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2 months ago

What is the best alternative to safely cooling foods if you do not have a blast chiller?

Chemistry

1 answer:

VMariaS [2.9K]2 months ago

4 0

Answer:

Store it in the refrigerator

Explanation:

A refrigerator keeps food and beverages cool within a household. Storing food that is not intended for immediate consumption in the refrigerator prevents it from spoiling.

It’s important to understand that higher temperatures accelerate the growth of microbes that lead to spoilage. Conversely, lower temperatures significantly slow down microbial activity, which helps prevent food deterioration. Refrigeration effectively cools food items, thereby reducing the risk of spoilage.

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Two hypothetical ionic compounds are discovered with the chemical formulas XCl2 and YCl2, where X and Y represent symbols of the

Tems11 [2777]

Answer:

THE MOLAR MASS OF XCL2 IS 400 g/mol

THE MOLAR MASS OF YCL2 IS 250 g/mol.

Explanation:

We derive the molar mass of XCl2 and YCl2 by recalling the molar mass formula when both mass and the number of moles are known.

Number of moles = mass / molar mass

Molar mass = mass / number of moles.

For XCl2,

mass = 100 g

number of moles = 0.25 mol

Thus, molar mass = mass / number of moles

Molar mass = 100 g / 0.25 mol

Molar mass = 400 g/mol.

For YCl2,

mass = 125 g

number of moles = 0.50 mol

Molar mass = 125 g / 0.50 mol

Molar mass = 250 g/mol.

Accordingly, the molar masses for XCl2 and YCl2 are 400 g/mol and 250 g/mol, respectively.

3 0

1 month ago

in a typical person, the level of glucose is about 85 mg/ 100 mL of blood. if an average body contains about 11 pt of blood, how

Anarel [2989]

It's important to remember that 1 pint equals 473.1765 mL, therefore 11 pints amounts to 5204.9415 mL.

We can formulate a proportion based on the problem statement

(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose

5 0

15 days ago

Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

eduard [2782]

Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Starting temperature of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Next, determine the entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, the entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Thus, the total entropy is the sum of the entropy changes of both blocks.

= -554.12 J/K + 647.49 J/K $\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

6 0

27 days ago

For the reaction NH4Cl (s)→NH3 (g) + HCl (g) at 25°C, ΔH = 176 kJ/moland ΔS = 0.285 kJ/(mol - K).

castortr0y [3046]

C: 91 kJ/mol; no.

4 0

18 days ago

Put the following elements into five pairs of elements that have similar chemical reactivity: F, Sr, P, Ca, O, Br, Rb, Sb, Li, S

Anarel [2989]

Explanation:

Elements provided:

F, Sr, P, Ca, O, Br, Rb, Sb, Li, S

Elements sharing similar reactivity belong to the same group in the periodic table, indicating that those in the same column exhibit comparable reactivity. Here are the identified groupings:

Li and Rb are alkali metals in group 1

Ca and Sr are alkaline earth metals in group 2

F and Br are halogens in group 7

O and S belong to group 6

P and Sb are classified in group 5 of the periodic table

Thus, these classifications illustrate elements with the same chemical characteristics.

7 0

1 month ago