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3 years ago

Oil and fat containing food items are flushed with nitrogen. Why?

Chemistry

2 answers:

Vesnalui [34]3 years ago

6 0

Answer:

Nitrogen is used to flush the foods because it does not interact with the foods and the oils and fats within them.

Explanation:

Nitrogen serves as an antioxidant to remove oxygen from the food and prevent it from being oxidized. Typically, it is also used to keep the fats and oils from binding with it and ruining the integrity and structure of the food.

Andreas93 [3]3 years ago

3 0

Answer:

oil and fat containing food terms items flushed with nitrogen because nitrogen acts as an antioxidant and it prevents them from being oxidised.

Explanation:

Nitrogen is an inert gas that is flushed in food items as it does not react with food items and removes all the oxygen present in food. Oxygen present in food products is then rancid by the presence of oils and fats. Nitrogen is chemically non-reactive with oils and fats and therefore preserves oily food products.

hope this helped!

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1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

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3 years ago

7. Calculate the amount of energy required heat 100.g to H2O(s) changes to H2O(l) at 0°C

Brrunno [24]

Answer:

33300J

Explanation:

Given parameters:

Mass of ice = 100g

Unknown:

Amount of energy = ?

Solution:

This is a phase change process from solid to liquid. In this case, the latent heat of melting of ice is 3.33 x 10⁵ J/kg.

So;

H = mL

m is the mass

L is the latent heat of melting ice

Now, insert the parameters and solve;

H = mL

mass from gram to kilogram;

100g gives 0.1kg

H = 0.1 x 3.33 x 10⁵ = 33300J

8 0

3 years ago

A container with 3.0 moles of gas has a volume of 60.0L with a temperature at 400.K what is the pressure

iris [78.8K]

Answer: P= 1.64 atm

Explanation: solution attached.

Use Ideal gas law

PV= nRT

Derive for P

P= nRT/V R= 0.08205 L.atm/mol.K

Substitute the values.

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3 years ago

A breeder nuclear reactor is a reactor in which nonfissile U-238 is converted into fissile Pu-239. The process involves bombardm

VARVARA [1.3K]

Answer: The nuclear equations for the given process is written below.

Explanation:

The chemical equation for the bombardment of neutron to U-238 isotope follows:

$_{92}^{238}\textrm{U}+n\rightarrow _{92}^{239}\textrm{U}$

Beta decay is defined as the process in which neutrons get converted into an electron and a proton. The released electron is known as the beta particle.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$