Answer : Option D) far left.
Explanation : If an unknown element displays many extremely strong metallic properties, where is it most likely to be located on the far left of the modern periodic table.
In the modern periodic table, the metallic character of the elements decreases across the periods in the periodic table from left to right. This occurs because the elemental atoms more readily accept electrons to fill their valence shell than losing them to remove the unfilled shell. Also, the metallic character increases down the groups in the periodic table. So, it is clear that the elements found on the far left are having high metallic properties in them.
Answer:
306.43 K
Explanation:
- Use Charle's law and rearrange formula (V1/T1=V2/T2)
- Hope this helped! Let me know if you would like me to show you step-by-step how to do these types of problems.
About 1.051 moles (depends on how u round it)
Answer:
109.7178g of H2O
Explanation:
First let us generate a balanced equation for the reaction. This is illustrated below:
2C3H8O + 9O2 —> 6CO2 + 8H2O
Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:
Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.
Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g
Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol
Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g
From the equation,
120.19184g of C3H8O produced 144.12224g of H20.
Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O
<h3>
Answer:</h3>
128 g HCl
<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>
<u>Stoichiometry</u>
- Reaction Mole Ratios
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Unbalanced] Mg (s) + HCl (aq) → MgCl (aq) + H₂ (g)
↓
[RxN - Balanced] 2Mg (s) + 2HCl (aq) → 2MgCl (aq) + H₂ (g)
[Given] 3.25 mol Mg
[Solve] x g HCl
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Mg → 2 mol HCl
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of Cl - 35.45 g/mol
Molar Mass of HCl - 1.01 + 35.45 = 36.46 g/mol
<u>Step 3: Stoich</u>
- [S - DA] Set up:
- [S - DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
127.61 g HCl ≈ 128 g HCl
