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4 years ago

What is the chemical composition for diet sodas?

1 answer:

5 0

Well here are the Main Chemicals, Compounds and Components

- the two main components are carbonated water and caramel color
- molecular formula for carbonated water is H2CO3
- Molecular formula for caramel color is C18H9N Na2O8S2
- Carbonated water is water that contains carbon dioxide gas, the gas causes there to be small bubbles in the water
- Caramel color is the most consumed food coloring ingredient in the world, it results from controlled heat treatments of carbohydrates

You might be interested in

Calculate the pH of a 0.2 M * 4 solution for which Kb = 1.8*10^-5 at 26 c . The equation for the reaction

lara31 [8.8K]

<u>Answer:</u> The pH of the solution is 11.24

<u>Explanation:</u>

We are given:

Molarity of ammonia = 0.2 M

$K_b=1.8\times 10^{-5}$

The given chemical equation follows:

$NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

I: 0.2

C: -x +x +x

E: 0.2-x x x

The expression for equilibrium constant follows:

$K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}$

Putting values in above expression, we get:

$1.8\times 10^{-5}=\frac{x^2}{0.2-x}\\\\1.8\times 10^{-5}(0.2-x)=x^2\\\\x^2+(1.8\times 10^{-5}x)-(0.36\times 10^{-5})=0\\\\x=1.88\times 10^{-3}, 1.9\times 10^{-3}$

Neglecting the negative value of x as concentration cannot be negative.

So, $[OH^-]=x=1.88\times 10^{-3}M$

pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution.

$pOH=-\log [OH^-]$

Putting values in above equation, we get:

$pOH=-\log (1.88\times 10^{-3})\\\\pOH=2.76$

We know:

$pH+pOH=14\\\\pH=14-2.76\\\\pH=11.24$

Hence, the pH of the solution is 11.24

6 0

3 years ago

A sample of argon gas is at a pressure of 1.725x10^4 kPa and a temperature of 24°C in a rigid 20 L tank. How many moles of argon

Paraphin [41]

To find how many moles of the gas you can use ideal gas formula. Remember to change the temperature unit to Kelvin. Since the pressure using kpa, the constant used would be 8.314 kpa*L / mol*K
PV=nRT
n= PV/RT
n= 17,250 kpa * 20 / 8.314 * (24+273.15)K
n=139.64moles

4 0

4 years ago

The diagram and data below represent gas and the conditions of pressure, volume, and temperature of the gas in a rigid cylinder

Sedaia [141]

Answer:

2.3 L

Explanation:

when you have this eqaution you use P1V1T2=P2V2T1.

since you are looking for volume of gas cylinder that means you are looking for V2.

so you have your given P= 1.0 atm, V=2.5L, T=298K.

They ask you to use STP or standard pressure and temperature.

STP P= 1atm, STP T=273 K

so you are looking for V2 meaning you are going to get V alone so your new equation is P1*V1*TS/ P2* T1= V2

so you now substitue so its now 1*273*2.5/ 1* 298= V

682.5/ 298= V

Lastly V= 2.2902 and you round to tenths so now you have

V= 2.3 L

5 0

3 years ago

A 17.0 L helium tank is pressurized to 22.0 atm. When connected to this tank, a balloon will inflate because the pressure inside

mel-nik [20]

Answer:

The volume of the balloon would be 374 L.

Explanation:

Initially, He occuplies a volume of 17.0 L (V₁) at a pressure of 22.0 atm (P₁). If pressure decreases to 1.00 atm (P₂), volume is expected to increase. Assuming ideal behavior of the gas and constant temperature, we can calculate the final volume (V₂) using Boyle's Law.

$P_{1} \times V_{1} = P_{2} \times V_{2}\\V_{2} = \frac{P_{1} \times V_{1}}{P_{2}} =\frac{22.0atm \times 17.0L}{1.00atm} = 374 L$

8 0

4 years ago

a. If 42.5 g of CH3OH reacts with 22.8 L of O2 at 27°C and a pressure of 2.00 atm, calculate the number of grams of water vapor

Korvikt [17]

Answer:

The mass of water vapor is 44.46 grams

The volume of water is 30.37 L

Explanation:

Step 1: Data given

Mass of CH3OH =42.5 grams

Molar mass CH3OH = 32.04 g/mol

Volume of O2 = 22.8 L

Pressure = 2.00 atm

Step 2: The balanced equation

2CH3OH + 3O2 → 2CO2 + 4H2O

Step 3: Calculate moles CH3OH

Moles CH3OH = mass CH3OH / molar mass CH3OH

Moles CH3OH = 42.5 grams / 32.04 g/mol

Moles CH3OH = 1.326 moles

Step 4: Calculate moles O2

p*V = n*R*T

⇒with p = the pressure = 2.00 atm

⇒with V = the volume of O2 = 22.8 L

⇒with n = the moles of O2 = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature = 27 °C = 300 K

n = (p*V) / (R*T)

n = (2.00 * 22.8) / (0.08206*300)

n = 1.85 moles

Step 5: Calculate the limiting reactant

For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2 and 4 H2O

O2 is the limiting reactant. It will completely be consumed ( 1.85 moles). CH3OH is in excess. There will react 2/3*1.85 = 1.233 moles. There will remain 1.326 - 1.233 = 0.093 moles

Step 6: Calculate moles products

For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2 and 4 H2O

For 1.85 moles O2 we'll have 1.233 moles CO2 and 2.467 moles H2O

Step 7: Calculate mass H2O

Mass H2O = moles H2O * molar mass H2O

Mass H2O = 2.467 moles * 18.02 g/mol

Mass H2O = 44.46 grams

Step 8: Calculate volume H2O

p*V = n*R*T

⇒with p = the pressure = 2.00 atm

⇒with V = the volume of H2O = TO BE DETERMINED

⇒with n = the moles of H2O = 2.467 moles

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature = 27 °C = 300 K

V = (n*R*T)/p

V = (2.467 * 0.08206 * 300) / 2.00

V = 30.37 L

The mass of water vapor is 44.46 grams

The volume of water is 30.37 L

3 0

3 years ago