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3 years ago

How does the size change when an atom forms a cation and when an atom forms an anion?

1 answer:

8 0

A cation is an atom that loses a valence electron. When a valence electron is released there is one electron less to create a repulsive force. The loss of a repulsive force will allow the atom to pull tighter together. An anion would therefore be larger in size due to increased repulsion of the valence electrons.

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The ph of a solution for which [OH]= 1.0 x 10^-4 is? A 10 B 14 C 11 D 9 E 7

zhenek [66]

To find pH, use the following formula ---> pH= - log [H+]

so first we need to calculate the [H+] concentration using the OH concentration. to do this, we need to use this formula--> 1.0x10-14= [H+] X [OH-], so we solve for H+ and plug in

[H+]= 1.0X10-14/[OH-]---> 1.0 x 10-14/ 1.0 x 10-4= 1.0 x 10-10

now that we have the H+ concentration, we can solve of pH

pH= -log (1.0x10-10)= 10

answer is A

8 0

3 years ago

How many grams of CO2 are used when 6.0 g of O2 are produced? Express your answer with the appropriate units.

Ipatiy [6.2K]

Answer:

that is why co2 is in the power of 2ik

8 0

3 years ago

URGENT CHEMISTRY EXPERT!

vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

$\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl$

(b) Moles of CuCl₂

$\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} = \text{3.340 mmol CuCl}_{2}$

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

$\text{Moles of Na$_{3}$PO}_{4} = \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}$

(d) Volume of Na₃PO₄

$V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}$

Part 2. Net ionic equation

(a) Molecular equation

$\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})$

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0

2 years ago

What is the formula, when rubidium reacts with tellurium?

natita [175]

Answer:

Like other alkali metals, rubidium metal reacts violently with water. As with potassium (which is slightly less reactive) and caesium (which is slightly more reactive), this reaction is usually vigorous enough to ignite the hydrogen gas it produces.

Explanation:

hope it helps

5 0

2 years ago

Read 2 more answers

Saul converts 2.392 hectoliters to liters. What should his new number be?

Lostsunrise [7]

2.392 hectoliters = 239.2 liters. 1 hectoliter = 100 liters.

3 0

2 years ago