a 0.050 m solution of the salt nab has a ph of 9.00. calculate the ph of a 0.010 m solution of hb

The component of chi square for the epidural / not breastfeeding cell comes out to be 166.58.

Chi-square is a statistical test that examines the differences between categorical variables from a random sample in order to determine whether the expected and observed results are well-fitting.

To calculate expected frequency for Epiduarl Yes and no breast feeding

=E(2,1) = row total *column total/Grand total

          = 488*412/1205

          = 166.8515

          = 166.85

Thus, the expected frequency comes out to be 166.85.

df=(r-1)(c-1)

r-->no of rows

c-->no of columns

df=(2-1)(2-1)

df=1

in excel for the chi sq statistic and df p value is

=CHISQ.DIST.RT(11.28,1)

=0.000783

=0.001(rounded to 3 decimals)

p=0.001

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Complete question-

The table shows the results of a study investigating whether aftereffects of epidurals administered during childbirth might interfere with successful breastfeeding. A researcher is planning to do a chi-square test. Assume the conditions for inference are met. Complete parts a) through c). Breastfeeding at 6 months? Epidural? Yes No Yes 218 499 No 194 294 Total 412 793 Total 717 488 1205 What is the component of chi-square for the epidural / no breastfeeding cell? 4.42 (Round to two decimal places as needed.) b) For this test, x? = 11.28. What's the P-value? P-value = 0.001 (Round to three decimal places as needed.) c) State your conclusion. (Assume a significance level of 0.05.) Choose the correct answer below. O A. Reject the null hypothesis. There is evidence that having an epidural and success in breastfeeding are not independent. B . Fail to reject the null hypothesis. There is evidence that having an epidural and success in breastfeeding are not independent. O C. Reject the null hypothesis. There is no evidence that having an epidural and success in breastfeeding are not independent. D. Fail to reject the null hypothesis. There is no evidence that having an epidural and success in breastfeeding are not independent. O

The reasoning for spreading ammonia around the top of the bottle containing acryloyl chloride is to neutralize any hydrochloric acid that may be present in the bottle.

Acryloyl chloride is a highly reactive and volatile compound that can release hydrochloric acid when exposed to moisture or air. By adding ammonia to the bottle, the hydrochloric acid is converted into ammonium chloride, a less hazardous compound, which makes it safer to handle and transport the acryloyl chloride. Additionally, the ammonia can also prevent the formation of unwanted byproducts or impurities that may affect the quality of the acryloyl chloride.

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The temperature at which the pressure of the gas will be reduced to 600 mmHg is 4.7°

The temperature (in °C) at which the pressure of a gas with an initial pressure of 700 mmHg will be reduced to 600 mmHg, with no change in volume or amount of gas.

To find the temperature at which the pressure of the gas will be reduced to 600 mmHg, we can use the combined gas law which relates the pressure, volume, and temperature of a gas. The formula for the combined gas law is:

(P₁V₁/T₁) = (P₂V₂/T₂)

where P₁, V₁, and T₁ are the initial pressure, volume, and temperature of the gas, and P₂ and V₂ are the final pressure and volume of the gas.

We are given P₁ = 700 mmHg, P₂ = 600 mmHg, and T₁ = 107 °C. We need to find T₂. Since there is no change in volume or amount of gas, we can assume that V₁ = V₂.

Rearranging the equation, we get:

T₂ = (P₂/T₁) * (V₁/P₁)

Substituting the given values, we get:

T₂ = (600 mmHg / 107°C) * (V₁ / 700 mmHg)

Since V₁ = V₂, we can assume that V₁/V₂ = 1. Substituting this, we get:

T₂ = (600 mmHg / 107°C) * (1 / 700 mmHg) = 0.0078°C/mmHg

Therefore, the temperature at which the pressure of the gas will be reduced to 600 mmHg is:

T₂ = 0.0078°C/mmHg * 600 mmHg = 4.7°C

So, the temperature at which the pressure of the gas will be reduced to 600 mmHg is 4.7°

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40.2 kJ of heat energy is required to vaporize 102 g of benzene at its boiling point.

First, we need to calculate the number of moles of benzene present in 102 g of benzene.

The molar mass of benzene, [tex]C6H6[/tex] is 6(12.01 g/mol) + 6(1.01 g/mol) = 78.11 g/mol.

