Buffer capacity denotes how much acid or base a buffer solution can integrate before alterations in pH becomes significant. It is crucial in maintaining physiological activities, particularly in blood pH regulation. The substance absorbing the ions is typically a weak acid/base and their conjugates.
Buffer capacity is the amount of acid or base a buffer solution can accommodate before the pH is significantly pushed outside of the buffer range. Solutions that contain sizable quantities of a weak conjugate acid-base pair are known as buffer solutions. These usually experience only slight changes in pH when small amounts of acid or base are added.
A large enough addition of these substances can exceed the buffer capacity, consuming most of the conjugate pair and leading to a drastic change in pH. In living organisms, a variety of buffering systems exist to maintain the pH of blood and other fluids within a strict range between pH 7.35 and 7.45, ensuring normal physiological functioning.
The substance that absorbs the ions is usually a weak acid, which absorbs hydroxyl ions, or a weak concentrate base, which absorbs hydrogen ions. The buffer capacity is greater in solutions that contain more of this weak acid/base and their conjugates.
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Buffer capacity refers to the amount of acid or base that a buffer solution can absorb before experiencing a significant shift in pH, commonly by one pH unit.
Buffer capacity is the amount of acid or base a buffer can handle before pushing the pH outside of the buffer range. Essentially, it is a measure of a buffer's resistance to pH change upon the addition of an acid or base. Buffer capacity depends on the concentrations of the weak acid and its conjugate base present in the mixture. For instance, a solution with higher concentrations of acetic acid and sodium acetate will have a greater buffer capacity than a more dilute solution of the same components. The buffer's capacity is directly proportional to its ability to absorb strong acids or bases before there's a significant change in pH, typically defined as a shift by one pH unit.
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Answer:
predators are controlling the population of the species who are below them in the food pyramid . Also if the population of the preys decrease it will alternatively reduce the predator population .therefore the predator prey relationship balance an eco system.
When C-C is having a triple bond the hybridization is sp. But I am not sure how to relate that to the linear shape.
Answer:
1. 176 × 10^12 W ; 78400000000
Explanation:
Given the following :
Fall rate = 2,400,000kg/s
Average height of fall = 50m
Gravitational Potential of falling water = mgh = mass × acceleration due to gravity × height =
How many 15 W LED light bulbs could it power?
Recall : power = workdone / time
Workdone = gravitational potential energy
Mass of water = density * volume
Density of water = 1 * 10^3kg/m^3
Rate of fow = volume / time = 2400000
Hence,
Power = 1000 * 2,400,000 * 9.8 * 50
Power = 1176000000000
Power = 1. 176 × 10^12 W
How many 15 W LED light bulbs could it power?
1176000000000 / 15 = 78400000000
= 78400000000 15 W bulbs
Answer : The volume at STP will be 0.2944 L
Solution : Given,
Initial volume = 500 ml
Initial temperature =
Initial pressure = 500 mmHg =
At STP,
Temperature = 298 K
Pressure = 1 atm
Formula used :
where,
= initial pressure
= pressure at STP
= initial temperature
= temperature at STP
= initial volume
= volume at STP
Now put all the given values in this formula, we get
By rearranging the terms, we get the volume at STP
(1 L = 1000 ml)
Therefore, the volume at STP will be 0.2944 L
To find the volume of nitrogen at STP, we need to use the ideal gas law equation.
To find the volume of nitrogen at STP, we need to use the ideal gas law equation. The ideal gas law equation is PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the universal gas constant, and T is temperature in Kelvin. At STP, the pressure is 1 atm and the temperature is 273 K. Plug in the given values and solve for V:
V = (500 mL * 500 mmHg * (1 atm / 760 mmHg))/(0.0821 L·atm/mol·K * 333 K)
V = 162.6 mL
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Answer:
1a. The balanced equation is given below:
2NO + O2 → 2NO2
The coefficients are 2, 1, 2
1b. 755.32g of NO2
2a. The balanced equation is given below:
2C6H6 + 15O2 → 12CO2 + 6H2O
The coefficients are 2, 15, 12, 6
2b. 126.25g of CO2
Explanation:
1a. Step 1:
Equation for the reaction. This is given below:
NO + O2 → NO2
1a. Step 2:
Balancing the equation. This is illustrated below:
NO + O2 → NO2
There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by putting 2 in front of NO and 2 in front of NO2 as shown below:
2NO + O2 → 2NO2
The equation is balanced.
The coefficients are 2, 1, 2
1b. Step 1:
Determination of the limiting reactant. This is illustrated below:
2NO + O2 → 2NO2
From the balanced equation above, 2 moles of NO required 1 mole of O2.
Therefore, 16.42 moles of NO will require = 16.42/2 = 8.21 moles of O2.
From the calculations made above, there are leftover for O2 as 8.21 moles out of 14.47 moles reacted. Therefore, NO is the limiting reactant and O2 is the excess reactant.
1b. Step 2:
Determination of the maximum amount of NO2 produced. This is illustrated below:
2NO + O2 → 2NO2
From the balanced equation above, 2 moles of NO produced 2 moles of NO2.
Therefore, 16.42 moles of NO will also produce 16.42 moles of NO2.
1b. Step 3:
Conversion of 16.42 moles of NO2 to grams. This is illustrated below:
Molar Mass of NO2 = 14 + (2x16) = 14 + 32 = 46g/mol
Mole of NO2 = 16.42 moles
Mass of NO2 =?
Mass = number of mole x molar Mass
Mass of NO2 = 16.42 x 46
Mass of NO2 = 755.32g
Therefore, the maximum amount of NO2 produced is 755.32g
2a. Step 1:
The equation for the reaction.
C6H6 + O2 → CO2 + H2O
2a. Step 2:
Balancing the equation:
C6H6 + O2 → CO2 + H2O
There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by 6 in front of CO2 as shown below:
C6H6 + O2 → 6CO2 + H2O
There are 6 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 3 in front of H2O as shown below:
C6H6 + O2 → 6CO2 + 3H2O
There are a total of 15 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 15/2 in front of O2 as shown below:
C6H6 + 15/2O2 → 6CO2 + 3H2O
Multiply through by 2 to clear the fraction.
2C6H6 + 15O2 → 12CO2 + 6H2O
Now, the equation is balanced.
The coefficients are 2, 15, 12, 6
2b. Step 1:
Determination of the mass of C6H6 and O2 that reacted from the balanced equation. This is illustrated below:
2C6H6 + 15O2 → 12CO2 + 6H2O
Molar Mass of C6H6 = (12x6) + (6x1) = 72 + 6 = 78g/mol
Mass of C6H6 from the balanced equation = 2 x 78 = 156g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 15 x 32 = 480g
2b. Step 2:
Determination of the limiting reactant. This is illustrated below:
From the balanced equation above,
156g of C6H6 required 480g of O2.
Therefore, 37.3g of C6H6 will require = (37.3x480)/156 = 114.77g of O2.
From the calculations made above, there are leftover for O2 as 114.77g out of 126.1g reacted. Therefore, O2 is the excess reactant and C6H6 is the limiting reactant.
2b. Step 3:
Determination of mass of CO2 produced from the balanced equation. This is illustrated belowb
2C6H6 + 15O2 → 12CO2 + 6H2O
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 12 x 44 = 528g
2b. Step 4:
Determination of the mass of CO2 produced by reacting 37.3g of C6H6 and 126.1g O2. This is illustrated below:
From the balanced equation above,
156g of C6H6 produced 528g of CO2.
Therefore, 37.3g of C6H6 will produce = (37.3x528)/156 = 126.25g of CO2