Answer:
d= 14.007 amu
Explanation:
Abundance of N¹⁴ = 99.63%
Abundance of N¹⁵ = 0.37%
Atomic mass of N¹⁴ = 14.003 amu
Atomic mass of N¹⁵ = 15.000 amu
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (14.003 × 99.63)+(15.000× 0.37) /100
Average atomic mass = 1395.12 + 5.55 / 100
Average atomic mass = 1400.67/ 100
Average atomic mass = 14.007 amu.
Answer:
12:7
Explanation:
To Determine: To calculate the formula of the unknown binary compound of oxygen and nitrogen and to ratio in the next compound of the series using given data.Answer:The formula of the unknown compound .The ratio for atoms of oxygen to nitrogen inis .The mass ratio of oxygen to nitrogen in next member of the series of compounds is .Explanation:The mass ratio of toinis .The mass ratio of toin theunknown compound is .The atomic mass of the oxygen is .The atomic mass of the nitrogen is .The mass ratio of to in a molecule can be represented as: …… (1)Here,is the number of atoms of the oxygen in the molecule.is the mass of an atom of oxygen.is the number of atoms of the nitrogen in the molecule.is the mass of an atom of nitrogen.For unknown compound, substitutefor , for and for mass ratio in equation (1).The ratio of atoms of oxygen to nitrogen is , therefore, the expected formulas of the unknown compound is and . If the unknown compound is in a series with then the formula of the unknown compound and the formula for the next member of the series is .The ratio for atoms of oxygen to nitrogen inis .For , substitute 5 for , for , 2 for and for in equation (1).The mass ratio of oxygen to nitrogen in next member of the series of the compounds is .
Answer: The water (in a subtle sort of way) will now be attracted towards the balloon
I attached a picture to help make sense
Use the state equation for ideal gases: pV = nRT
Data:
V = 88.89 liter
n = 17 mol
T = 67 + 273.15 = 340.15 K
R = 0.0821 atm * liter / (K*mol)
=> p = nRT / V = 17 mol * 0.0821 (atm*liter / K*mol) * 340.15 K / 88.89 liter
p = 5.34 atm
Answer: p = 5.34 atm
Answer:
Na₂₆F₁₁
Explanation:
We find the moles of the substance assuming 100 g of the substance is present. Why do we take 100 g? Because then the percent of sodium/fluorine, would be the g of sodium/fluorine respectively:
74.186 g Sodium | 1 mol Sodium/23 g => 3.2255 mol Na
25.814 g Fluorine | 1 mol Fluorine/19 g => 1.3586 mol F
Divide each by smallest number of moles:
3.2255/1.3586 = 2.37
1.3586/1.3586 = 1
Multiply by common number to get a smallest whole number:
2.37*11 = 26,
1*11 = 11
The empirical formula is Na₂₆F₁₁
