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2 years ago

How long does vacuum sealed raw meat last in the freezer?

Chemistry

2 answers:

beks73 [17]2 years ago

6 0

Answer: Six months or three years

Explanation: Frozen meats, poultry, and fish can last in the freezer for up to six months and when they are vacuum sealed, frozen meats can stay fresh and free from freezer burn for up to 3 years.

Allushta [10]2 years ago

4 0

Answer:

Frozen Meats – Frozen raw meats that are properly vacuum sealed can be stored in the freezer from 1-3 years depending on the type of meat. However, raw meat that is not vacuum sealed will only last 1-12 months depending on the meat.

Explanation:

hop this helps

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Ozone gas is a form of elemental oxygen containing molecules with three oxygen atoms, O3. Ozone is produced from atmospheric oxy

adelina 88 [10]

<u>Answer:</u> The unbalanced chemical equation is written below

<u>Explanation:</u>

Unbalanced chemical equation does not follow law of conservation of mass.

In an unbalanced chemical equation, total number of individual atoms on the reactant side will not be equal to the total number of individual atoms on the product side.

The chemical equation for the formation of ozone gas from oxygen gas:

$O_2(g)\rightarrow O_3(g)$

Hence, the unbalanced chemical equation is written above.

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3 years ago

Read 2 more answers

In a galvanic cell, electrons are transferred from one half cell to the other as the redox reaction progresses. what happens in

Sidana [21]

The half cell in which the electrode gains electrons is where reduction occurs, and the half cell in which the electrode loses electrons is where oxidation occurs.

<h3><u>What is a Galvanic cell ?</u></h3>

Voltaic or galvanic cells are electrochemical devices that use spontaneous oxidation-reduction events to generate electricity. In order to balance the overall equation and highlight the actual chemical changes, it is frequently advantageous to divide the oxidation-reduction reactions into half-reactions while constructing the equations.

Two half-cells make up most electrochemical cells. The half-cells allow electricity to pass via an external wire by separating the oxidation half-reaction from the reduction half-reaction.

<h3><u>Oxidation:</u></h3>

The anode is located in one half-cell, which is often shown on the left side of a figure. On the anode, oxidation takes place. In the opposite half-cell, the anode and cathode are linked.

<h3><u>Reduction:</u></h3>

The second half-cell, cathode, which is frequently displayed on a figure's right side. The cathode is where reduction happens. The circuit is completed and current can flow by adding a salt bridge.

To know more about processes in Galvanic cell, refer to:

brainly.com/question/13031093

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7 0

1 year ago

Mi One should know the properties of the components of the mixture to separate it. Explain with an example.

Alekssandra [29.7K]

One should know that the properties of the components of the mixture.
Suppose a mixture contains Sodium or potassium .If you put them in open air it may catch fire.
Suppose a mixture has flammable components like kerosene,spirit .If you put them in fire it may harm you .

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3 years ago

Which gas law would apply to the following scenario.

chubhunter [2.5K]

The ansewer is Boyle’s law

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3 years ago

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0

2 years ago