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1 month ago

Why does 5060 have three significant figures?

Chemistry

1 answer:

Anarel [2.7K]1 month ago

3 0

5060 has three significant figures: Below is the clarification

Explanation:

Significant figures

Significant figures (also referred to as significant digits and decimal places) in a number are those digits that carry substantial meaning.

These include all digits except: leading zeros.

Guidelines for determining significant figures

1. All non-zero digits are counted as significant. For instance, the number 23 has two significant figures.

2. Zeros located between two non-zero digits are significant; for example, 202.1201 contains seven significant figures.

3. Zeros preceding the significant figures are not significant. For example,.000021 has two significant figures, with zeros being non-contributory.

4. Zeros following the significant figures are significant.

This explains why the number 5060 has three significant figures.

Answer;

Considering the types of bonds being created and severed in the transition state, the stability of this temporary structure is comparatively high.

Explanation;

The reaction can be expressed as follows; NO(g)+F2(g)→NOF(g)+F(g)
All chemical reactions, including exothermic ones, require activation energy to initiate. The activation energy is the least amount of energy needed for the reactants to come together, overcome opposing forces, and begin breaking bonds.
When molecules encounter each other, their kinetic energy may be sufficient to stretch, bend, and eventually break bonds, resulting in chemical reactions.

3 0

14 days ago

A 0.784 g sample of magnesium is added to a 250 ml flask and dissolved in 150 ml of water. magnesium hydroxide obtained from the

Anarel [2728]

Although multiple values are given, our focus is on HCl.

<span>We have 215 mL (0.215 L) of 0.300 M HCl fully consumed in the reaction. It's important to recall that the number of moles is found by multiplying volume by molarity:</span>

moles = 0.215 L × 0.300 M

<span>moles = 0.0645 moles of HCl</span>

4 0

1 month ago

What types of compounds are CaCl2, Cu, C2H6, respectively.

KiRa [2853]

Response:

Ionic, metal, organic

Clarification:

For this scenario, we should examine each compound:

-) $CaCl_2$

In this compound, there is a non-metal atom (Cl) paired with a metal atom (Ca). This leads to a significant difference in electronegativity, indicating that an ionic bond will form. Ions can be generated:

$CaCl_2~->~Ca^+^2~+~2Cl^-$

The positive ion would be $Ca^+^2$ while the negative ion is $Cl^-$ . Thus, we have an ionic compound.

-) $Cu$

Here, we are looking at a single atom. Consulting the periodic table shows that this atom belongs to the transition metals section (central part of the periodic table). Hence, Cu (Copper) is identified as a metal.

-) $C_2H_6$

Within this molecule, carbon and hydrogen are linked by single bonds. The difference in electronegativity between C and H is insufficient to lead to ion formation. Therefore, we have covalent bonds. This property is typical of organic compounds. (Refer to figure 1)

5 0

18 days ago

66.667 mL of 3.000 M H2SO4 (aq) solution was neutralized by the stoichiometric amount of 4.000 M Al(OH)3 solution in a coffee cu

eduard [2645]

Answer:

$\large \boxed{\Delta_{\textbf{r}}H =\text{-4600 J$\cdot$ mol}^{-1}}$

Explanation:

This scenario is unrealistic since Al(OH)₃ is not soluble in water.

The question consists of two parts:

A. Stoichiometry — where we determine volumes, masses, and moles for the products

B. Calorimetry — where we assess the enthalpy of the reaction.

A. Stoichiometry

1. Determine the volume of Al(OH)₃

(a) The balanced chemical equation:

2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 6H₂O

M/V: 66.667

c/mol·L⁻¹: 4.000 3.000

(b) Moles of H₂SO₄

$\rm \text{66.667 mL H$_{2}$}SO_{4} \times \dfrac{\text{3.000 mmol H$_{2}$SO}_{4}}{\text{1 mL H$_{2}$SO}_{4}} = \text{200.00 mmol H$_{2}$SO}_{4}$

(c) Moles of Al(OH)₃

The molar ratio stands at 2 mmol Al(OH)₃: 3 mmol H₂SO₄

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{2 mmol Al(OH)}_{3}}{\text{3 mmol H$_{2}$SO}_{4}}\\\\= \text{133.33 mmol Al(OH)}_{3}$

(d) Volume of Al(OH)₃

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{1 mL Al(OH)}_{3}}{\text{4 mmol H$_{2}$SO}_{4}} = \text{50.000 mL Al(OH)}_{3}$

B. Calorimetry

This reaction has two energy exchanges.

q₁ = heat from the reaction

q₂ = heat used to heat the calorimeter

q₁ + q₂ = 0

nΔH + mCΔT = 0

Data:

Moles of Al₂(SO₄)₃ = 0.066 667 mol

C = 1.10 J°C⁻¹g⁻¹

T_initial = 22.3 °C

T_final = 24.7 °C

Calculations

(a) Mass of solution

Assume solutions are as dense as water (though not realistic).

Mass of sulfuric acid solution = 66.667 g

Mass of aluminium hydroxide solution = 50.000

TOTAL = 116.667 g

(b) ΔT

ΔT = T_final - T_initial = 24.7 °C - 22.3 °C = 2.4°C

(c) ΔH

$\begin{array}{ccccl}n\Delta H & +& mC \Delta T& = &0\\\text{0.066 667 mol }\times \Delta H& + & \text{116.667 g} \times 1.10 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 2.4 \, ^{\circ}\text{C} & = & 0\\0.066667 \Delta H \text{ mol} & + & \text{310 J} & = & 0\\&&0.066667 \Delta H \text{ mol} & = & \text{-310 J} & & \\\end{array}\\$

$\begin{array}{ccccl}& &\Delta H & = & \dfrac{\text{-310 J}}{\text{0.066667 mol}}\\\\& &\Delta H & = & \textbf{-4600 kJ/mol}\\\end{array}\\\large \boxed{\mathbf{\Delta_{\textbf{r}}H} =\textbf{-4600 J$\cdot$ mol}^{\mathbf{-1}}}$

This result appears nonsensical, but it is derived from your given figures.

6 0

1 month ago