July.

Total direct labor hours for July = Number of units to be produced in July * Direct labor hours per unit = 12,000 units * 2 hours per unit = 24,000 hours.

Total estimated direct labor cost for July = Total direct labor hours for July * Direct labor wage rate = 24,000 hours * $13 per hour = $312,000.

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Answer:

A physical property is a characteristic of matter that is not associated with a change in its chemical composition. Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity.

Explanation:

The concept Arrhenius equation is used here to determine the activation energy for the isomerization of methyl isocyanide. The activation energy is 1.892 kJ / mol.

According to Arrhenius equation, lower the activation energy of a reaction, the higher is the rate of the reaction. The equation can be represented as:

\(k = A. e^{-} ^{E_{a} } / RT\)

For two different temperature and rate constants, the equation is:

log K₂/K₁ = Eₐ / 2.303 R [ T₂ - T₁ / T₁T₂]

log 5.54 × 10⁻² / 4.30 × 10⁻³ = Eₐ / 2.303 × 8.314 [ 503 - 472 / 503 × 472]

Eₐ  = 1.892 kJ / mol

Thus the activation energy for the isomerization of methyl isocyanide is 1.892 kJ / mol.

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Out of all the given subshells it is 2d shell which doesn't have real existence as an atom.

(i) If d-subshell l=2, the values of sof l in the second shell (jn=2) are 0 and 1. Consequently, the third shell cannot contain any d-subshell. As a result, we can conclude that the second subshell does not actually exist as atom.

(ii) In the third shell (n=3), the values of l are 0, 1, and 2 in the case of the f-subshell, where l=3. There is no actual existence of 3f-subhsell because there cannot be any f-subshell in this shell.

(iii) In the fourth shell (n=4), where l=4, the values of l are 0, 1, 2, and 3. There is no actual 4g subshell because there cannot be any g sub shell in this shell.

(iv) The fifth shell (n=5) has l values of 0, 1, 2, 3, and 4, which proves that the d-sub shell is actually present

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To determine the specific volume of a sample of dry air, we can use the ideal gas law. The ideal gas law equation is:

PV = nRT

Where:

P = Pressure (in Pascal)

V = Volume (in cubic meters)

n = Number of moles of the gas

R = Ideal gas constant (8.314 J/(mol·K))

T = Temperature (in Kelvin)

To convert the given pressure and temperature to the SI units (Pascal and Kelvin), we have:

Pressure = 50 kPa = 50,000 Pa

Temperature = 263 K

Now, we can rearrange the ideal gas law equation to solve for volume (V):

V = (nRT) / P

Since we're dealing with a specific volume, we need to determine the volume per unit mass. Therefore, we'll consider one kilogram (kg) of dry air.

To calculate the number of moles (n) of air in one kilogram, we need to know the molar mass of dry air. The molar mass of dry air can be approximated as 28.97 g/mol.

1 kg of air = 1000 g

Number of moles (n) = (mass of air) / (molar mass of air)

n = (1000 g) / (28.97 g/mol)

Now we can substitute the values into the equation:

V = [(1000 g) / (28.97 g/mol)] * (8.314 J/(mol·K)) * (263 K) / (50,000 Pa)

V/n ≈ 0.0434 m³/mol is the specific volume of the dry air sample.

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Answer:

Li is more reactive then Be.

Explanation:

As you go to the farther left and down on the periodic table, the higher that element reactivity is.

Answer:

Lichens can be used as air pollution indicators, especially of the concentration of sulphur dioxide in the atmosphere. Lichens are organisms that grow in exposed places such as rocks or tree bark. They need to be very efficient at absorbing water and nutrients to grow there.

Answer:

[Au] = 0.171 M

Explanation:

For this question, we assume the rock is 100 % gold.

First of all, we determine the moles of gold

67.3 g . 1mol/ 196.97g = 0.342 moles

Molar concentration is defined as the moles of solute, contained in 1L of solution.

Our solution volume is 2L.

M = 0.342 mol / 2L = 0.171

Molar concentration, also called molarity of solution is the most typical unit of concentration.

Answer:airr

Explanation:

The approximate resistance value at 45 degrees Celsius is around 5.8 ohms.

To calculate the value of the resistance temperature coefficient at 45 degrees Celsius using linear approximation, we can use the formula:

R(T) = R0 + α(T - T0),

where R(T) is the resistance at temperature T, R0 is the resistance at a reference temperature T0, α is the resistance temperature coefficient, and (T - T0) is the temperature difference.

Given that the resistance at 25 degrees Celsius is 5 ohms (R0 = 5) and the resistance at 50 degrees Celsius is 6 ohms (R(T) = 6), we can calculate the value of α.

