A mole of CO2 = 2 moles of O2
8 CO moles x 2 =
16 moles
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
The mass of hydrogen atoms that is measured at 54 u given the relationship is 89.64×10¯²⁴ g
<h3>Conversion scale </h3>
1 u = 1.66×10¯²⁴ g
<h3>How to determine the mass of hydrogen atoms </h3>
- Mass of Hydrogen (u) = 54 u
- Mass of Hydrogen (g) =?
1 u = 1.66×10¯²⁴ g
Therefore
54 u = 54 × 1.66×10¯²⁴ g
54 u = 89.64×10¯²⁴ g
Thus, the mass of the hydrogen atoms measured at 54 u is 89.64×10¯²⁴ g
Learn more about conversion:
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Answer:
D.) H-O
Explanation:
Polarity is determined based on the difference in electronegativity of the atoms. The greater the difference, the more polar the bond. The general trend is that the atoms in the top-right corner of the periodic table are the most electronegative.
A.) is incorrect because H-H has no electronegativity difference, making it nonpolar.
B.) and C.) are incorrect because their electronegativity differences are not the greatest.
D.) is correct because the electronegativity difference between the H and O is the greatest.