Number of moles of benzene = mass of benzene/molar mass of benzene

= 102 g / 78.11 g/mol

= 1.3078 mol

Next, we can use the formula for heat of vaporization to calculate the amount of energy required to vaporize this amount of benzene:

Heat energy = n x ΔHvap

where n is the number of moles of benzene and ΔHvap is the heat of vaporization of benzene.

Heat energy = 1.3078 mol x 30.8 kJ/mol

= 40.2 kJ

Therefore, 40.2 kJ of heat energy is required to vaporize 102 g of benzene at its boiling point.

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The third ionization energy is the energy required to remove a third electron from an atom. Zinc (Zn) has a greater nuclear charge than scandium (SC) due to its higher atomic number, which means that its electrons are held more tightly by the nucleus.

Therefore, it takes more energy to remove a third electron from Zn than from SC, making the statement true.

the third ionization energy of Zn (Zinc) is higher than Sc (Scandium). This is because Zn has a completely filled 3d10 subshell, making it more stable and harder to remove an electron, while Sc has a less stable 3d1 configuration.

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L-glucose is the enantiomer of D-glucose. An enantiomer is a pair of molecules that are mirror images of each other but are not superimposable.

L-glucose and D-glucose have the same molecular formula but differ in the configuration of the chiral centers. In this case, the chiral centers are the carbons with four different substituents attached. The term "C-5 epimer" refers to a pair of sugars that differ in the configuration of only one chiral center, which is not the case for L-glucose and D-glucose, as they differ in all their chiral centers.

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Calculating the new concentrations of the conjugate acid and base following the addition is necessary in order to determine the pH that will be produced when a solid NAOH is added to the buffer solution.

The reaction between NAOH and  C₆H₅NH₂ has a balanced equation that looks like this:

NaOH +  C₆H₅NH₂ NaC₆H₅NH₂ + H₂O

According to the stoichiometry of this reaction, one mole of  C₆H₅NH₂ is consumed for every mole of NAOH added, resulting in one mole of NaC₆H₅NH₂.

Introductory convergences of the support parts:

[C₆H₅NH₃Cl] = 0.100 mol/0.200 L = 0.500 M

[C₆H₅NH₂] = 0.500 M

After the expansion of 0.040 mol of NAOH, the grouping  C₆H₅NH₂ will diminish by 0.040 mol/0.200 L = 0.200 M. The convergence  NaC₆H₅NH₂ will be 0.040 mol/0.200 L = 0.200 M.

The response between C₆H₅NH₂ and water creates C₆H₅NH₃+ Gracious particles. C₆H₅NH₂ 's Kb expression is as follows:

Kb = [C₆H₅NH₃+][OH-]/[C₆H₅NH₂]

We can expect that the convergence  [OH-] is equivalent to the grouping of NAOH added, which is 0.040 mol/0.200 L = 0.200 M. The grouping  C₆H₅NH₃+ can be determined utilizing the Henderson-Hasselbalch condition:

pH = pKa + log([C₆H₅NH₂]/[C₆H₅NH₃+])

The pKa of C₆H₅NH₃+ isn't given, however, we can utilize the estimation pKa = 9.24 C₆H₅NH₃+. Thus, pH equals 9.24 times log(0.500/0.300) = 9.64.

As a result, the pH of the buffer solution after adding 0.040 mol of NAOH is 9.64.

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The balanced redox reaction is: Fe³⁺ + NO₂⁻ + H₂O → Fe²⁺ + H⁺ + NO₃⁻

In this redox reaction, we need to balance the number of atoms and charges on both sides of the equation. First, we need to determine the oxidation state of each element in the reactants and products.

Fe³⁺ is being reduced to Fe²⁺, which means it is gaining electrons. NO₂⁻ is being oxidized to NO₃⁻, which means it is losing electrons. To balance the electrons, we need to add 2 electrons to the left side of the equation:

Fe³⁺ + NO₂⁻ + 2e⁻ + H₂O → Fe²⁺ + H⁺ + NO₃⁻

Next, we balance the charges by adding 2 H⁺ ions to the left side of the equation:

Fe³⁺ + NO₂⁻ + 2e⁻ + 2H⁺ + H₂O → Fe²⁺ + H⁺ + NO₃⁻

Finally, we balance the number of atoms by adding a water molecule to the right side of the equation:

Fe³⁺ + NO₂⁻ + 2e⁻ + 2H⁺ + H₂O → Fe²⁺ + H⁺ + NO₃⁻ + H₂O

This is the balanced redox reaction.

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At equilibrium, the concentration of ethene is 0.282 M, the concentration of hydrogen is 0.473 M, and the concentration of ethane is 0.1058 M.

