# Calculate the percent saturated fat in the total fat in butter

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50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH solution is added, the pH in the titration flask will be a. 2.17
b. 3.35
c. 2.41
d. 1.48
e. 7.00

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid])     (Eq. 01)

Where

pKa = -log(Ka)        (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2    (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles       (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

The pH value in the titration flask after 25.00 mL of the 0.10 M KOH solution is added to 50.00 mL of 0.10 M HNO2 solution is 3.35.

### Explanation:

The subject of this question is titration, which is a method used in chemistry to measure the concentration of an unknown solution. Given 50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 10-4), titrated with 0.10 M KOH (potassium hydroxide), we need to calculate the pH after 25.00 mL of the KOH solution is added.

First, we need to find the moles of the HNO2 and the KOH. Moles equals Molarity times Volume. So, for HNO2, it is 0.10 M * 0.050 L which equals 0.005 moles. For KOH, it is 0.10 M * 0.025 L which equals 0.0025 moles.

Then, subtract the moles of OH- from the moles of HNO2 to determine the concentration of HNO2 left, which is 0.005 moles - 0.0025 moles = 0.0025 moles. Divide this by the total volume of the solution (50.00 mL + 25.00 mL = 75.00 mL or 0.075 L to determine the new concentration of HNO2, 0.0025 moles / 0.075 L = 0.033 M. Then use the given Ka value with the equation [H+] = sqrt(Ka * [HNO2]) to get [H+].

To find acids' pH, we use the formula pH = -log[H+]. Use the calculated [H+] to find the pH.

Upon performing these calculations, the resulting pH value should be approximately 3.35 after 25.00 mL of the KOH solution is added, so the answer is (b) 3.35.

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Name the following : [Ni(NH3) 4(H2O)2(NO3)2​

Any member of the family of chemicals known as coordination compounds has a core metal atom that is surrounded by nonmetal atoms or groups of atoms, known as ligands, that are connected to it by chemical bonds. The name of the compound is tetraaminodiaquanickel (II)nitrate.

The additional molecular compounds known as coordination compounds are those that are stable in both the solid and dissolved states. In these compounds, ions or molecules connected by coordinate bonds connect the main metal atom or ion.

Coordination compounds are used in both vital catalytic processes that lead to the polymerization of organic molecules like polyethylene and polypropylene as well as hydrometallurgical processes that remove metals like nickel, cobalt, and copper from their ores.

To know more about Coordination compounds, visit;

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What happens if more product is added to a system at equilibrium

More reactants will be produced

Explanation:

Le Chatelier's principle; adding additional product or reactant will move the equilibrium left or right to compensate and come back to equilibrium

By adding more product to your system at equilibrium, the equilibrium will shift towards reactants, more reactants will be produced

Answer: Liquid molecules forming a gas and gas molecules forming a liquid are equal in number

Explanation: :/

How much heat energy would be needed to raise the temperature of a 223 g sample of aluminum [(C=0.895 Jig Cy from 22.5°C to 55 0°C? Η Ο Ο Ο ΟΟ 10x10) not enough information given Prov 40 25 11 Next >

Answer : The heat energy needed would be, 6486.5125 J

Explanation :

To calculate the change in temperature, we use the equation:

where,

q = heat needed = ?

m = mass of aluminum = 223 g

c = specific heat capacity of aluminum =

= change in temperature

= initial temperature =

= final temperature =

Putting values in above equation, we get:

Therefore, the heat energy needed would be, 6486.5125 J

A weather balloon is inflated to a volume of 27.3 L at a pressure of 738 mmHg and a temperature of 26.9 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 375 mmHg and the temperature is -15.6 ∘C.Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

The new volume of the balloon is 46.1 L

Explanation:

Step 1: Data given

Initial volume of the balloon = 27.3 L

Initial pressure in the balloon = 738 mmHg = 0.97105 atm

Initial temperature in the balloon = 26.9 °C = 300.05 K

The pressure decreases to 375 mmHg = 0.493421 atm

The temperature lowers to -15.6 °C = 257.55 K

Step 2: Calculate the volume

P1*V1 / T1 = P2 *V2 / T2

⇒with P1 = the Initial pressure = 738 mmHg = 0.97105 atm

⇒with V1 = Initial volume of the balloon = 27.3 L

⇒with T1 = Initial temperature in the balloon = 26.9 °C = 300.05 K

⇒with P2 = the decreased pressure = 0.493421 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the lowered temperature = 257.55 K

0.97105 * 27.3 / 300.05 = 0.493421*V2 / 257.55

V2 = 46.1 L

The new volume of the balloon is 46.1 L

Lithium (Li) has a charge of +1, and oxygen has a charge of -2. Which is the chemical formula?

Considering the formation of a chemical formula, the chemical formula is Li₂O.

### Ionic compounds

Cations (positivelycharged ions) and anions (negatively charged ions) combine to form ionic compounds, which must be electrically neutral. Therefore, the cations and anions must combine in such a way that the net charge contributed by the total number of cations exactly cancels the net charge contributed by the total number of anions.

### Formation of a chemical formula

To form the chemical formula:

• It is first necessary to place the chemical symbol of the element that is a cation, and then the negative part (anion).
• Since the net charge of the compound formed must be zero (that is, add zero), when cations and anions with different charges are joined, it is necessary to cross them and place them as subscripts of the other element. In this way, balanced loads are obtained, that is, the algebraic sum is equal to zero.

### Chemical formula in this case

Lithium (Li) has a charge of +1, and oxygen has a charge of -2. Taking into account the above, the chemical formula is Li₂O.