Answer:
The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is
.
Explanation:
The combined gas equation is,
![\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
where,
= initial pressure of gas = 104 kPa
= final pressure of gas = 52 kPa
= initial volume of gas = ![2.0m^3](https://tex.z-dn.net/?f=2.0m%5E3)
= final volume of gas = ?
= initial temperature of gas = ![21.1^oC=273+21.1=294.1K](https://tex.z-dn.net/?f=21.1%5EoC%3D273%2B21.1%3D294.1K)
= final temperature of gas = ![-5.0^oC=273+(-5.0)=268 K](https://tex.z-dn.net/?f=-5.0%5EoC%3D273%2B%28-5.0%29%3D268%20K)
Now put all the given values in the above equation, we get:
![\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}](https://tex.z-dn.net/?f=%5Cfrac%7B104%20kPa%5Ctimes%202.0m%5E3%7D%7B294.1%20K%7D%3D%5Cfrac%7B52%20kPa%5Ctimes%20V_2%7D%7B268%20K%7D)
![V_2=3.64 m^3](https://tex.z-dn.net/?f=V_2%3D3.64%20m%5E3)
The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is
.
<h3>
Answer:</h3>
1.2 × 10⁻⁸ mol Pb
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 7.2 × 10¹⁵ atoms Pb
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:
![\displaystyle 7.2 \cdot 10^{15} \ atoms \ Pb(\frac{1 \ mol \ Pb}{6.022 \cdot 10^{23} \ atoms \ Pb})](https://tex.z-dn.net/?f=%5Cdisplaystyle%207.2%20%5Ccdot%2010%5E%7B15%7D%20%5C%20atoms%20%5C%20Pb%28%5Cfrac%7B1%20%5C%20mol%20%5C%20Pb%7D%7B6.022%20%5Ccdot%2010%5E%7B23%7D%20%5C%20atoms%20%5C%20Pb%7D%29)
- [DA] Multiply/Divide [Cancel out units]:
![\displaystyle 1.19562 \cdot 10^{-8} \ mol \ Pb](https://tex.z-dn.net/?f=%5Cdisplaystyle%201.19562%20%5Ccdot%2010%5E%7B-8%7D%20%5C%20mol%20%5C%20Pb)
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
1.19562 × 10⁻⁸ mol Pb ≈ 1.2 × 10⁻⁸ mol Pb
Using extension or flexible cords improperly
Explanation:
A client undergoing therapy in a whirlpool unit plugged into an extension cord demonstrates a wrong and improper use of extensions and flexible cords. This is not a safe way to connect an electrical circuit.
- Unless a Ground Fault Circuit Interrupter is used, the client is at a risk of suffering from electrocution.
- The Ground Fault Circuit Interrupter is able to detect leakages in the circuit and breaks it off.
learn more:
electric current brainly.com/question/4438943
#learnwithBrainly
Answer:
The value of the equilibrium constant for reaction asked is
.
Explanation:
![CO_2(g)\rightleftharpoons C(s)+O_2(g)](https://tex.z-dn.net/?f=CO_2%28g%29%5Crightleftharpoons%20C%28s%29%2BO_2%28g%29)
![K_{goal}=?](https://tex.z-dn.net/?f=K_%7Bgoal%7D%3D%3F)
![K_{goal}=\frac{[C][O_2]}{[CO_2]}](https://tex.z-dn.net/?f=K_%7Bgoal%7D%3D%5Cfrac%7B%5BC%5D%5BO_2%5D%7D%7B%5BCO_2%5D%7D)
..[1]
![K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BCH_3COOH%5D%5BO_2%5D%5E2%7D%7B%5BCO_2%5D%5E2%5BH_2O%5D%5E2%7D)
..[2]
![K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2%5D%5E2%5BO_2%5D%7D)
..[3]
![K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_3%3D%5Cfrac%7B%5BC%5D%5E2%5BH_2%5D%5E2%5BO_2%5D%7D%7B%5BCH_3COOH%5D%7D)
[1] + [2] + [3]
![2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)](https://tex.z-dn.net/?f=2CO_2%28g%29%5Crightleftharpoons%202C%28s%29%2B2O_2%28g%29)
( on adding the equilibrium constant will get multiplied with each other)
![K=K_1\times K_2\times K_3](https://tex.z-dn.net/?f=K%3DK_1%5Ctimes%20K_2%5Ctimes%20K_3)
![K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}](https://tex.z-dn.net/?f=K%3D5.40%5Ctimes%2010%5E%7B-16%7D%5Ctimes%201.06%5Ctimes%2010%5E%7B10%7D%5Ctimes%202.68%5Ctimes%2010%5E%7B-9%7D)
![K=1.53\times 10^{-14}](https://tex.z-dn.net/?f=K%3D1.53%5Ctimes%2010%5E%7B-14%7D)
![K=\frac{[C]^2[O_2]^2}{[CO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC%5D%5E2%5BO_2%5D%5E2%7D%7B%5BCO_2%5D%5E2%7D)
On comparing the K and
:
![K^2=K_{goal}](https://tex.z-dn.net/?f=K%5E2%3DK_%7Bgoal%7D)
![K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}](https://tex.z-dn.net/?f=K_%7Bgoal%7D%3D%5Csqrt%7BK%7D%3D%5Csqrt%7B1.53%5Ctimes%2010%5E%7B-14%7D%7D%3D1.24%5Ctimes%2010%5E%7B-7%7D)
The value of the equilibrium constant for reaction asked is
.