Answer:
614 034 kg
Explanation:
n = m/Mm
m = n * Mm
Mm(MgSO4) = 1 * 24.3 * 1 * 32.1 * 4 * 16 = 49921.92
m = 12.3 * 49921.92
m = 614 034 kg
The number of mole of nitrogen that occupies 1.2 L under the same condition is 0.6 mole
<h3>Data obtained from the question </h3>
- Initial mole (n₁) = 0.2 mole
- Initial volume (V₁) = 0.4 L
- Final volume (V₂) = 1.2 L
- Final mole (n₂) =?
<h3>How to determine the final mole </h3>
The final mole can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
Divide both side n
PV / n = RT
Divide both side by P
V / n = RT / P
RT / P = constant
V / n = constant
Thus,
V₁ / n₁ = V₂ / n₂
0.4 / 0.2 = 1.2 / n₂
2 = 1.2 / n₂
Cross multiply
2 × n₂ = 1.2
Divide both side by 2
n₂ = 1.2 / 2
n₂ = 0.6 mole
Learn more about ideal gas equation:
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Answer:
The stronger electrolyte is the HCl
Explanation:
Stronger electrolyte are the ones, that in water, completely dissociates.
HCl(aq) → H⁺(aq) + Cl⁻(aq)
HCl(aq) + H₂O(l) → H₃O⁺ (aq) + Cl⁻(aq)
Both are acids, they bring protons to medium but the hydrochloric completely dissociates.
HF (aq) + H₂O(l) ⇄ H₃O⁺(aq) + F⁻(aq) Ka
In the dissociation of weak electrolytes, they ionize but at the same time they bond again, so the reaction is always kept in equilibrium.
Answer:
The radius of tantalum (Ta) atom is 
Explanation:
From the Body-centered cubic (BBC) crystal structure we know that a unit cell length <em>a </em>and atomic radius <em>R </em>are related through
So the volume of the unit cell 
is
We can compute the theoretical density ρ through the following relationship
where
n = number of atoms associated with each unit cell
A = atomic weight
= volume of the unit cell
= Avogadro’s number (
atoms/mol)
From the information given:
A = 180.9 g/mol
ρ = 16.6 g/cm^3
Since the crystal structure is BCC, n, the number of atoms per unit cell, is 2.
We can use the theoretical density ρ to find the radio <em>R</em> as follows:
Solving for <em>R</em>
![\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}\\\frac{64\sqrt{3}R^3}{9}=\frac{nA}{\rho N_{a}}\\R^3=\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}} \\R=\sqrt[3]{\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%7BnA%7D%7B%28%5Cfrac%7B64%5Csqrt%7B3%7DR%5E3%7D%7B9%7D%29N_%7Ba%7D%7D%5C%5C%5Cfrac%7B64%5Csqrt%7B3%7DR%5E3%7D%7B9%7D%3D%5Cfrac%7BnA%7D%7B%5Crho%20N_%7Ba%7D%7D%5C%5CR%5E3%3D%5Cfrac%7BnA%7D%7B%5Crho%20N_%7Ba%7D%7D%5Ccdot%20%5Cfrac%7B1%7D%7B%5Cfrac%7B64%5Csqrt%7B3%7D%7D%7B9%7D%7D%20%5C%5CR%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BnA%7D%7B%5Crho%20N_%7Ba%7D%7D%5Ccdot%20%5Cfrac%7B1%7D%7B%5Cfrac%7B64%5Csqrt%7B3%7D%7D%7B9%7D%7D%7D)
Substitution for the various parameters into above equation yields
![R=\sqrt[3]{\frac{2\cdot 180.9}{16.6\cdot 6.023 \times 10^{23}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}\\R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm](https://tex.z-dn.net/?f=R%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B2%5Ccdot%20180.9%7D%7B16.6%5Ccdot%206.023%20%5Ctimes%2010%5E%7B23%7D%7D%5Ccdot%20%5Cfrac%7B1%7D%7B%5Cfrac%7B64%5Csqrt%7B3%7D%7D%7B9%7D%7D%7D%5C%5CR%20%3D%201.43%20%5Ctimes%2010%5E%7B-8%7D%20%5C%3Acm%20%3D%200.143%20%5C%3Anm)