Assuming the reaction is as follows:

C₂H₂(g) +H₂ (g) ⇌C₂H₆ (g)

We can use the equilibrium constant expression to determine the concentrations of each substance at equilibrium:

Kc = [C2₂H₆] / ([C₂H₄] * [H₂])

where Kc is the equilibrium constant and the square brackets denote concentration in units of mol/L.

At equilibrium, the reaction will have reached a state where the rate of the forward reaction is equal to the rate of the reverse reaction. This means that the concentration of each substance will no longer be changing with time.

Let x be the change in concentration of C₂H₂ and H₂ and 2x be the change in concentration of C₂H₂ (based on the stoichiometry of the reaction). Then, the equilibrium concentrations can be expressed in terms of the initial concentrations and the change in concentration:

[C₂H₄]eq = 0.335 - x

[H₂]eq = 0.526 - x

[C₂H₆]eq = 2x

We can substitute these expressions into the equilibrium constant expression:

Kc = [C₂H₆]eq / ([C₂H₄]eq * [H₂]eq)

Solving for x and substituting the given values for the initial concentrations and the equilibrium constant (Kc = 3.5), we get:

3.5 = (2x) / ((0.335 - x) * (0.526 - x))

x = 0.0529 mol/L

Substituting this value back into the expressions for the equilibrium concentrations, we get:

[C₂H₄]eq = 0.335 - 0.0529 = 0.282 mol/L

[H₂]eq = 0.526 - 0.0529 = 0.473 mol/L

[C₂H₆]eq = 2(0.0529) = 0.1058 mol/L

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The density of water is approximately 1 g/mL. The liquid with a density of 4 g/mL would definitely sink in water since the liquid's density (4 g/mL) is greater than the density of water.

The density of the liquid can be calculated by dividing its mass by its volume. So, the density of the liquid is 4 g/mL (200 g ÷ 50 mL).

Whether the liquid would float on water or not depends on the density of water. If the density of water is less than 4 g/mL, then the liquid would sink in water. However, if the density of water is more than 4 g/mL, then the liquid would float on water. The density of water is approximately 1 g/mL, so the liquid with a density of 4 g/mL would definitely sink in water.

Alternatively, to find the density of the liquid, we will use the formula:

Density = Mass / Volume

Given the mass of the liquid is 200 grams and the volume is 50 milliliters, we can plug these values into the formula:

Density = 200 grams / 50 milliliters = 4 grams per milliliter (g/mL)

Now, to determine if the liquid would float on water, we need to compare its density to that of water. The density of water is approximately 1 g/mL. Since the liquid's density (4 g/mL) is greater than the density of water, it will not float on water, and will instead sink.

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The net kinetic energy of the two curling stones is 160 J, and their net momentum is 0 kg·m/s.

The kinetic energy of an object is given by the formula KE = 1/2mv^2, where m is the mass of the object and v is its velocity. Since the two curling stones have the same mass and speed, their individual kinetic energies are given by:

KE1 = 1/2(20 kg)(4 m/s)²= 160 J

KE2 = 1/2(20 kg)(4 m/s)² = 160 J

The net kinetic energy of the two stones is simply the sum of their individual kinetic energies, which is:

KE net = KE1 + KE2 = 160 J + 160 J = 320 J

The momentum of an object is given by the formula p = mv, where m is the mass of the object and v is its velocity. Since the two curling stones are moving in opposite directions with the same speed, their individual momenta are equal in magnitude but opposite in direction.

Therefore, their net momentum is zero:

p1 = (20 kg)(4 m/s) = 80 kg·m/s (in the positive direction)

p2 = -(20 kg)(4 m/s) = -80 kg·m/s (in the negative direction)

p net = p1 + p2 = 0 kg·m/s

As a result, the two stones' net kinetic energy is 320 J, and their net momentum is 0 kgm/s.

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Therefore, the concentration of potassium nitrate in the coolant is 2,068.7 ppm.

First, let's find the mass of the mixture of propyl glycol and potassium nitrate:

Mass of propyl glycol = density x volume = 965 kg/m3 x 15 L = 14475 g

Mass of potassium nitrate = 30 g

Total mass of mixture = 14475 g + 30 g = 14505 g

Next, we can calculate the concentration of potassium nitrate in parts per million (ppm):

1 ppm = 1 mg/L = 0.001 g/L

Concentration of potassium nitrate = (mass of potassium nitrate / total mass of mixture) x 10⁶

= (30 g / 14505 g) x 10⁶

= 2,068.7 ppm

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The Chornobyl nuclear disaster led to significant environmental problems due to the release of radioactive isotopes like iodine-131 and cesium-137. Given a soil sample near Chornobyl containing 187kBq/m2 of cesium-137 and its half-life of approximately 30 years, we can calculate the amount remaining after 90 years.

