Answer:
The atoms on left side are larger than the atoms on the right side of the periodic table because those on the right have more proton's.
Explanation:
As we travel along a period in a periodic table then the atomic radii decreases
This is because as we travel along a period we have that the atomic number of the atoms increases which means the the number of proton's increased
But the electron's add to the same outer shell throughout the period , which means the effective nuclear charge increases which pulls the outer electrons toward's the nucleus and the size decreases.
Therefore the atoms on left side are larger than the atoms on the right side of the periodic table because those on the right have more proton's.
The answer is 37.5 moles
The explanation:
1- when we have the mass of SO3 in Kg so, first we have to convert it to grams:
mass = 3 * 1000 = 3000 g
2- we need to get the molar mass of SO3 :
molar mass of SO3 = 32 + (16*3)
= 80 g/mol
3- then we can use this formula to get number of moles:
moles = mass / molar mass
= 3000 g / 80 g/mol
= 37.5 moles
Somewhat false
observations can be made of a model of the statue of liberty, say, or in real line
The outermost energy shell of an atom
because they are involved in forming bonds
0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.
<h3>Explanation:</h3>
Based on the reaction
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
1 mole of CaCl₂ reacts per mole of Na₂CO₃
we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g
- We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
- These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
- Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:
<h3>
Moles CaCl₂.2H₂O:</h3>
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
Moles Na₂CO₃:
0.0102 moles Na₂CO₃
Mass Na₂CO₃:
0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present
Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.
To learn more about stoichiometric quantities visit:
<h3>
brainly.com/question/28174111</h3>
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