Write the complete and balanced chemical equation for this precipitation reaction. feso4(aq)+ba(oh)2(aq)→

The concentration of La3+ in a saturated solution of La(IO3)3 can be found by setting up and solving a solubility product (Ksp) expression, using the given Ksp value and assuming [La3+] = x and [IO3-] = 3x based on the dissolution stoichiometry of La(IO3)3.

The question relates to the solubility product of lanthanum iodate, denoted by the compound formula La(IO3)3. The solubility product (Ksp) expression is [La3+][IO3-]3 = Ksp. As the dissolution stoichiometry of La(IO3)3 shows that for each formula unit that dissolves, one La3+ ion and three IO3- ions are produced, it can be assumed that [La3+] = x and [IO3-] = 3x. Solving for x in the Ksp expression using these assumptions and the given Ksp value of 1.0 x 10-11 will yield the concentration of La3+ in a saturated solution of lanthanum iodate.

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after 3.7 days, approximately [tex]\( 4.064 \text{ mg} \)[/tex] of the sodium-24 sample will remain.

To solve this problem, we can use the formula for radioactive decay:

[tex]\[ N_t = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \][/tex]

Where:

-[tex]\( N_t \)[/tex] is the final amount of the substance after time \( t \)

- [tex]\( N_0 \)[/tex] is the initial amount of the substance

- ( t ) is the time that has passed

- [tex]\( T_{1/2} \)[/tex] is the half-life of the substance

Given:

- Initial amount of sodium-24 [tex](\( N_0 \))[/tex] = 260.1 mg

- Half-life of sodium-24 ([tex]\( T_{1/2} \)[/tex]) = 14.8 hours

- Time that has passed [tex](\( t \))[/tex] = 3.7 days

First, let's convert the time that has passed from days to hours:

[tex]\[ 3.7 \text{ days} \times 24 \text{ hours/day} = 88.8 \text{ hours} \][/tex]

Now, we can substitute the given values into the formula for radioactive decay to find [tex]\( N_t \)[/tex], the amount of sodium-24 remaining after 3.7 days:

[tex]\[ N_t = 260.1 \text{ mg} \times \left( \frac{1}{2} \right)^{\frac{88.8 \text{ hours}}{14.8 \text{ hours}}} \][/tex]

Now, let's calculate:

[tex]\[ N_t = 260.1 \text{ mg} \times \left( \frac{1}{2} \right)^{\frac{88.8}{14.8}} \][/tex]

[tex]\[ N_t = 260.1 \text{ mg} \times \left( \frac{1}{2} \right)^{6} \][/tex]

Since [tex]\( \left( \frac{1}{2} \right)^6 = \frac{1}{64} \),[/tex] we have:

[tex]\[ N_t = 260.1 \text{ mg} \times \frac{1}{64} \][/tex]

[tex]\[ N_t = \frac{260.1}{64} \text{ mg} \][/tex]

[tex]\[ N_t \approx 4.064 \text{ mg} \][/tex]

So, after 3.7 days, approximately [tex]\( 4.064 \text{ mg} \)[/tex] of the sodium-24 sample will remain.

The first element that has exactly ten p electrons is Neon. However, the first element to have 10 p-electrons in the ground state (including core electrons) is Silicon.

The element of lowest atomic number that has exactly ten p electrons is neon. This is because the periodic table allows us to understand the electron configuration of elements. In the case of Neon (atomic number 10), it fills the 1s, 2s, and 2p orbitals; the 1s and 2s orbitals can hold two electrons each while the 2p can hold six, all for a total of 10 electrons, with the last six being p electrons.

However, if we're seeking the first element to have 10 p electrons in its ground state (including those beyond the Neon atom), the element would be Silicon (atomic number 14). Its electron configuration is [Ne]3s²3p², meaning it has 10 total p-electrons in the ground state: six from the Neon core and four more in the next energy level.

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Lone pairs and bonding domains differ in the amount of space they occupy due to repulsion effects. Lone pairs occupy a larger area compared to bonding pairs, and are often positioned to minimize repulsions. This impacts the geometry of the molecule.

The trend that distinguishes lone pairs from bonding domains comes from the spatial arrangement of electrons. Due to repulsions, a lone pair of electrons tends to occupy a larger region of space compared to electrons in a bond. The repulsion order from the largest to smallest is: lone pairs > triple bond > double bond > single bond.

Consider a case where a central atom has two lone pairs and four bonding regions, which results in an octahedral electron-pair geometry. The lone pairs are positioned on opposite sides, leading to a square planar molecular structure. This placement minimizes lone pair-lone pair repulsions.

Space must be provided for each pair of electrons, whether they are in a bond or are lone pairs. This concept contributes to the formation of different molecular structures.

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