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Nuetrik [128]
3 years ago
9

Which of the following conditions would speed up the rate of decomposition? (Select

Chemistry
1 answer:
Anna11 [10]3 years ago
4 0

Answer:

E and B

Explanation:

the warmer the weather the faster it will decompose, this why is why sometimes bodies last longer frozen, or in cool temperatures.

that statement takes a out.

An illness shouldn't effect a decomposing body

Weight also shouldn't effect a decomposing body

Being buried can make bodies decompose fast because of all the bugs and other animals that will get down there and decompose it faster

Answer: B,E

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A direct relationship between 2 quantities or variables may be a description of a specific approach that the 2 quantities or variables act with each other. once a amount or variable encompasses a direct relationship with another amount or variable, we are able to describe it employing a mathematical expression.

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Which statement is correct about the law of conversation of energy
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Which orbitals form a pi bond? A. The s orbital and three p orbitals B. The s orbital and two p orbitals C. Overlapping p orbita
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C.  overlapping  p orbitals.

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3 years ago
The instruction booklet for your pressure cooker indicates that its highest setting is 12.8 psi . You know that standard atmosph
lubasha [3.4K]

Answer:At 237.57^oC food will cook in the pressure cooker set on high.

Explanation:

Standard atmospheric pressure P_1= 14.7 Psi = 1.0001 atm (1 Psi=0.06804 atm)

Standard temperatureT_1 = 273.15 K

Highest pressure offered by pressure cooker: 12.8 Psi = 0.8709 atm

Pressure (P_2) inside the pressure cooker on the highest setting at temperatureT_2 :

P_2 =  1.0001 atm + 0.8709 atm = 1.8710 atm

According to Gay lussac law  :

(pressure)\propto (temperature) (at constant Volume)

\frac{P_1}{T_1}=\frac{P_2}{T_2}

T_2=\frac{P_2\times T_1}{P_1}=\frac{1.8710 atm\times 273.15 K}{1.0001 atm}=510.73 K= 237.57^oC (T(^oC)=T-273.15 K)

At 237.57^oC  food will cook in the pressure cooker set on high.

8 0
4 years ago
Read 2 more answers
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