90 years is equal to three half-lives (90 / 30 = 3). After each half-life, the amount of cesium-137 is reduced by half. So, we can apply the following formula:
Remaining cesium-137 = Initial amount * (1/2)^number of half-lives
Remaining cesium-137 = 187kBq/m2 * (1/2)^3
Remaining cesium-137 = 187kBq/m2 * (1/8)
Remaining cesium-137 = 23.38kBq/m2
After 90 years, 23.38 kilo-becquerels per square meter of cesium-137 will remain in the soil sample.

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The solubility of calcium sulfate in grams solute per 100 grams of solvent is 0.67 g/100g. The correct answer is (A) CaSO₄(s) ⇌ Ca²+(aq) + SO4²⁻(aq); (B) 0.67 g/100g

(A) The solubility equilibrium of calcium sulfate can be represented by the following chemical equation:

CaSO₄(s) ⇌ Ca²+(aq) + SO4²⁻(aq)

(B) To calculate the solubility of calcium sulfate in grams solute per 100 grams of solvent, we need to first calculate the molar solubility of calcium sulfate, which is the number of moles of calcium sulfate that dissolve in one liter of solvent at equilibrium.

The solubility product constant (K_sp) of calcium sulfate is given by:

K_sp = [Ca²⁺][SO4²⁻]

At equilibrium, let x be the molar solubility of calcium sulfate, then:

[Ca²⁺] = x

[SO₄²⁻] = x

Substituting these values into the expression for K_sp, we get:

K_sp = x²

Solving for x, we get:

x = sqrt(K_sp) = sqrt(2.4 × 10⁻⁵) = 0.0049 M

The solubility of calcium sulfate in grams solute per 100 grams of solvent can be calculated using the following equation:

solubility (g/100g) = (molar mass of solute) × (molar solubility) ÷ (density of solvent) × 100

Assuming water as the solvent (which has a density of 1 g/mL), we get:

solubility (g/100g) = (136 g/mol) × (0.0049 mol/L) ÷ (1 g/mL) × 100 = 0.67 g/100g

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The Vrms, in meters per second, for helium atoms at 5.25 k is found to be 1233.9 m/s.

Following equation gives the root-mean-square (rms) speed of gas molecules:

v(rms) = √[(3kT) / (m)], Boltzmann constant is k, temperature in Kelvin is T, and molar mass of the gas in kilograms per mole is m. The molar mass of helium is 4.003 g/mol, or 0.004003 kg/mol. We can convert the temperature of 5.25 K to Kelvin by adding 273.15 K, giving a temperature of 278.4 K.

Plugging in the values, we get,

v(rms) = √[(3kT) / (m)]

v(rms) = √[(3 × 1.38 × 10⁻²³ J/K × 278.4 K) / (0.004003 kg/mol)]

v(rms) = 1233.9 m/s (rounded to four significant figures)

Therefore, the rms speed of helium atoms at 5.25 K is approximately 1233.9 m/s.

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The correct option is A, The percentage of total fluorescence captured by the objective is 37.5%,

% fluorescence captured = (excitation light power x fluorescence emitted x transmittance) / (2π x objective NA x oil refractive index x area of sphere x one-photon absorption cross section x exposure time)

Plugging in the given values, we get:

% fluorescence captured = (57 kW/cm x 0.5 x 0.4 x 0.9 x 0.99 x 0.9 x 0.4) / (2π x 1.3 x 1.5 x 4π x [tex]10^{-16}[/tex] cm² x 5 ms)

% fluorescence captured = 37.5%

Fluorescence is a phenomenon that occurs when a substance absorbs light of a specific wavelength and then emits light of a longer wavelength. This emission of light is known as fluorescence. Fluorescence is commonly observed in certain chemicals, dyes, and biological molecules, such as proteins and nucleic acids.

When a molecule is excited by absorbing light of a specific wavelength, it enters an excited state. The excited state is unstable, and the molecule quickly returns to its ground state by releasing the excess energy as light of a longer wavelength. The emitted light can be detected using a fluorometer, which measures the intensity and wavelength of the emitted light.

