<span>There are few main factors affecting the atomic radii, the outermost electrons and the protons in the nucleus and also the shielding of the internal electrons. I would speculate that the difference in radii is given by the electron clouds since the electrons difference in these two elements is in the d orbital and both has at least 1 electron in the 4s (this 4s electron is the outermost electron in all the transition metals of this period). The atomic radio will be mostly dependent of these 4s electrons than in the d electrons. Besides that, you can see that increasing the atomic number will increase the number of protons in the nucleus decreasing the ratio of the atoms along a period. The Cu is an exception and will accommodate one of the 4s electrons in the p orbital.
</span><span>Regarding the density you can find the density of Cu = 8.96g/cm3 and vanadium = 6.0g/cm3. This also correlates with the idea that if these two atoms have similar volume and one has more mass (more protons; density is the relationship between m/V), then a bigger mass for a similar volume will result in a bigger density.</span>
Answer:
Sulfur
Explanation:
You find the identity by looking at the number of protons. The number of protons never change for an element.
Answer:
tri
Explanation:
1-nothing
2-di
3-tri
- Hope that helps! Please let me know if you need further explanation.
Using PV=nRT or the ideal gas equation, we substitute n= 15.0 moles of gas, V= 3.00L, R equal to 0.0821 L atm/ mol K and T= 296.55 K and get P equal to 121.73 atm. The Van der waals equation is (P + n^2a/V^2)*(V-nb) = nRT. Substituting a=2.300L2⋅atm/mol2 and b=0.0430 L/mol, P is equal to 97.57 atm. The difference is <span>121.73 atm- 97.57 atm equal to 24.16 atm.</span>
The number of grams in 1.70 moles of Ca(NO₃)₂ is 384.2 grams
<h3>How to determine the mass of Ca(NO₃)₂</h3>
The mole of a substance is related to it's mass and molar mass according to the following equation:
Mole = mass / molar mass
With the above formula, we can determine the mass of Ca(NO₃)₂ as illustrated below:
- Mole of Ca(NO₃)₂ = 1.70 moles
- Molar mass of Ca(NO₃)₂ = 40 + 3[14 + (16 × 3)] = 40 + 3[14 + 48] = 40 + 3(62) = 40 + 186 = 226 g/mol
- Mass of Ca(NO₃)₂ = ?
Mole = mass / molar mass
1.70 = Mass of Ca(NO₃)₂ / 226
Cross multiply
Mass of Ca(NO₃)₂ = 1.70 × 226
Mass of Ca(NO₃)₂ = 384.2 grams
Thus, the mass of 1.70 moles of Ca(NO₃)₂ is 384.2 grams
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