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17 days ago

1. Adakah benar bahawa anda tidak boleh membasuh rambut, meminum air sejuk dan

Chemistry

1 answer:

KiRa [976]17 days ago

7 0

Answer:

Is it true that you shouldn't wash your hair, drink cold water, eat ice cream, or exercise during your period? Please explain your answer.

No, this is not accurate; doing any of these activities is perfectly fine. None of them affects us because they are not connected to our bodily systems. Also, I apologize for any language errors as I utilized Google Translate.

I hope this is helpful :)

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"The compound K2O2 also exists. A chemist can determine the mass of K in a sample of known mass that consists of either pure K2O

lorasvet [960]

Answer:

Indeed, the chemist is capable of identifying the compound present in the sample.

Explanation:

In one mole of K₂O, potassium has a mass of 2 × 39.1 g = 78.2 g, while the total mass of K₂O is 94.2 g. The mass ratio of K compared to K₂O is calculated as 78.2 g / 94.2 g = 0.830.

For 1 mole of K₂O₂, potassium's mass remains the same at 78.2 g, but the total mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ then equates to 78.2 g / 110.2 g = 0.710.

When the chemist measures the mass of K in relation to the overall sample, the mass ratio can be computed.

If the mass ratio is 0.830, then it indicates a pure K₂O compound.
If the mass ratio is 0.710, it indicates a pure K₂O₂ compound.
If the mass ratio falls outside of 0.830 or 0.710, the sample is assessed to be a mixture.

6 0

14 days ago

A 3.81-gram sample of NaHCO3 was completely decomposed in an experiment. 2NaHCO3 → Na2CO3 + H2CO3 In this experiment, carbon dio

Anarel [852]

Answer:

The yield percentage of H_2CO_3 is 24.44%

Explanation:

5 0

12 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 [944]

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

4 days ago

To save time you can approximate the initial volume of water to ±1 mL and the initial mass of the solid to ±1 g. For example, if

castortr0y [927]

Answer:

The correct options include choice 2, 3, and 6.

Explanation:

Density is identified as the mass of a substance per unit volume occupied by that substance.

$Density=\frac{Mass}{Volume}$

The density remains constant for a given substance, regardless of variations in mass and volume hence it is considered an intensive property.

2. 20.2 g of silver in 21.6 mL of water and 12.0 g of silver also in 21.6 mL of water.

3. 15.2 g of copper in 21.6 mL of water and 50.0 g of copper in 23.4 mL of water.

6. 11.2 g of gold in 21.6 mL of water and 14.9 g of gold in 23.4 mL of water.

The same metals in both instances will yield consistent densities due to the fixed density of the metal.

7 0

3 days ago

How many protons neutrons and electrons are there in a neutral atom of 43k (potassium-43)?

castortr0y [927]

Answer:

Protons: 19

Neutrons: 25

Electrons: 19

Explanation:

Protons:

The atomic number determines the number of protons in an atom. Consequently, with Potassium's atomic number being 19, it contains 19 protons.

Neutrons:

The formula to find neutrons is:

# of Neutrons = Atomic Mass - # of Protons

Given:

Atomic Mass = 43

# of Protons = 19

Thus,

# of Neutrons = 43 - 19

# of Neutrons = 24

Electrons:

In a neutral atom, the quantity of electrons matches that of protons. Therefore, a neutral Potassium atom with 19 protons must equally have 19 electrons.

3 0

19 hours ago