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If 275 L of methane (measured at STP) is converted to synthesis gas, 74.36 g of hydrogen gas will be formed.

The balanced chemical equation for the conversion of methane to synthesis gas (a mixture of carbon monoxide and hydrogen) is:

CH₄ + H₂O → CO + 3H₂

From the balanced equation, we can see that one mole of methane reacts with one mole of water to produce one mole of carbon monoxide and three moles of hydrogen.

At STP (standard temperature and pressure), one mole of gas occupies 22.4 L. Therefore, 275 L of methane at STP is equivalent to:

moles of methane = volume of methane at STP / molar volume at STP

moles of methane = 275 L / 22.4 L/mol

moles of methane = 12.29 moles

Using the stoichiometry of the balanced equation, we can calculate the number of moles of hydrogen that will be produced:

moles of H₂ = 3 x moles of methane

moles of H₂ = 3 x 12.29 moles

moles of H₂ = 36.87 moles

Finally, we can convert moles of hydrogen to grams using its molar mass:

molar mass of H₂ = 2.016 g/mol

mass of H₂ = moles of H₂ x molar mass of H₂

mass of H₂ = 36.87 moles x 2.016 g/mol

mass of H₂ = 74.36 g

Therefore, if 275 L of methane is transformed to synthesis gas (as measured at STP), 74.36 g of hydrogen gas is produced.

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To calculate the density of rhodium, we need to use the formula:
Density = (number of atoms per unit cell) x (atomic mass) / (volume of unit cell)
For a face-centered cubic unit cell, there are 4 atoms per unit cell. The volume of a face-centered cubic unit cell can calculated using the formula:
Volume = (4/3) x π x (r^3)
Where r is the atomic radius of rhodium, which is given as 135 pm (or 1.35 Å, since 1 Å = 100 pm).
Plugging in the values, we get:
Volume = (4/3) x π x (1.35 Å)^3
Volume = 3.602 Å^3
Converting to cm^3, we get:
Volume = 3.602 x 10^-23 cm^3
Now we can calculate the density:
Density = (4 atoms per unit cell) x (102.9 g/mol) / (3.602 x 10^-23 cm^3)
Density = 1.120 x 10^5 g/cm^3
Therefore, the density of rhodium is 1.120 x 10^5 g/cm^3.

To determine the density of rhodium, we need to consider its atomic mass, crystalline structure, and atomic radius. Rhodium has an atomic mass of 102.9 g/mol, crystallizes in a face-centered cubic (FCC) unit cell, and has an atomic radius of 135 pm.

In an FCC unit cell, there are 4 atoms per cell. The edge length of the cell can be found using the formula: edge length = 2√2 * atomic radius. Plugging in the given values:
Edge length = 2√2 * 135 pm = 380.8 pm = 0.3808 nm
The volume of the unit cell is edge length^3:
Volume = (0.3808 nm)^3 = 0.0551 nm³ = 55.1 cm³/mol (1 nm³ = 1 x 10⁻²¹ cm³)
Now, we can determine the density using the formula: density = mass/volume:
Density = (4 * 102.9 g/mol) / (55.1 cm³/mol) = 7.50 g/cm³
Therefore, the density of rhodium is approximately 7.50 g/cm³.

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The main cause of the increase in the amount of CO₂ in Earth's atmosphere over the past 170 years is the burning of fossil fuels and deforestation (option B). The Industrial Revolution, which began in the late 18th century, led to a significant rise in the use of fossil fuels like coal, oil, and natural gas to power machines, vehicles, and factories. The combustion of these fuels releases large amounts of carbon dioxide, a greenhouse gas, into the atmosphere.

Deforestation, particularly in tropical regions, also contributes to the increase in atmospheric CO₂ levels. Trees and plants act as carbon sinks, absorbing CO₂ during photosynthesis and storing it in their biomass. When forests are cut down or burned, the stored carbon is released back into the atmosphere, and the capacity of the ecosystem to absorb CO₂ is reduced.

These human activities have disrupted the natural balance of the carbon cycle, leading to a significant increase in atmospheric CO₂ concentrations. This rise in CO₂ levels contributes to global warming, as CO₂ traps heat within the Earth's atmosphere, increasing the greenhouse effect and raising average global temperatures.

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Answer: a) 3500.16 grams of O2 are required to react with 1000 g of octane.

b) Approximately 9254.24 grams of CO2 are produced per gallon of gasoline burned.

