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4 days ago

A blue circle labeled proton and A overlaps a pink circle labeled electron and C. The overlap is labeled B. The Venn diagram com

pares protons with electrons. Which shared property belongs in the region marked "B”? Is found in the nucleus Has mass of 1 amu Orbits the nucleus Is electrically charged

Chemistry

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0.036549 round to three sig figs

Alekssandra [3086]

3 first significant figure
6 second significant figure
5 third significant figure
4 cannot exceed 5, so retain 5 instead of increasing it to 6

0.0365

6 0

2 months ago

Read 2 more answers

For the reaction below, Kp 5 1.16 at 800.8C. CaCO3(s) 34 CaO(s) 1 CO2(g) If a 20.0-g sample of CaCO3 is put into a 10.0-L contai

Tems11 [2777]

The percentage of calcium carbonate that reacted is 2.5%. The reaction in question allows us to determine the equilibrium Kp: Kp = the partial pressure of carbon dioxide, since the other components are solids. We'll apply the ICE table to the provided equilibrium. At the start, we have 0.2 for calcium carbonate with no initial moles of other substances. As the reaction progresses, we set the changes to be -x for calcium carbonate, +x for carbon dioxide, and +x for the other product, leading us to an equilibrium of 0.2-x for calcium carbonate while both other products are at x. Using Kp = Kc(RT)ⁿ, where n represents the mole difference of gaseous products and reactants, we find n to equal 1 for this reaction. With R as the gas constant (8.314 J/mol K) and the temperature at 800 °C (1073 K), we substitute the values accordingly. Upon calculation, we find x = 0.005, which indicates the amount of calcium carbonate that dissociated or reacted, leading us to the reacted percentage.

7 0

1 month ago

Suppose a 20.0 g gold bar at 35.0°C absorbs 70.0 calories of heat energy. Given that the specific heat of gold is 0.0310 cal/g °

VMariaS [2998]

The change in temperature can be expressed as:

$T_2-T_1=\dfrac{q}{mC_p(Gold)}$

By substituting in the known values, we arrive at:

$T_2-T_1=\dfrac{70\ cal}{20\ g\times 0.0310\ cal/g^o\ C}\\\\T_2-T_1=112.90^oC\\\\T_2-35^oC=112.90^oC\\\\T_2=(112.90+35)^oC\\\\T_2=147.9^oC$

Thus, we obtain the required answer.

6 0

2 months 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 [2782]

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

3 months ago

Chlorine atoms from CFC molecules speed the breakdown of stratospheric O3 in a process that is affected by both homogeneous and

eduard [2782]

False, it's solely heterogeneous. Explanation: The degradation of the ozone layer caused by CFC molecules happens in the gaseous state since it does not involve liquids or solids at stratospheric conditions. Additionally, the reaction occurs independently as ozone is chemically unstable, eliminating the need for a catalyst.

7 0

2 months ago