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2 years ago

How does body temperature, energy level, and water level affect a person’s ability to survive?

Chemistry

1 answer:

kifflom [539]2 years ago

4 0

U won’t get hyperthermia or a heat stroke when ur hot u use up more energy but when ur cold u use up less

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A balloon at 25°C has 30 L. What will the balloon's volume at 35°C?

anygoal [31]

The balloon's volume at 35°C : V₂=31.01 L

<h3>Further explanation</h3>

Given

T₁ = 25°C+273 = 298 K

V₁ = 30 L

T₂ = 35 °C + 273 = 308 K

Required

The new volume (V₂)

Solution

Charles's Law

When the gas pressure is kept constant, the gas volume is proportional to the temperature

$\tt \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

Input the value :

V₂=(V₁.T₂)/T₁

V₂=(30 x 308)/298

V₂=31.01 L

8 0

2 years ago

How should data be organized?

natima [27]

It should be organized in people

6 0

2 years ago

Read 2 more answers

Consider the following reaction at constant pressure. Use the information provided below to determine the value of ΔS at 473 K.

Elis [28]

Answer:

The reaction will be spontaneous

Explanation:

To determine if the reaction will be spontaneous or not at this temperature, we need to calculate the Gibbs's energy using the following formula:

$\Delta G= \Delta H - T * \Delta S$

<u>If the Gibbs's energy is negative, the reaction will be spontaneous, but if it's positive it will not.</u>

Calculating the $\Delta G= -1267 - 473 K* \Delta S$ :

$\Delta G= -1267 - 473 K* \Delta S$

Now, other factor we need to determine is the sign of the S variation. When talking about gases, the more moles you have in your system the more enthropic it is.

In this reaction you go from 7 moles to 8 moles of gas, so you can say that you are going from one enthropy to another higher than the first one. This results in: $\Delta S>0[/tex}Back to this expression: [tex]\Delta G= -1267 - 473 K* \Delta S$

If the variation of S is positive, the Gibbs's energy will be negative always and the reaction will be spontaneous.

4 0

3 years ago

A sample of gas has an initial pressure of 1.5 atm, an initial volume of 3.0 L, and an initial temperature of 293K. If the final

Nataliya [291]

Answer:

1.9 L

Explanation:

Step 1: Given data

Initial pressure (P₁): 1.5 atm

Initial volume (V₁): 3.0 L

Initial temperature (T₁): 293 K

Final pressure (P₂): 2.5 atm

Final volume (V₂): ?

Final temperature (T₂): 303 K

Step 2: Calculate the final volume of the gas

If we assume ideal behavior, we can calculate the final volume of the gas using the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

V₂ = P₁ × V₁ × T₂ / T₁ × P₂

V₂ = 1.5 atm × 3.0 L × 303 K / 293 K × 2.5 atm = 1.9 L

6 0

2 years ago

A gas with a volume of 4.0 L at a pressure of 2.02 atm is allowed to expand to a volume of 12.0

Maurinko [17]

<h3>Answer:</h3>

P₂ = 0.67 atm

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

Boyle's Law: P₁V₁ = P₂V₂

P₁ is pressure 1
V₁ is volume 1
P₂ is pressure 2
V₂ is volume 2

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

<u>Step 2: Solve</u>

Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
[Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
[Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
[Pressure] Rewrite: P₂ = 0.673333 atm

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>

0.673333 atm ≈ 0.67 atm

4 0

2 years ago