Explanation:

a) To determine how many grams of O2 are required to react with 1000 g of octane, we need to use the balanced chemical equation for the combustion of octane:

C8H18(l) + 12.5 O2(g) -> 8 CO2(g) + 9 H2O(I)

From the balanced equation, we can see that 12.5 moles of O2 are required to react with 1 mole of octane. To convert grams of octane to moles, we need to divide the given mass by the molar mass of octane:

1000 g / 114.23 g/mol = 8.75 mol

So we need:

12.5 mol O2 / 1 mol C8H18 x 8.75 mol C8H18 = 109.38 mol O2

Finally, to convert moles of O2 to grams, we multiply by the molar mass of O2:

109.38 mol O2 x 32 g/mol = 3500.16 g O2

Therefore, 3500.16 grams of O2 are required to react with 1000 g of octane.

b) To determine how many grams of CO2 are produced per gallon of gasoline burned, we need to know the molar mass of gasoline. Since gasoline is a mixture of hydrocarbons with different molecular weights, we cannot determine its exact molar mass. However, we can estimate it based on the average molar mass of a hydrocarbon, which is around 114 g/mol (similar to octane).

Assuming a gallon of gasoline weighs 3000 grams (as stated in the question), we can estimate the number of moles of gasoline burned:

3000 g / 114 g/mol = 26.32 mol gasoline

From the balanced equation for the combustion of gasoline (which is similar to the equation for octane), we can see that 8 moles of CO2 are produced for every mole of gasoline burned:

CnHm + (n + m/4) O2 -> n CO2 + (m/2) H2O

So the number of moles of CO2 produced is:

8 mol CO2 / 1 mol gasoline x 26.32 mol gasoline = 210.56 mol CO2

Finally, to convert moles of CO2 to grams, we multiply by the molar mass of CO2:

210.56 mol CO2 x 44 g/mol = 9254.24 g CO2

Therefore, approximately 9254.24 grams of CO2 are produced per gallon of gasoline burned.

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The Kb of hydroxylamine, NH₂OH, is 1.10×10^−8. A buffer solution is prepared by mixing 110 mL of a 0.35 M hydroxylamine solution with 60 mL of a 0.28 M HCl solution. The pH of the resulting buffer solution is approximately 5.36.

The pH of the resulting buffer solution, we first need to calculate the concentrations of NH₂OH and NH₃+ in the solution after the reaction with HCl.

The balanced chemical equation for the reaction between NH₂OH and HCl is:

NH₂OH + HCl → NH₃+ + Cl- + H₂O

The amount of HCl used to neutralize the NH₂OH can be determined using the stoichiometry of the reaction. Since the reaction is a 1:1 reaction, the amount of HCl used is equal to the amount of NH₂OH present in the solution.

moles of NH₂OH = M x V = 0.35 M x 0.11 L = 0.0385 mol

moles of HCl used = 0.0385 mol

The moles of NH₃+ formed in the reaction is also equal to the moles of HCl used, as per the balanced equation. Thus, the new concentration of NH₃+ in the buffer solution is:

[C(NH₃+)] = moles of NH₃+ / total volume of solution

          = 0.0385 mol / (0.11 L + 0.06 L)

          = 0.385 M

The concentration of NH₂OH remaining in the buffer solution can be calculated by subtracting the moles of HCl used from the initial moles of NH₂OH:

moles of NH₂OH remaining = initial moles of NH₂OH - moles of HCl used

                        = (0.35 M x 0.11 L) - 0.0385 mol

                        = 0.01265 mol

The new concentration of NH₂OH is therefore:

[C(NH₂OH)] = moles of NH₂OH / total volume of solution

           = 0.01265 mol / (0.11 L + 0.06 L)

           = 0.127 M

Now we can use the Henderson-Hasselbalch equation to find the pH of the buffer solution:

pH = pKa + log([base]/[acid])

The acid in this case is HCl, which is completely dissociated in water and does not contribute to the buffer. The base is NH₃+, which is the conjugate base of NH₂OH.

The pKa of NH₂OH can be calculated using the Kb value:

Kb = Kw/Ka

Ka = Kw/Kb

Ka = 1.0 x 10⁻¹⁴ / 1.10 x 10⁻⁸

Ka = 9.09 x 10⁻⁷

pKa = -log(Ka)

pKa = -log(9.09 x 10⁻⁷)

pKa = 6.04

Substituting the values into the Henderson-Hasselbalch equation, we get:

pH = 6.04 + log(0.127/0.385)

pH = 6.04 - 0.681

pH ≈ 5.36

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The mol fraction of Argon is 0.00892 [tex]CO_{2}[/tex] is 0.00038.

