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2 years ago

Does temperature have the same effect on gases as it does on liquids when it comes to viscosity? Explain.

Chemistry

1 answer:

tensa zangetsu [6.8K]2 years ago

4 0

Answer: Explanation:

With an increase in temperature, there is typically an increase in the molecular interchange as molecules move faster in higher temperatures. The gas viscosity will increase with temperature. ... With high temperatures, viscosity increases in gases and decreases in liquids, the drag force will do the same.

The increase in temperature causes the kinetic or thermal energy to increase and the molecules become more mobile. The attractive binding energy is reduced and therefore the viscosity is reduced.

liquids show a reduction in viscosity with increasing temperature. With high temperatures, viscosity increases in gases and decreases in liquids, the drag force will do the same.

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The Planet Venus is surrounded by a thick layer of gases. In fact the atmospheric preassure on Venus is over 90 times grater tha

SCORPION-xisa [38]

B. water will boil at a lower temperature on Venus than on Earth.
Explanation:
Because temperature and pressure are directly proportional... increase in pressure will increase in the atmospheric temperature.... Venus has higher pressure than earth that also means it has higher atmospheric temperature and as the temperature is already high the water will require lesser heat to boil

7 0

2 years ago

Brainliest Available! Thank you in advance!

liraira [26]

Answer:

3.0×10⁻¹³ M

Explanation:

The solubility product Ksp is the product of the concentrations of the ions involved. This relation can be used to find the solubility of interest.

<h3>Equation</h3>

The power of each concentration in the equation for Ksp is the coefficient of the species in the balanced equation.

Ksp = [Al₃⁺³]×[OH⁻]³

<h3>Solving for [Al₃⁺³]</h3>

The initial concentration [OH⁻] is that in water, 10⁻⁷ M. The reaction equation tells us there are 3 OH ions for each Al₃ ion. If x is the concentration [Al₃⁺³], then the reaction increases the concentration [OH⁻] by 3x.

This means the solubility product equation is ...

Ksp = x(10⁻⁷ +3x)³

For the given Ksp = 3×10⁻³⁴, we can estimate the value of x will be less than 10⁻⁸. This means the sum will be dominated by the 10⁻⁷ term, and we can figure x from ...

3.0×10⁻³⁴ = x(10⁻⁷)³

Then x = [Al₃⁺³] will be ...

$[\text{Al}_3^{\,+3}]=\dfrac{3.0\times10^{-34}}{10^{-21}}\approx \boxed{3.0\times10^{-13}\qquad\text{moles per liter}}$

We note this value is significantly less than 10⁻⁷, so our assumption that it could be neglected in the original Ksp equation is substantiated.

<em>Additional comment</em>

The attachment shows the solution of the 4th-degree Ksp equation in x. The only positive real root (on the bottom line) rounds to 3.0×10^-13.

6 0

1 year ago

Read 2 more answers

What is the [h3o+] in a solution that consists of 1.0 m nh3 and 2.5 m nh4cl? [kb (nh3) = 1.8 ×10-5]a) 1.1 × 10-5

Archy [21]

Answer:

[H₃O⁺] = 1.4 × 10⁻⁹ M.

Explanation:

NH₄Cl is a salt that dissolves well in water. The 2.5 M NH₄Cl will give an initial NH₄⁺ concentration of 2.5 M.

NH₃ is a weak base. It combines with water to produce NH₄⁺ and OH⁻. The opposite process can also take place. NH₄⁺ combines with OH⁻ to produce NH₃ and H₂O. The final H₃O⁺ concentration can be found from the OH⁻ concentration. What will be the final OH⁻ concentration?

Let the increase in OH⁻ concentration be x. The initial OH⁻ concentration at room temperature is 10⁻⁷ M.

Construct a RICE table for the equilibrium between NH₃ and NH₄⁺:

$\begin{array}{c|ccccccc}\text{R}&\text{NH}_3 &+&\text{H}_2\text{O}&\rightleftharpoons &{\text{NH}_4}^{+}&+&\text{OH}^{-}\\\text{I}&1.0&&&&2.5&&1.0\times 10^{-7}\\\text{C}& -x &&&& +x &&+x\\\text{E} &1.0 - x &&&&2.5+x&&1.0\times 10^{-7}+x\end{array}$ .

The $\text{K}_b$ value for ammonia is small. The value of x will be so small that at equilibrium, $1.0 - x \approx 1.0$ and $2.5- x \approx 2.5$ .

$\displaystyle \text{K}_b = \frac{[{\text{NH}_4}^{+}]\cdot [{\text{OH}}^{-}]}{[\text{NH}_3]} \approx \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0}$ .

$\displaystyle \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0} = 1.8\times 10^{-5}$ .

$\displaystyle [\text{OH}^{-}] = x+1.0\times 10^{-7} = 1.8\times 10^{-5} /\left(\frac{2.5}{1.0}\right) = 7.2\times 10^{-6}\;\text{mol}\cdot\text{L}^{-}$ .

Again, $\text{K}_w = 1.0\times 10^{-14}$ at room temperature.

$\displaystyle [\text{H}_3\text{O}^{+}] = \frac{\text{K}_w}{[\text{OH}^{-}]}=\frac{1.0\times 10^{-14}}{7.2\times 10^{-6}} = 1.4\times 10^{-9} \;\text{mol}\cdot\text{L}^{-1}$

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2 years ago

For water to change from a liquid to solid, thermal energy needs to be <br><br> .

telo118 [61]

answer: dispersed from the liquid so cold air can take its place

6 0

2 years ago

Use the Periodic Table of the Elements to answer

Juliette [100K]

Answer:

b. lithium

Explanation:

Li the least likely, to lose an electron.

6 0

3 years ago