6 = 5 + α(50 - 25),

Simplifying the equation:

1 = 25α,

Therefore, α = 1/25 = 0.04 ohm/degree Celsius.

Using the linear approximation, we can approximate the value of the resistance at 45 degrees Celsius:

R(45) = 5 + 0.04(45 - 25) = 5 + 0.04(20) = 5 + 0.8 = 5.8 ohms.

Therefore, the value of the resistance at 45 degrees Celsius is approximately 5.8 ohms.

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When performing research, you should never believe everything a source says without checking it against another source of information.

Even very trustworthy sources can occasionally contain information that is inaccurate or biased. A good way to corroborate the information is checking it against another source. If several sources agree on the information, it is likely that this is trustworthy.

Accuracy is the nearness of observed values or measured values to the true value. The difference between the measured and observed value gives the error incurred during the experimental procedure.

Therefore, When performing research, you should never believe everything a source says without checking it against another source of information.

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Answer:

Just click on the answer, if it is an external link DO NOT CLICK IT. it may contain dangerous stuff including h*cked stuff.

Explanation:

The reason the reaction produced H20 and not H2O2 is because Oxygen is a diatomic particle which needs both of the hydrogen atoms.

Examining the electron configurations of hydrogen and oxygen can help us understand the process. While hydrogen has one valence electron, oxygen has six. When the two elements come together, the hydrogen atoms share an electron with an oxygen atom to create a covalent connection.

To establish two additional covalent connections, the two oxygen atoms share two of their valence electrons with one of the hydrogen atoms.

This is why the balanced equation is:

2H2 + O2 → 2H2O

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Answer:

what is an element?

element is a pure substance which cannot be broken down by chemical means

what is it represented by?

By its symbol and atomic number

what is it made up off?

It is made up of it's owm type of atom.

Example

carbon,oxygen,hydrogen,gold,silver,iron

Answer:

Oxides of Nitrogen (NOx)

NOx is a collective term used to refer to two species of oxides of nitrogen: nitric oxide (NO) and nitrogen dioxide (NO2).

Annual mean concentrations of NO2 in urban areas are generally in the range 10-45 ppb (20-90 µgm-3). Levels vary significantly throughout the day, with peaks generally occurring twice daily as a consequence of "rush hour" traffic. Maximum daily and one hourly means can be as high as 200 ppb (400 µgm-3) and 600 ppb (1200 µgm-3) respectively.

Globally, quantities of nitrogen oxides produced naturally (by bacterial and volcanic action and lightning) far outweigh anthropogenic (man-made) emissions. Anthropogenic emissions are mainly due to fossil fuel combustion from both stationary sources, i.e. power generation (21%), and mobile sources, i.e. transport (44%). Other atmospheric contributions come from non-combustion processes, for example nitric acid manufacture, welding processes and the use of explosives.

Sulphur Dioxide (SO2)

SO2 is a colourless gas. It reacts on the surface of a variety of airborne solid particles, is soluble in water and can be oxidised within airborne water droplets.

Annual mean concentrations in most major UK cities are now well below 35 ppb (100 µgm-3) with typical mean values in the range of 5-20 ppb (15-50 µgm-3). Hourly peak values can be 400-750 ppb (1000-2000 µgm-3) on infrequent occasions. Natural background levels are about 2 ppb (5 µgm-3).

The most important sources of SO2 are fossil fuel combustion, smelting, manufacture of sulphuric acid, conversion of wood pulp to paper, incineration of refuse and production of elemental sulphur. Coal burning is the single largest man-made source of SO2 accounting for about 50% of annual global emissions, with oil burning accounting for a further 25-30%.

Carbon Monoxide (CO)

Carbon Monoxide is a colourless, odourless, tasteless gas that is slightly lighter than air.

Natural background levels of CO fall in the range of 10-200 ppb. Levels in urban areas are highly variable, depending upon weather conditions and traffic density. 8-hour mean values are generally less than 10 ppm (12 mgm-3) but have been known to be as high as 500 ppm (600 mgm-3).

CO is an intermediate product through which all carbon species must pass when combusted in oxygen (O2). In the presence of an adequate supply of O2 most CO produced during combustion is immediately oxidised to carbon dioxide (CO2). However, this is not the case in spark ignition engines, especially under idling and deceleration conditions. Thus, the major source of atmospheric CO is the spark ignition combustion engine. Smaller contributions come from processes involving the combustion of organic matter, for example in power stations and waste incineration.

Ozone (O3)

O3 is the tri-atomic form of molecular oxygen. It is a strong oxidising agent, and hence highly reactive.