The  mol fraction of all the remaining gases present in air can be calculated by subtracting the partial pressures of nitrogen and oxygen from the total pressure of air at sea level, and then dividing each remaining gas's partial pressure by the resulting value.

1. Subtract the partial pressures of nitrogen and oxygen from the total pressure of air at sea level:

760.0 torr - 593.5 torr ([tex]N_{2}[/tex]) - 159.2 torr ([tex]O_{2}[/tex]) = 7.3 torr

2. Divide each remaining gas's partial pressure by the resulting value:

- Carbon dioxide ([tex]CO_{2}[/tex]): 0.04 x 7.3 torr = 0.292 torr

- Argon (Ar): 0.93 x 7.3 torr = 6.789 torr

- Trace gases (Ne, He, Kr, Xe): collectively make up less than 0.01% of air, so their partial pressures are negligible

Therefore, the mol fraction of all the remaining gases present in air is:

- [tex]CO_{2}[/tex]: 0.00038 (0.292 torr / 760.0 torr)

- Ar: 0.00892 (6.789 torr / 760.0 torr)

To calculate the mol fraction of all the remaining gases present in air, you need to subtract the partial pressures of nitrogen and oxygen from the total pressure of air at sea level, and then divide each remaining gas's partial pressure by the resulting value.

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The total change in enthalpy of this reaction is A, 25 KJ

The total change in enthalpy of the reaction is equal to the difference between the enthalpy of the products and the enthalpy of the reactants.

From the potential energy diagram, so see that the enthalpy of the products is 55 kJ and the enthalpy of the reactants is 30 kJ.

Therefore, the total change in enthalpy is:

ΔH = enthalpy of products - enthalpy of reactants

ΔH = 55 kJ - 30 kJ

ΔH = 25 kJ

So the answer is 25 kJ.

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The beta-pleated sheet is a secondary structure found in proteins that is characterized by the orientation of hydrogen bonds between adjacent strands. The correct answer to the question is (2) H bonds perpendicular to the molecular axis.

The correct answer is option 2.

In a beta-pleated sheet, the strands of the protein backbone are extended and oriented in a zigzag pattern, forming a flat sheet-like structure. The hydrogen bonds between adjacent strands occur between the carbonyl oxygen of one amino acid and the amide hydrogen of an adjacent amino acid, with the bonds running perpendicular to the axis of the strands. This arrangement allows for maximum stability and strength of the structure, as the hydrogen bonds provide strong interactions between adjacent strands. The orientation of the hydrogen bonds also creates a characteristic "pleated" appearance in the sheet, as the strands are forced to bend slightly to accommodate the perpendicular arrangement of the bonds. Overall, the beta-pleated sheet is an important structural motif in proteins, contributing to the overall stability and function of the molecule.

The correct answer is option 2.

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a) One glycolytic enzyme that catalyzes the severing of a carbon-carbon bond is aldolase.

b) One glycolytic enzyme whose substrate molecule and product molecule have precisely the same molecular weight is triose phosphate isomerase.

c) One glycolytic enzyme that catalyzes a dehydration reaction is enolase.

d) One glycolytic enzyme whose substrate in the forward direction of glycolysis does not contain a phosphate group is hexokinase.

e) One glycolytic enzyme that "salvages" a 3-carbon ketone fuel that otherwise would not go forward through glycolysis is glyceraldehyde-3-phosphate dehydrogenase.

f) One glycolytic enzyme whose product has a phosphate group linked to a carboxyl group is phosphoglycerate kinase.

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a) Enolase is a glycolytic enzyme that catalyzes the severing of a carbon-carbon bond.

Glycolytic is a metabolic pathway that converts glucose into two molecules of pyruvate, resulting in the production of energy in the form of ATP. This process occurs in the absence of oxygen, and is thus referred to as anaerobic glycolysis. During glycolysis, enzymes break down the six-carbon sugar molecule glucose into two three-carbon molecules of pyruvate. In the process, two molecules of ATP are produced and two molecules of NADH are generated. The ATP and NADH molecules can be used to drive other cellular processes, while the pyruvate molecules can be used in other metabolic pathways, such as the Krebs cycle. Glycolysis provides the initial energy required for the production of ATP in cells and is the most important metabolic pathway in the body.

b) Hexokinase is a glycolytic enzyme whose substrate molecule and product molecule have precisely the same molecular weight.

c) Phosphoglycerate kinase is a glycolytic enzyme that catalyzes a dehydration reaction.

d) Hexokinase is a glycolytic enzyme whose substrate in the forward direction of glycolysis does not contain a phosphate group.

e) Pyruvate carboxylase is a glycolytic enzyme that "salvages" a 3-carbon ketone fuel that otherwise would not go forward through glycolysis.

f) Phosphoglycerate kinase is a glycolytic enzyme whose product has a phosphate group linked to a carboxyl group.

