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3 years ago

Guyz help me !!! what is pasteurization please don't copy from google

Chemistry

1 answer:

Svet_ta [14]3 years ago

5 0

Answer:

<u><em>Pasteurization:</em></u>

It is the process in which any beverage such as milk or any other liquid is heated at a certain temperature to kill the bacteria and microorganisms that can cause spoilage or fermentation or any other disease.

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Bismuth-210 has a half life of 5.0 days if you had a sample of 60.00g of Bismuth-210 how much would be left after 15 days? (Show

Zielflug [23.3K]

Answer:

7.5 gm left

Explanation:

Bismuth-210 has a half life of 5 days

15 days is 15/5 = 3 half lives

since half the amount is left in 5 days or 1 half life

(1/2) x (1/2) x (1/2) the staring amount would be left in

3 half lives. so 1/8 is left

(1/8) x 60.0 = 7.5 gm left

5 0

3 years ago

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. Cac

Tanya [424]

Answer:

The equation that gives the overall equilibrium in terms of the equilibrium constants K and Ky is K1 = K^6 * Ky

Explanation:

we have the following balanced reaction:

CaC2 + 2H2O = C2H2 + Ca(OH)2

the value of K for this reaction will be equal to:

K = ([C2H2] * [Ca(OH)2])/([CaC2] * [H2O]^2)

if we multiply the reaction by the value of 6, we have:

6CaC2 + 12H2O = 6C2H2 + 6Ca(OH)2

Again, the value of K for this reaction will be equal to:

K,´ = ([C2H2] ^6 * [Ca(OH)2]^6)/([CaC2]^6 * [H2O]^12) = K^6

For the second reaction:

6C2H2 + 3CO2 + 4H2O = 5CH2CHCO2H

The value of K for this reaction:

K2 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^4)

we also have:

K1 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^16)

Thus:

K1 = K^6 * Ky

4 0

3 years ago

Does liquid evaporate with adding heat or losing it

Marysya12 [62]

Adding because evaporation speed is increased with increased temperature

6 0

3 years ago

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is: