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3 years ago

The atomic number of sulfur is 16, which indicates that a sulfur atom contains

1 answer:

8 0

The atomic number of an element indicates only how many protons are in that specific element. The atomic number of sulfur is 16, so it means that in one sulfur atom, there are 16 protons. The number of protons is also equal to the number of electrons in any given atom, so this number also indicates that there are 16 electrons in a sulfur atom.

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aleksklad [387]

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5 0

4 years ago

Read 2 more answers

Illustrate: The Avogadro constant is so large it is normally written in scientific notation. To

Licemer1 [7]

Answer:

602200000000000000000000

Explanation:

move the decimal place over 23 places to the right

5 0

3 years ago

Read 2 more answers

Calculate the mass-to-mass ratio of 25.0 g of salt in 105 g of water. Work must be shown in order to earn credit.

Sveta_85 [38]

Answer:

0.23

Explanation:

It is known that, the mass to mass ratio of the salt to water

= (mass of salt / mass of water)

= (25.0 g / 105.0 g)

= 0.23

So, the answer is 0.23

4 0

3 years ago

Compare the molar concentration of the crystal violet solution to that of the sodium hydroxide solution. approximately how much

romanna [79]

The molar concentration of the crystal violet solution is more concentrated than that of the sodium hydroxide solution. It is because the crystal violet solution has more solute in it compared to the sodium hydroxide.

6 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

3 years ago