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3 years ago

In what 3 ways are isotopes similar

Chemistry

2 answers:

Fittoniya [83]3 years ago

7 0

Answer:

same atomic number

same proton number

same electron number

shows same chemical properties

MAVERICK [17]3 years ago

6 0

An isotope is one of two or more forms of the same chemical element. Different isotopes of an element have the same number of protons in the nucleus, giving them the same atomic number, but a different number of neutrons giving each elemental isotope a different atomic weight.

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After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108. Expre

Marina CMI [18]

Answer:

$\large \boxed{1.77 \times 10^{-5}\text{ mol/L}}$

Explanation:

Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.

1. Moles of Ni²⁺

$n = \text{135 mL} \times \dfrac{\text{0.0147 mmol}}{\text{1 mL}} = \text{1.984 mmol}$

2. Moles of NH₃

$n = \text{190 mL} \times \dfrac{\text{0.250 mmol}}{\text{1 mL}} = \text{47.50 mmol}$

3. Initial concentrations after mixing

(a) Total volume

V = 135 mL + 190 mL = 325 mL

(b) [Ni²⁺]

$c = \dfrac{\text{1.984 mmol}}{\text{325 mL}} = 6.106 \times 10^{-3}\text{ mol/L}$

(c) [NH₃]

$c = \dfrac{\text{47.50 mmol}}{\text{325 mL}} = \text{0.1462 mol/L}$

3. Equilibrium concentration of Ni²⁺

The reaction will reach the same equilibrium whether it approaches from the right or left.

Assume the reaction goes to completion.

Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺

I/mol·L⁻¹: 6.106×10⁻³ 0.1462 0

C/mol·L⁻¹: -6.106×10⁻³ 0.1462-6×6.106×10⁻³ 0

E/mol·L⁻¹: 0 0.1095 6.106×10⁻³

Then we approach equilibrium from the right.

Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺

I/mol·L⁻¹: 0 0.1095 6.106×10⁻³

C/mol·L⁻¹: +x +6x -x

E/mol·L⁻¹: x 0.1095+6x 6.106×10⁻³-x

$K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}$

Kf is large, so x ≪ 6.106×10⁻³. Then

$K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}$

3 0

4 years ago

2 H2(g) + O2(g) → 2 H2O(g)

alexgriva [62]

Answer:

77 L of water can be made.

Explanation:

Molar mass of $O_{2}$ = 32 g/mol

So, 55 g of $O_{2}$ = $\frac{55}{32}$ mol of $O_{2}$ = 1.72 mol of $O_{2}$

As hydrogen is present in excess amount therefore $O_{2}$ is the limiting reagent.

According to balanced equation, 1 mol of $O_{2}$ produces 2 mol of $H_{2}O$ .

So, 1.72 mol of $O_{2}$ produce $(2\times 1.72)$ mol of $H_{2}O$ or 3.44 mol of $H_{2}O$ .

Let's assume $H_{2}O$ gas behaves ideally at STP.

Then, $P_{H_{2}O}.V_{H_{2}O}=n_{H_{2}O}.R.T$ , where P, V, n, R and T represents pressure, volume, no. of moles, gas constant and temperature in kelvin scale respectively.