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The equatorial isomer will react more rapidly in an E2 reaction. (C)

In E2 reactions, the reaction rate is determined by the steric hindrance around the carbon that is undergoing elimination. The equatorial isomer has a more favorable geometry for elimination because the leaving group and the beta-hydrogen are in the same plane, allowing for optimal orbital overlap during the elimination process.

In contrast, the cis and trans isomers have the leaving group and beta-hydrogen in different planes, leading to increased steric hindrance and a slower rate of reaction. The axial isomer is also hindered due to the large substituents in the axial positions, which leads to unfavorable steric interactions and a slower rate of reaction.

Therefore, the equatorial isomer is the most reactive towards E2 elimination due to its favorable geometry and lower steric hindrance.(C)

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The correct statement is that glycogen is nearly undetectable in meats. While glycogen is abundant in some types of food such as liver and certain seafood, it is not present in large quantities in meat. This is because glycogen is primarily stored in the muscles and liver of animals, and meat typically only contains small amounts of these tissues.

Therefore, if you are looking to consume glycogen in your diet, it may be more beneficial to seek out other sources such as certain types of seafood or organ meats.

glycogen is nearly undetectable in meats. Glycogen is a carbohydrate that serves as an energy source for animals, but it's usually broken down into glucose soon after an animal's death. Therefore, you won't find significant amounts of glycogen in meats.

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In CF4, the hybridization of carbon is sp3. In H2CO, the hybridization of carbon is sp2. In CO2, the hybridization of carbon is sp. The hybridization of carbon in CF4, H2CO, and CO2 can be determined using the formula is given below.

Hybridization = Number of sigma bonds + Number of lone pairs on the central atom
For CF4, there are four sigma bonds between carbon and the four fluorine atoms. There are no lone pairs on the central carbon atom.  

For H2CO, there are three sigma bonds between carbon and the two hydrogen atoms and the oxygen atom. There is also one lone pair on the central carbon atom. Therefore, the hybridization of carbon in H2CO is sp2.For CO2, there are two sigma bonds between carbon and the two oxygen atoms. There are no lone pairs on the central carbon atom. Therefore, the hybridization of carbon in CO2 is sp.

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Calculated the built-in voltage, you can use it in the electron and hole concentration equations to obtain the values of n and p, respectively.

In an n-type silicon with steady-state illumination, the electron and hole concentrations can be calculated using the following equations:

1. Electron concentration:

n = ni² / N_A * exp(E_g / (k_B * T)) * (exp(q * V / (k_B * T)) - 1)

where

- ni is the intrinsic carrier concentration of silicon (approximately 1.45 x 10¹⁰ cm⁻³ at room temperature),

- N_A is the doping concentration of the n-type silicon,

- E_g is the energy gap of silicon (approximately 1.12 eV),

- k_B is the Boltzmann constant,

- T is the temperature in Kelvin,

- q is the electron charge,

- V is the voltage across the semiconductor.

2. Hole concentration:

p = ni² / N_A * exp(-E_g / (k_B * T)) * (exp(q * V / (k_B * T)) - 1)

where all the parameters are the same as in the electron concentration equation, except that p represents the hole concentration.

Note that the voltage V in both equations is the built-in voltage of the n-type semiconductor under illumination.

The value of the built-in voltage V can be calculated using the following equation:

V = (k_B * T / q) * ln(N_A / n_i)

where all the parameters are the same as in the electron concentration equation.

Calculated the built-in voltage, you can use it in the electron and hole concentration equations to obtain the values of n and p, respectively.

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The correct name for the compound [tex]FeSO_{4}-6H_{2}O[/tex] is Iron II Sulfate hexahydrate. This is because the compound contains iron in its +2 oxidation state (hence the "II" in the name), and the sulfate ion ([tex]SO_{4}[/tex]) has a -2 charge.

The "hexahydrate" part of the name indicates that there are six water molecules associated with each formula unit of the compound.

Therefore, the correct name for this compound is Iron II Sulfate hexahydrate, and this name accurately reflects its chemical composition.

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