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Aleks04 [339]
2 years ago
13

When you exercise strenuously, your body produces excess heat. Describe what your body does to help prevent your temperature fro

m rising excessively and
explain why your body's response effective.
Chemistry
2 answers:
postnew [5]2 years ago
8 0

Answer:

you sweat which makes your body cool down so that you don't overheat and faint.

Explanation:

VARVARA [1.3K]2 years ago
7 0

Answer:

Your body can cool itself by sweating. When sweat evaporates, it lowers your temperature

Explanation:

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A sample of Neon is in a sealed container held under isothermic conditions. The initial pressure and volume are 2.7 atm and 4.5
Marina CMI [18]

Answer:

The final volume in mL is 7.14 mL or 7.1 mL.

Explanation:

1.Use Boyle's Law(P_{1} V_{1}= P_{2} V_{2}). Re-arrange to solve for V_{2}<em> for the final volume.</em>

<em />

<em>2. Plug in values. </em>V_{2} =\frac{(2.7 atm)(4.5 mL)}{(1.7 atm)}  = 7.14 mL

3 0
2 years ago
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
3 years ago
The beakers in the picture above are labeled according to the ionic solutions they contain. What metal or metals could replace c
tangare [24]

Explanation:

When a metal replaces another metal in solution, we say such a reaction has undergone a single displacement reaction.

In such a reaction, metal higher up in the activity series replaces another one due to their position.

To known the metal or metals that will replace the given copper, we need to reference the activity series of metals.

Every metal higher than copper in the series will displace copper from the solution.

So, there metals are: potassium, sodium, lithium, barium, strontium etc.

8 0
3 years ago
For the reaction below, initially the partial pressure of all 3 gases is 1.0atm. . 2NH3(g)--&gt; N2(g) + 3H2(g) K, 0.83 1. When
erma4kov [3.2K]

Answer:

The reaction would shift toward the reactants

When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm

Explanation:

For the reaction:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

Where K is defined as:

K = \frac{P_{N_{2}}*P_{H_2}^3}{P_{NH_3}^2} = 0.83

As initial pressures of all 3 gases is 1.0atm, reaction quotient, Q, is:

Q = \frac{1atm*{1atm}^3}{1atm^2} = 1

As Q > K, <em>the reaction will produce more NH₃ until Q = K consuming N₂ and H₂.</em>

Thus, there are true:

<h3>The reaction would shift toward the reactants</h3><h3>When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm</h3>

<em />

4 0
3 years ago
A 5.00 L sample of air at 0 C is warmed to 100.0 C. What is the new volume of the air? First, identify V1.
Paladinen [302]

Answer : The new volume of the air is, 6.83 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=5.00L\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=100^oC=(100+273)K=373K

Putting values in above equation, we get:

\frac{5.00L}{273K}=\frac{V_2}{373K}\\\\V_2=6.83L

Therefore, the new volume of the air is, 6.83 L

6 0
4 years ago
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