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3 years ago

A pupil adds 5 cm of 12 mol.dm sulphuric acids to make a 250 cmº solution. Calculate the

Chemistry

1 answer:

klemol [59]3 years ago

8 0

Answer:

0.500 mol/dm³

Explanation:

Using the formula below;

CaVa = CbVb

Where;

Ca = concentration of acid (mol/dm³)

Cb = concentration of base (mol/dm³)

Va = volume of acid (cm³)

Vb = volume of base (cm³)

In accordance to the information provided in this question is;

Va = 5cm³

Vb = 250 cm³

Ca = 12 mol/dm³

Cb = ?

Using CaVa = CbVb

12 × 5 = Cb × 250

60 = 120Cb

Cb = 60/120

Cb = 0.500 mol/dm³

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Answer:

The answer to your question is below

Explanation:

1) number of molecules = x

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Molar mass CoCl₂ = 59 + 71 = 130 g

130 g ------------------ 6.023 x 10²³ molecules

23 g ----------------- x

x = (23 x 6.023 x 10²³) / 130

x = 1.066 x 10²³ molecules

3.2 moles of H₃PO₄ to grams

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x = (3.2 x 98)/1

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Molar mass = (14 x 2) + (8 x 1)

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x ---------------- 2.2 moles

x = (2.2 x 36)/1

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= 82 g

82 g of NH₄SO₂ --------------- 6.023 x 10²³ molecules

4.5 g ---------------- x

x = (4.5 x 6.023 x 10²³) / 82

x = 3.3 x 10²³ molecules

Liters of N₂ = ? 7.9 x 10³⁰ particles

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3 years ago

Describe hydrogen's atomic structure and its molecular structure.

Kryger [21]

Explanation:

A hydrogen atom is an atom of the chemical element hydrogen. The electrically neutral atom contains a single positively charged proton and a single negatively charged electron bound to the nucleus by the Coulomb force. Atomic hydrogen constitutes about 75% of the baryonic mass of the universe.

Names: hydrogen atom, H-1, protium, ¹H

Protons: 1

Neutrons: 0

Symbol: 1H

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3 years ago

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Usimov [2.4K]

Answer:

A). Atomic number

Explanation:

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3 years ago

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san4es73 [151]

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3 years ago

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90 POINTSSSS!!!

Katarina [22]

Answer:

$\large \boxed{\text{-2043.96 kJ/mol}}$

Explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn: C₃H₈(g) + O₂(g) ⟶ CO₂(g) + H₂O(g)

Balanced eqn: C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

(a) Table of enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}$

(b)Total enthalpies of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\= \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}$

ΔᵣH° is negative, so the reaction is exothermic.

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3 years ago