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19 days ago

Of elements N, O, Cl, Na, and Which two would likely have similar chemical properties and why

1 answer:

5 0

Oxygen and Nitrogen are the most alike among the listed elements due to their proximity in the periodic table. While this reasoning may not be particularly robust, these two elements share certain similarities. They are both classified as non-metals, exhibit high electronegativity, exist as diatomic gases in their elemental forms, contain a similar number of valence electrons, and generally act as oxidizing agents. Although Oxygen and Chlorine also show similarities, they are not as closely related as Oxygen and Nitrogen.

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An infant acetaminophen suspension contains 80 mg/0.80 mL suspension. The recommended dose is 15 mg/kg body weight.

KiRa [2711]

Response:

0.8853 mL

Clarification:

Initially, we convert 13 lb to kg, remembering that 1 lb = 0.454 kg:

13 lb * $\frac{0.454kg}{1lb}$ = 5.902 kg

Next, we determine the required mg of acetaminophen to administer, applying the recommended dosage and infant's weight:

15 mg/kg * 5.902 kg = 88.53 mg

Finally, we compute the necessary mL of suspension, utilizing its concentration:

88.53 mg ÷ (80 mg/0.80 mL) = 0.8853 mL

8 0

1 month ago

gypsum is insoluble in water. you are asked to purify a sample of gypsum that is contaminated with a soluble salt.

KiRa [2711]

Response:

a. To purify a gypsum sample, you will need the following equipment: Bunsen burner, beaker, filter funnel, stirring rod, and filter paper.

b. Gypsum, a sulfate mineral consisting of calcium sulfate dihydrate, can be purified by following these steps:

1. Add water to the gypsum in a beaker.

2. Stir the mixture thoroughly with the stirring rod.

3. Use the filter paper and filter funnel to remove excess solids from the mixture.

4. Heat the filtered mixture on the Bunsen burner to evaporate the remaining water.

5. After cooling, filter again through the filter paper to obtain pure gypsum.

7 0

27 days ago

What is the mass of 22.4 L of H2 at STP?

eduard [2520]

A. 1.01 is the accurate result

Because

The formula used is Pv= nRT

P=1 atm

V= 22.4 L

N= x

R= 0.0821

T= 273 K (since it’s standard temperature)

Thus, (1)(22.4)=(x)(0.0821)(273)

X= 1.001

7 0

1 month ago

Read 2 more answers

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 [2520]

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

A gas sample of argon, maintained at constant temperature, occupies a volume of 500. l at 4.00 atm. what is the new volume if th

lorasvet [2529]

Boyle's law describes the relationship between gas pressure and volume.
It asserts that at a constant temperature, pressure is inversely proportional to gas volume.
PV = k
where P represents pressure, V denotes volume, and k is a constant.
P1V1 = P2V2
where the parameters for the initial condition are on the left, and the parameters for the second condition appear on the right side of the formula.
By substituting values into the equation: 4.00 atm x 500 L = 8.0 atm x V
V calculates to 250 L.
Thus, the new volume becomes 250 L.

6 0

1 month ago

Read 2 more answers