V1/T1=V2/T2
V2=(V1)(T2)/T1
Plug in values given (for the temp you can either turn 300K to 27°C or turn 132°C into kelvin
V2= 4400 mL= 4.4L
Element at Extreme Left In Periodic Table:
The elements of Group I-A (1) are present at extreme left of the periodic table. They are called as Alkali Metals. Alkali Metals are strong metals. These elements can easily loose their valence electron. The valence shell electronic configuration of these elements is,
ns¹
where n is principle quantum number, which shows main energy level or shell. These metals can gain Noble gas configuration (stable configuration) either by loosing one electron or by gaining seven or more electrons. As it is quite reasonable to loose one electron instead of gaining seven or more electrons so these element easily loose one electron to gain noble as configuration. The Metallic character decreases along the period from left to right. So Group II-A (2) are second most metallic elements and so on. These metals at extreme left mainly exist in solid form.
Element at Extreme Right In Periodic Table:
Elements present at extreme right of the periodic table lacks the properties of metallic character and act as non-Metals. They have almost complete outermost shell or have the deficiency of one or two electrons. They are not as hard as metallic elements and they exist with complete octet like in Noble gases, or deficient with one electron (Halogens) or two electrons (oxygen group). These elements tend to gain or accept electron if their valence shell is deficient with required number of elements. Like the valence electronic configuration of Halogens is,
ns², np⁵
So, Halogens readily accept one electron and attain noble gas configuration. Elements at extreme left exist mainly in gas phase.
Missing question:
Nitrogen: <span>2.0 L; </span>1.0 atm; 25°<span>C.
Oxygen: 3</span>.0 L; 2.0 atm; 25°C.
<span>When the valve between the two containers is opened, nitrogen gas moves from one container to another container and gases are mixed together, total volume of nitrogen is than:
V(nitrogen) = 2,0 L + 3,0 L = 5,0 L.</span>
<h3>
Answer:</h3>
3.01 × 10²⁵ molecules H₂O
<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>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
50.0 mol H₂O
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
<u />
= 3.011 × 10²⁵ molecules H₂O
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
3.011 × 10²⁵ molecules H₂O ≈ 3.01 × 10²⁵ molecules H₂O
