Answer:
1) Maximun ammount of nitrogen gas:
2) Limiting reagent:
3) Ammount of excess reagent:
Explanation:
<u>The reaction </u>
Moles of nitrogen monoxide
Molecular weight:
Moles of hydrogen
Molecular weight:
Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess
1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted
2) <u>Limiting reagent</u>:
3) <u>Ammount of excess reagent</u>:
Answer:
8.13 ×10²³ atoms
Explanation:
Given data:
Mass of magnesium = 32.45 g
Number of atoms = ?
Solution:
Number of moles of Mg:
Number of moles = mass/molar mass
Number of moles = 32.45 g/ 24 g/mol
Number of moles = 1.35 mol
Number of atoms:
1 mole contain 6.022×10²³ atoms
1.35 mol × 6.022×10²³ atoms/ 1mol
8.13 ×10²³ atoms
Answer:
Except for mature red blood cells, all human cells contain a complete genome. DNA in the human genome is arranged into 24 distinct chromosomes—physically separate molecules that range in length from about 50 mil- lion to 250 million base pairs.
Answer:
Fluorine
General Formulas and Concepts:
<u>Chemistry</u>
- Reading a Periodic Table
- Periodic Trends
- Electronegativity - the tendency for an element to attract an electron to itself
- Z-effective and Coulomb's Law, Forces of Attraction
Explanation:
The Periodic Trend for Electronegativity is up and to the right of the Periodic Table.
Fluorine is Element 9 and has 9 protons. Radium is Element 88 and has 88 protons. Therefore, Radium has a bigger Zeff than Flourine.
However, since Radium is in Period 7 while Fluorine is in Period 2, Radium has more core e⁻ than Fluorine does. This will create a much larger shielding effect, causing Radium's outermost e⁻ to have less FOA between them. Fluorine, since it has less core e⁻, the FOA between the nucleus and outershell e⁻ will be much stronger.
Therefore, Fluorine would attract an electron more than Radium, thus bringing us to the conclusion that Fluorine has a higher electronegativity.
Answer:
KBr is limiting reactant.
Explanation:
Given data:
Mass of KBr =4g
Mass of Cl₂ = 6 g
Limiting reactant = ?
Solution:
Chemical equation:
2KBr + Cl₂ → 2KCl + Br₂
Number of moles of KBr:
Number of moles = mass/molar mass
Number of moles = 4 g/ 119 gmol
Number of moles = 0.03 mol
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 6 g/ 70 gmol
Number of moles = 0.09 mol
Now we will compare the moles of reactant with product.
KBr : KCl
2 : 2
0.03 : 0.03
KBr : Br₂
2 : 1
0.03 : 1/2×0.03= 0.015
Cl₂ : KCl
1 : 2
0.09 : 2/1×0.09 = 0.18
Cl₂ : Br₂
1 : 1
0.09 : 0.09
Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂ is present in excess.