Background levels of O3 in Europe are usually less than 15 ppb but can be as 100 ppb during summer time photochemical smog episodes. In the UK ozone occurs in higher concentrations during summer than winter, in the south rather than the north and in rural rather than urban areas.

Most O3 in the troposphere (lower atmosphere) is formed indirectly by the action of sunlight on nitrogen dioxide - there are no direct emissions of O3 to the atmosphere. About 10 - 15% of tropospheric O3 is transported from the stratosphere where it is formed by the action of ultraviolet (UV) radiation on O2. In addition to O3, photochemical reactions involving sunlight produce a number of oxidants including peroxyacetyl nitrate (PAN), nitric acid and hydrogen peroxide, as well as secondary aldehydes, formic acid, fine particulates and an array of short lived radicals. As a result of the various reactions that take place, O3 tends to build up downwind of urban centres where most of NOx is emitted from vehicles.

Explanation:

The amount of CO2 dissolved in water can be calculated using Henry's Law, which states that the amount of gas dissolved in a liquid is proportional to the partial pressure of the gas above the liquid.

The equation for Henry's Law is:C = kH * P

where C is the concentration of the gas in the liquid (in mol/L), kH is the Henry's Law constant (in bar), and P is the partial pressure of the gas above the liquid (in bar).

We can convert the pressure used in the carbonation process from bar to atm (atmospheres) by dividing by the conversion factor of 1.01325 bar/atm:

P = 3.4 bar / 1.01325 bar/atm = 3.352 atm

We can then use Henry's Law to calculate the concentration of CO2 in the water:

C = kH * P = (1.65 * 10^3 bar) * (3.352 atm) = 5.53 mol/L

To convert this to grams of CO2 per liter of water, we need to multiply by the molar mass of CO2 (44.01 g/mol) and density of water (998 kg/m^3 or 0.998 g/mL):

5.53 mol/L * 44.01 g/mol * 0.998 g/mL = 244 g/L

Therefore, the amount of CO2 dissolved in a 1.00 L bottle of carbonated water at 298 K is 244 grams.

There are 246 grams of CO₂ dissolved in the 1.00 L bottle of carbonated water.

The amount of CO₂ dissolved in water can be calculated using Henry's Law, which states that the concentration of a gas dissolved in a liquid is proportional to the pressure of the gas above the liquid. The Henry's Law constant for CO₂ in water at 298 K is 1.65 x \(10^3\) bar.

The equation for Henry's Law is:

C = k * P

where C is the concentration of the gas in the liquid (in mol/L), k is the Henry's Law constant (in bar), and P is the partial pressure of the gas (in bar).

First, we need to calculate the partial pressure of CO₂ in the carbonated water bottle. The pressure used in the carbonation process was 3.4 bar, so we assume that the partial pressure of CO₂ in the bottle is also 3.4 bar.

Next, we can use Henry's Law to calculate the concentration of CO₂ in the water:

C = k * P

C = 5.61

Now we can calculate the mass of CO₂ in the bottle:

mass = concentration * volume * molar mass

The molar mass of CO₂ is 44.01 g/mol.

mass = (5.61) * (1.00 L) * (44.01 g/mol)

mass = 246 g

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The formula for calculating the concentration of a solution in mg/L is as follows:Concentration in mg/L = (weight of solute in mg / volume of solution in L). Concentration in mg/L = (0.5 g × 1000 mg/g) / 0.5 L. Concentration in mg/L = 1000 mg/L.

(b) The formula for calculating the concentration of a solution in ppm is as follows:

Concentration in ppm = (weight of solute in mg / volume of solution in ml) × 10³.

Concentration in ppm = (0.5 g × 1000 mg/g) / 500 ml × 10³.

Concentration in ppm = 1000 ppm.

(c)The formula for calculating the concentration of a solution in ppb is as follows:

Concentration in ppb = (weight of solute in mg / volume of solution in ml) × 10⁶.

Concentration in ppb = (0.5 g × 1000 mg/g) / 500 ml × 10⁶.

Concentration in ppb = 1000000 ppb.

(d) Moles/lThe molecular weight of KCl is 74.55 g/mol.

To calculate the concentration of KCl in moles/liter, we must first convert 0.5 g to moles.

Number of moles = weight / molecular weight.

Number of moles of KCl = 0.5 g / 74.55 g/mol.

Number of moles of KCl = 0.0067 moles.

Concentration in moles/liter = number of moles / volume in L.

Concentration in moles/liter = 0.0067 moles / 0.5 L.

Concentration in moles/liter = 0.013 M (M = moles/L).

