All categories

12 hours ago

For which applications would you choose a liquid over a gas or solid?

Chemistry

2 answers:

Alekssandra [968]12 hours ago

5 0

Solution:

Washing Clothes & Dissolving Sugar

Clarification:

Consider each scenario:

1) For washing clothes, water is essential; without it, washing is ineffective.

2) Connecting brake pedals to brake pads requires solids, not liquids.

3) To deodorize a space, one would likely reach for an aerosol spray, which is a gas.

4) Sculpting involves solid tools and a solid medium.

5) Dissolving sugar necessitates a liquid to be effective!

6) While one might assert that paint is a liquid, it still might not fit the category; I would categorize this application as solid.

7) Gears employed in machinery are solid components!

alisha [964]12 hours ago

4 0

Solution:

The suitable applications include washing clothes, connecting a brake pedal to the brake pads on a vehicle, dissolving sugar, and painting a wall.

Clarification:

The response at the top marked as expert verified is incorrect, as my choices were wrongly evaluated.

Hope this assists! Please show support and mark as the best answer!

You might be interested in

At what temperature would the volume of a gas be 0.550 L if it had a volume of 0.432 L at –20.0 o C?

castortr0y [923]

To find the temperature at which the volume of the gas would be 0.550 L, given that it is 0.432 L at -20.0 °C, apply Charles’s Law.

The formula is v1/T1 = v2/T2
Known values:
V1 = 0.550 L
T1 = ?
T2 = -20°C + 273 = 253 K
V2 = 0.432 L

Rearranging for T1:
T1 = (V1 × T2) / V2

Calculating:
T1 = (0.55 L × 253) / 0.432 L = 322.11 K or 49.11°C

8 0

13 days ago

Read 2 more answers

Combustion analysis of an unknown compound containing only carbon and hydrogen produced 0.2845 g of co2 and 0.1451 g of h2o. wha

VMariaS [1037]

CxHy + (x+0.25)O₂ → xCO₂ + 0.5yH₂O

m(CO₂)/{xM(CO₂)}=m(H₂O)/{0.5yM(H₂O)}

0.2845/{44.01x}=0.1451/{9.01y}

x/y=0.4=2:5

The empirical formula is C₂H₅.

7 0

20 hours 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

A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a

KiRa [971]

Answer:

There are 5.5668 moles of water for every mole of CuSO₄.

Explanation:

The mass of anhydrous CuSO₄ is:

23.403g - 22.652g = 0.751g.

mass of crucible + lid + CuSO₄ - mass of crucible + lid

Given that the molar mass of CuSO₄ is 159.609g/mol, we calculate the moles:

0.751g × $\frac{1mol}{159,609g}$ = 4.7052x10⁻³ moles CuSO₄

The mass of water in the initial sample is:

23.875g - 0.751g - 22.652g = 0.472g.

mass of crucible + lid + CuSO₄ hydrate - CuSO₄ - mass of crucible + lid

As the molar mass of H₂O is 18.02g/mol, we find the moles:

0.472g × $\frac{1mol}{18,02g}$ = 2.6193x10⁻² moles H₂O

The mole ratio of H₂O to CuSO₄ is:

2.6193x10⁻² moles H₂O / 4.7052x10⁻³ moles CuSO₄ = 5.5668

This indicates there are 5.5668 moles of water per mole of CuSO₄.

I hope this is helpful!

5 0

8 days ago

Find the age ttt of a sample, if the total mass of carbon in the sample is mcmcm_c, the activity of the sample is AAA, the curre

castortr0y [923]

Answer:

Explanation:

In a desert cave, an artifact has been discovered. The anthropologists investigating this artifact want to determine its age. They note that the current activity level of the artifact is 9.25 decays/s, and the carbon mass present is 0.100 kg. To ascertain the artifact's age, they will employ specific constants:

r=1.2

The formula for carbon 14 activity is

$A=A_0e^{\lambda t}$

where,

$A_0$ is the initial activity of the substance

Now, solve for t

$-\lambda t=In\frac{A}{A_0}$

$t=-\frac{1}{\lambda} In(\frac{A}{A_0} )$

$=-\frac{1}{\lambda} In(\frac{A}{\lambda r(\frac{m_c}{m_a} )} )$

since,

$A_0=\lambda r(\frac{m_c}{m_a} )$

$=-\frac{1}{\lambda} In(\frac{A\ m_a}{\lambda r m_c} )$

Thus, the age of the artifact is

$=-\frac{1}{\lambda} In(\frac{A\ m_a}{\lambda r m_c} )$

$=-\frac{1}{1.21\times 10^{-4}} In(\frac{(9.25)(2.32\times 10^{-26}}{1.21\times 10^{-4}(\frac{1}{3.15569\times10^7} )(1.2\times 10^{-12})(0.100)}} )\\\\=6303.4 \ years$

to two significant figures = 6300 years

4 0

6 days ago