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Mumz [18]
3 years ago
5

Some studies have found that when a person mimics an angry expression, their temperature actually rises.

Chemistry
1 answer:
posledela3 years ago
5 0
The correct answer is A) True
You might be interested in
A 50 L cylinder is filled with argon gas to a pressure of 10130.0 kPa at 300°C. How many moles of argon gas does the cylinder co
Vlad1618 [11]
To answer this question, we will use the general gas law which states that:
PV = nRT where:
P is the pressure of the gas = <span>10130.0 kPa
</span>V is the volume of the gas = 50 liters
n is the number of moles that we want to calculate
R is the gas constant = <span>8.314 L∙kPa/K∙mol
T is the temperature = 300+273 = 573 degree kelvin

Substitute with the givens in the equation to get the number of moles as follows:
</span><span>10130 * 50 = n * 8.314 * 573
506500 = 4763.922 n
n = </span>506500 / 4763.922
n = 106.3199 moles
7 0
3 years ago
At standard temperature and pressure. 0.500 mole of xenon gas occupies
ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Mole ratio
  • Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

0.500 mole Xe (g)

<u>Step 2: Convert</u>

  1. [DA] Set up:                                                                                                  \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)
  2. [DA] Evaluate:                                                                                               \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}

Topic: AP Chemistry

Unit: Stoichiometry

3 0
1 year ago
The reaction is first order in cyclopropane and has a measured rate constant of k=3.36×10−5 s−1 at 720 k. if the initial cyclopr
sergey [27]

Answer:

0.0277 M.

Explanation:

The integral rate law of a first order reaction:

<em>Kt = ln ([A₀]/[A]),</em>

where, k is the rate constant of the reaction <em>(k = 3.36 × 10⁻⁵ s⁻¹)</em>,

t is the time of the reaction <em>(t = 235.0 min = 14100 s)</em>,

[A₀] is the initial concentration of cyclopropane <em>([A₀] = 0.0445 M)</em>

<em>∵ Kt = ln ([A₀]/[A]),</em>

∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]

Taking the exponential of both sides:

1.6 = (0.0445 M)/[A]

<em>∴ [A] = (0.0445 M)/1.6 = 0.0277 M.</em>

<em />

6 0
2 years ago
Will chlorine gas effuse more quickly than ammonia gas
beks73 [17]
This statement is false due to the fact that the ammonia gas has the lower molar mass.
6 0
3 years ago
Read 2 more answers
Which shows an isomer of the molecule below?
kobusy [5.1K]

Answer:

B

Explanation:

3 0
2 years ago
Read 2 more answers
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