Therefore, the concentration of KCl is (a) 1000 mg/L, (b) 1000 ppm, (c) 1000000 ppb, and (d) 0.013 M.

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Explanation:

Our environment is very important to us because it is where we live and share resources with other species. ... Environmental science enlightens us on how to conserve our environment in the face of increasing human population growth and anthropogenic activities that degrade natural resources and ecosystems.

The cycle of citric acid Hydrogen atoms are transported to FAD during the conversion of succinate to fumarate. This process is catalyzed by a succinate dehydrogenase enzyme.

During the process of succinate dehydrogenation, succinate is oxidized to fumarate by losing electrons, which are transferred to FAD, reducing it to FADH2. This transfer of electrons from succinate to FAD is an oxidation-reduction reaction, also known as redox reaction. The energy released during this reaction is harnessed to generate ATP, which is an important energy currency for the cell. Succinate dehydrogenase is an enzyme complex that contains multiple subunits and cofactors, including flavin adenine dinucleotide (FAD) and iron-sulfur clusters. It is located in the inner mitochondrial membrane, which allows it to transfer electrons to the electron transport chain, leading to the production of ATP. It's important to note that the citric acid cycle is a crucial metabolic pathway that takes place in the mitochondria of eukaryotic cells, and it is also known as the TCA cycle or Krebs cycle. The cycle is a series of chemical reactions that converts acetyl-CoA, derived from carbohydrates, fats, and proteins, into carbon dioxide and water, releasing energy in the form of ATP, hydrogen atoms other high-energy molecules.

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Answer:

5.12 atm

Explanation:

Before you can use the Ideal Gas Law to find the pressure, you need to convert grams to moles (via molar mass).

Molar Mass (H₂): 2(1.008 g/mol)

Molar Mass (H₂): 2.016 g/mol

 1.8 grams H₂               1 mole
----------------------  x  ----------------------  =  0.893 moles H₂
                                 2.016 grams

The Ideal Gas Law equation looks like this:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas Constant (0.0821 L*atm/mol*K)

-----> T = temperature (K)

After converting Celsius to Kelvin, you can plug the given values into the equation and simplify to find the pressure.

P = ? atm                                R = 0.0821 L*atm/mol*K

V = 4.3 L                                T = 27 °C + 273.15 = 300.15 K

n = 0.893 moles

PV = nRT

P(4.3 L) = (0.893 moles)(0.0821 L*atm/mol*K)(300.15 K)

P(4.3 L) = 22.0021

P = 5.12 atm

**Based on my past experiences, I believe the constant (R) you provided may have been mistyped. Instead of 0.821, I used 0.0821.**

For preparing a 50 ml, 5% (v/v) aqueous solution of methanol (CH3OH), measure 2.5 ml of methanol and dilute it with distilled water in a 50 ml volumetric flask.

To prepare a 50 ml, 5% (v/v) aqueous solution of methanol (CH3OH), measure 2.5 ml of methanol using a graduated cylinder or pipette. Transfer the methanol to a 50 ml volumetric flask and fill it to the mark with distilled water. By following these steps, you can successfully prepare the desired 50 ml, 5% (v/v) aqueous solution of methanol.

To prepare a 50 ml, 5% (v/v) aqueous solution of methanol (CH3OH), follow these steps:

1. Calculate the volume of methanol needed:

  Volume of methanol = (desired concentration) * (desired final volume)

                    = 5% * 50 ml

                    = 0.05 * 50 ml

                    = 2.5 ml

2. Measure out 2.5 ml of methanol using a graduated cylinder or pipette.

3. Transfer the measured methanol into a 50 ml volumetric flask using a funnel, if necessary.

4. Fill the volumetric flask to the 50 ml mark with distilled water, ensuring the bottom of the meniscus aligns with the mark.

5. Gently swirl the volumetric flask to ensure thorough mixing of the methanol and water.

Now, you have prepared a 50 ml, 5% (v/v) aqueous solution of methanol (CH3OH).

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In an acid – base titration, the titration curve reflects the strengths of the corresponding acid and base.

If one reagent is a weak acid or base and the other is a strong acid or base, the titration curve is irregular, and the pH shifts less with small additions of titrant near the equivalence point

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Answer:

B. Water vapor transfers back to water droplets or ice crystals.

Explanation:

Its What I know ehe sorry in advance if its wrong :))

Answer:

ok here is you answer

Explanation:

Blood is a liquid because it is composed of cells and plasma that are suspended in a liquid state and can easily flow through the circulatory system, delivering oxygen and nutrients to cells and removing waste products.

Answer:

Deciduous forests

Explanation: