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3 years ago

In the system model, the direction of the arrow tells you that

Chemistry

2 answers:

nadya68 [22]3 years ago

6 0

Answer: option b

Step by step Explanation:

AlladinOne [14]3 years ago

4 0

Answer: Answer b

Explanation:

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How many moles of aluminum are in 95.0 grams of aluminum

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Edit the reaction by drawing all steps in the appropriate boxes and connecting them with reaction arrows. Add charges where need

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Edit the reaction by drawing all steps in the appropriate boxes and connecting them with reaction arrows. Add charges where needed. Electron flow arrows should start on the electron(s) of an atom or a bond and should end on an atom, bond, or location where a new bond should be created. Include all free radicals by right-clicking on an atom on the canvas and then using the Atom properties to select the monovalent radical.

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The oxidation number that the Alkaline Earth Metals normally have when forming compounds is

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Explanation:

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3 years ago

16. Which of the following aqueous solutions contains the greatest number of 10ns.

gavmur [86]

Answer:

$\boxed{\text{b) 300.0 mL of 0.10 mol/L CaCl}_{2}}$

Explanation:

a) 400.0 mL of 0.10 M NaCl

(i) Moles of NaCl

$\text{ Moles of NaCl } = \text{0.4000 L NaCl} \times \dfrac{\text{0.10 mol NaCl}}{\text{1 L NaCl}} = \text{0.040 mol NaCl}$

(ii) Moles of ions

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

We get 2 mol of ions from 1 mol of NaCl

$\text{Moles of ions } = \text{0.040 mol NaCl} \times \dfrac{\text{2 mol ions}}{\text{1 mol NaCl}} = \text{0.080 mol ions}$

b) 300.0 mL of 0.10 M CaCl₂

(i) Moles of CaCl₂

$\text{ Moles of CaCl}_{2} =\text{0.3000 L CaCl}_{2} \times \dfrac{\text{0.10 mol CaCl}_{2}}{\text{1 L CaCl}_{2}} = \text{0.030 mol CaCl}_{2}$

(ii) Moles of ions

CaCl₂(s) ⟶ Ca²⁺(aq) + 2Cl⁻(aq)

We get 3 mol of ions from 1 mol of CaCl₂

$\text{Moles of ions } = \text{0.030 mol CaCl}_{2} \times \dfrac{\text{3 mol ions}}{\text{1 mol CaCl}_{2}} = \text{0.090 mol ions}$

c) 200.0 mL of 0.10 M FeCl₃

(i) Moles of FeCl₃

$\text{ Moles of FeCl}_{3} =\text{0.2000 L FeCl}_{3} \times \dfrac{\text{0.10 mol FeCl}_{3}}{\text{1 L FeCl}_{3}} = \text{0.020 mol FeCl}_{3}$

(ii) Moles of ions

FeCl₂(s) ⟶Fe³⁺(aq) + 3Cl⁻(aq)

We get 4 mol of ions from 1 mol of FeCl₃

$\text{ Moles of FeCl}_{3} =\text{0.2000 L FeCl}_{3} \times \dfrac{\text{0.10 mol FeCl}_{3}}{\text{1 L FeCl}_{3}} = \text{0.020 mol FeCl}_{3}$

d) 200.0 mL of 0.10 M KBr

(i) Moles of KBr

$\text{ Moles of KBr} = \text{0.2000 L KBr} \times \dfrac{\text{0.10 mol KBr }}{\text{1 L KBr}} = \text{0.020 mol KBr}$

(ii) Moles of ions

KBr(s) ⟶ K⁺(aq) + Br⁻

We get 2 mol of ions from 1 mol of KBr

$\text{Moles of ions } = \text{0.020 mol KBr} \times \dfrac{\text{2 mol ions}}{\text{1 mol KBr}} = \text{0.040 mol ions}\\\\\text{We get the most ions ions from }\boxed{\textbf{300.0 mL of 0.10 mol/L CaCl}_{2}}$

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3 years ago