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Goryan [66]
3 years ago
14

Both the esophagus and the small intestine are involved in the digestion of food. The esophagus squeezes food into the stomach b

y wave-like muscle contractions. Peptidase enzymes in the small intestine break food molecules into smaller molecules.
Which statement best describes changes to food during digestion?

A. The muscle contractions result in physical changes, while the action of the peptidase results in chemical changes.

B. The muscle contractions result in chemical changes, while the peptidase results in physical changes.

C. The muscle contractions and the action of the peptidase both result only in physical changes.

D. The muscle contractions and the action in the peptidase both result only in chemical changes.
Chemistry
1 answer:
UkoKoshka [18]3 years ago
3 0

Answer:

The answer is: A

Explanation:

The wave-like contractions are to be considered a physical change as they do not take part in the breaking-down of the food. The peptidase is a chemical change because it is used to chemically break down the food to retrieve nutrients and such.

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A sample of neon has a volume of 40.81 m3 at 23.5C. At what temperature, in Kelvins, would the gas occupy 50.00 cubic meters? As
mezya [45]

At  \fbox{\begin \\363 K \end{minispace}}  temperature, a sample of neon gas will occupy 50.00 \text{ m}^{3} volume.

Further Explanation:

The given problem is based on the concept of Charles’ law. Charles’ law states that “at constant pressure and fixed mass the volume occupied an ideal gas is directly proportional to the Kelvin temperature.”

Mathematically the law can be expressed as,

\fbox{ \begin \\ V \propto T \end{minispace}}

Or,

\frac{V}{T}=k

Here, <em>V</em> is the volume of the gas, <em>T</em> is Kelvin temperature, and <em>k</em> is proportionality constant.

Given information:

The initial volume of neon gas is 40.81 \text{ m}^{3} .

The final volume of neon gas is  50.00 \text{ m}^{3}.

The initial temperature value is 23.5 \text{ } ^{\circ} \text{C} .

To calculate:

The final temperature

Given Condition:

  • The pressure is constant.
  • Mass of gas is fixed.

Solution:

Step 1: Modify the mathematical expression for Charles’ law for two different temperature and volume values as follows:

\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

Here,

  • V_{1}is the initial volume of the gas.
  • V_{2} is the final volume of the gas.
  • T_{1} is the initial temperature of the gas.
  • T_{2} is the final temperature of the gas.

Step 2: Rearrange equation (2) for .

\fbox {\begin \\T_{2}=\frac{(V_{2}) \times (T_{1})}{V_{1}}\\\end{minispace}}                                                                  …… (2)

Step 3: Convert the given temperature  from degree Celsius to Kelvin.

The conversion factor to convert degree Celsius to Kelvin is,

T(\text{K}) = T(^{\circ}\text{C}) + 273.15                                      …… (3)

Substitute 23.5\text{ }^{\circ} \text{C} for T(^{\circ}\text{C})  in equation (3) to convert temperature from degree Celsius to Kelvin.

T(\text{K}) = 23.5 \text{ } ^{\circ} \text{C} + 273.15\\T(\text{K})= 296.65 \text{ K}

Step 4: Substitute 40.81 \text{ m}^{3}  for V_{1} ,  50.00 \text{ m}^{3} for V_{2}  and  296.65 \text{ K} for T_{1}  in equation (2) and calculate the value of T_{2} .

T_{2}=\frac{(50.00 \text{ m}^{3}) \times (296.65 \text{ K})}{40.81 \text{ m}^{3}}\\T_{2}=363.45 \text{ K}\\T_{2} \approx 363 \text{ K}

Important note:

  • The temperature must be in Kelvin.
  • The condition of fixed mass and fixed pressure must be fulfilled in order to apply Charles’ law.

Learn More:

1. Gas laws brainly.com/question/1403211

2. Application of Charles’ law brainly.com/question/7434588

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: States of matter

Keywords: neon, volume, occupies, temperature, Kelvin, degree Celsius, Charle’s law, constant pressure, fixed mass, 40.81 m^3 , 50.00 m^3 , 23.5 degree C , celsius , 363 K , sates of matter, initial volume, final volume, initial temperature, final temperature, V1 , V2 , T1 , T2 .

5 0
3 years ago
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Suppose that a dependent quantity increases when the independent quantity is decreased. This kind of proportional relationship b
Anna71 [15]
I thinks it’s not this article we should cajnges it
3 0
3 years ago
How long would it take the same amount of gaseous i2 to effuse from the same container under identical conditions?
77julia77 [94]

(46 sec) x √ ((253.80894 g I2/mol) / (44.0128 g N2O/mol)) = 110 sec

7 0
3 years ago
What mass of sulfur has to burn to produce 4.5L SO2 at 300°C and 101 kPa in the following reaction?s(s)+O2(g) SO,(g)
Alex_Xolod [135]
<span>Answer: option B. 3.07 g

Explanation:

1) given reaction:

S(s) + O₂ (g) → SO(g)

2) Balanced chemical equation:

</span><span>2S(s) + O₂ (g) → 2SO(g)

3) Theoretical mole ratios:

2 mol S : 1 mol O₂ : 2 mol SO

3) number of moles of 4.5 liter SO₂ at</span><span> 300°C and 101 kPa

use the ideal gas equation:

pV = nRT

with V = 4.5 liter
p = 101 kPa
T = 300 + 273.15 K = 573.15 K
R = 8.314 liter×kPa / (mol×K)

=> n = pV / (RT) =

n =  [101 kPa × 4.5 liter] / [8.314 (liter×kPa) / (mol×K)  × 573.15 K ]

n = 0.0954 mol SO

4) proportion with the theoretical ratio S / SO

 2 mol S                   x
-------------- = ----------------------
 2 mol SO      0.0954 mol SO

=> x = 0.0954 mol S.

5) Convert mol of S to grams by using atomic mass of S = 32.065 g/mol

mass = number of moles × atomic mass

mass = 0.0954 mol × 32.065 g/mol = 3.059 g of S

6) Therefore the answer is the option B. 3.07 g
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8 0
3 years ago
Assuming the reaction is at STP if 2.25 moles of Li are exposed to 3.25 moles of water, which is the limiting reactant?
crimeas [40]

Answer: Lithium

Explanation:

The ratio for the reaction of Li and water is the same, but there are more moles of water than lithium. Therefore, lithium is the limiting reactant.

5 0
3 years ago
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