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3 years ago

In coulombs, how much positive charge is there IN a barium nucleus? On a barium ion?

Chemistry

1 answer:

faltersainse [42]3 years ago

4 0

Answer:

The amount of positive charge in a barium nucleus is approximately 8.972 × 10⁻¹⁸ coulomb

Explanation:

The amount of positive charge there is in a barium nucleus is found as follows;

The atomic number of an element is equal to the number of proton present in the atom

The atomic number of barium = 56 = The number of protons in an atom of barium

The positive charge of a single proton is known as an elementary charge, e, given by the formula, e = F/N $_A$

Where;

F = Faraday's constant

N $_A$ = The Avogadro's number

Therefore, e = 1.602 176 634 × 10⁻¹⁹ C

Which gives the positive charge, Q, of the protons of a barium nucleus as follows;

Q = Number of protons × e

Q = 1.602176634 × 10⁻¹⁹ × 56 ≈ 8.972 × 10⁻¹⁸ Coulomb per atom of barium.

The amount of positive charge in a barium nucleus ≈ 8.972 × 10⁻¹⁸ coulomb.

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Calculate the pH at the equivalence point when 22.0 mL of 0.200 M hydroxylamine, HONH2, is titrated with 0.15 M HCl. (Kb for HON

solniwko [45]

Answer:

pH = 3.513

Explanation:

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In this case, since this titration is carried out via the following neutralization reaction:

$HONH_2+HCl\rightarrow HONH_3^+Cl^-$

We can see the 1:1 mole ratio of the acid to the base and also to the resulting acidic salt as it comes from the strong HCl and the weak hydroxylamine. Thus, we first compute the required volume of HCl as shown below:

$V_{HCl}=\frac{22.0mL*0.200M}{0.15M}=29.3mL$

Now, we can see that the moles of acid, base and acidic salt are all:

$0.0220L*0.200mol/L=0.0044mol$

And therefore the concentration of the salt at the equivalence point is:

$[HONH_3^+Cl^-]=\frac{0.0044mol}{0.022L+0.0293L} =0.0858M$

Next, for the calculation of the pH, we need to write the ionization of the weak part of the salt as it is able to form some hydroxylamine as it is the weak base:

$HONH_3^++H_2O\rightleftharpoons H_3O^++HONH_2$

Whereas the equilibrium expression is:

$Ka=\frac{[H_3O^+][HONH_2]}{[HONH_3^+]}$

Whereas Ka is computed by considering Kw and Kb of hydroxylamine:

$Ka=\frac{Kw}{Kb}=\frac{1x10^{-14}}{9.10x10^{-9}} \\\\Ka=1.10x10^{-6}$

So we can write:

$1.10x10^{-6}=\frac{x^2}{0.0858-x}$

And neglect the x on bottom to obtain:

$1.10x10^{-6}=\frac{x^2}{0.0858}\\\\x=\sqrt{1.10x10^{-6}*0.0858}=3.07x10^{-4}M$

And since x=[H3O+] we obtain the following pH:

$pH=-log(3.07x10^{-4})\\\\pH=3.513$

Regards!

4 0

3 years ago

Determine balanced chemical equations for the following chemical reactions

Lena [83]

Answer:

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Explanation:

okay?.........................................

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3 years ago

Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: PCl5(g) → PCl3(g) + Cl2(g)

nalin [4]

Answer:

The value of $\Delta H$ for the desired reaction will be 249.75 KJ.

Explanation:

The desired reaction is shown below

$\textrm{PCl}_{5}\left ( g \right )\rightarrow \textrm{PCl}_{3}\left ( g \right )+\textrm{Cl}_{2}\left ( g \right )$

The desired reaction can be obtained by adding the given reactions and then dividing both sides by 4.

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