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prisoha [69]
3 years ago
13

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.

At STP, pressure is 1 atm and T is 273 K.

Here, n_{H_{2}O} = 3.44 mol and R = 0.0821 L.atm/(mol.K)

So, (1atm)\times V_{H_{2}O}=(3.44mol)\times (0.0821L.atm.mol^{-1}.K^{-1})\times (273K)

  \Rightarrow  V_{H_{2}O}=77L

Option (b) is correct.

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3 years ago
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Protons and neutrons i think , sorry if i’m wrong.
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There are four stages to the classical demographic transition model Pre-transitional Europe was characterized by high and
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The answer is D! Explaination:
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