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3 years ago

The sea of electrons in metals acts as

Chemistry

1 answer:

Marrrta [24]3 years ago

5 0

The answer is a bonding agent

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What is the new boiling point if 25 g of NaCl is dissolved in 1.0 kg of water

Inessa05 [86]

The correct answer for the question that is being presented above is this one:

Given that:
delta Tb = Kbm Kb H2O = 0.52 degrees C/m
delta Tf = Kfm Kf H2O = 1.86 degrees C/m

We need to know the formula for Molality.
molality = mol solute / kg solvent

We are given the amount of solute in grams
Since amount of solute is given in moles, we have to convert 25 g NaCl to moles. Divide by molar mass. 

25 g NaCl / 58.5 g/mol = 0.427 mol 

Then, use the formula for molality. 

molality = mol solute / kg solvent 
= 0.427 / 1 
= 0.427 m 

Use now the formula to get the boiling point.

delta Tb = Kbm 
= (0.52)(0.427) 
= 0.22C

8 0

3 years ago

When dilute HCL is added to zinc pieces taken in a test tube

Romashka-Z-Leto [24]

Answer:

zince chloride will be formed and hydrogen gas will be librated.

Explanation:

When dilute HCl is added to zinc pieces, a rection will takes place as follows :

$Zn+HCl\rightarrow ZnCl_2+H_2$

It means that zinc chloride will form when zinc reacts with dilute HCL. Also hydrogen gas will produced.

As zinc is more reactive than hydrogen, it displaces hydrogen from its solution and forms zinc chloride. The form product is white in color and H₂ is an odorless gas.

Hence, zince chloride will be formed and hydrogen gas will be librated.

6 0

3 years ago

What is the total probability of finding a particle in a one-dimensional box in level n = 4 between x = 0 and x = L/8?

Lubov Fominskaja [6]

Answer:

P = 1/8

Explanation:

The wave function of a particle in a one-dimensional box is given by:

$\psi = \sqrt \frac{2}{L} sin(\frac{n \pi x}{L})$

Hence, the probability of finding the particle in the one-dimensional box is:

$P = \int_{x_{1}}^{x_{2}} \psi^{2} dx$

$P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx$

$P = \frac{2}{L} \int_{x_{1}}^{x_{2}} (sin^{2}(\frac{n \pi x}{L}) dx$

Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:

$P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]$

$P = \frac{1}{8} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})$

Solving for n=4:

$P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi})$

$P = \frac{1}{8} (1 - \frac{sin (\pi)}{\pi})$

$P = \frac{1}{8}$

I hope it helps you!

7 0

3 years ago

The brain tells his stomach intestines and liver want to be active and want to rest. Messages to and from the brain decide when

Nana76 [90]

Answer:

Nervous System I believe

Explanation:

5 0

3 years ago

quien y en que cantidad sera el reactivo limite, si utilizamos 125 g de ácido, H (CH3COO) y 275 g de hidróxido , Al(OH)3

Deffense [45]

Answer:

The limiting reactant is acetic acid. All 125 g will react.

Explanation:

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

Mᵣ: 60.05 78.00

3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O

Mass/g: 125 275

2. Calculate the moles of each reactant

$\text{Moles of CH$_{3}$COOH} = \text{125 g CH$_{3}$COOH} \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{60.05 g CH$_{3}$COOH}} = \text{2.082 mol CH$_{3}$COOH }\\\\\text{Moles of Al(OH)}_{3} = \text{275 g Al(OH)}_{3} \times \dfrac{\text{1 mol Al(OH)}_{3}}{\text{78.00 g Al(OH)}_{3}} = \text{3.526 mol Al(OH)}_{3}$

3. Calculate the moles of (CH₃COO)₃Al from each reactant

$\textbf{From CH$_{3}$COOH:}\\\text{Moles of (CH$_{3}$COO)$_{3}$Al} = \text{2.082 mol CH$_{3}$COOH} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al}}{\text{3 mol CH$_{3}$COOH}}\\\\= \text{0.6939 mol (CH$_{3}$COO)$_{3}$)Al}\\\textbf{From Al(OH)}_{3}:\\\text{Moles of (CH$_{3}$COO)$_{3}$Al } =\text{3.526 Al(OH)}_{3} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al }}{\text{1 mol Al(OH)}_{3}}\\\\= \text{3.526 mol (CH$_{3}$COO)$_{3}$Al}$

$\text{Acetic acid is the $\textbf{limiting reactant}$ because it gives fewer moles of} \\\text{(CH$_{3}$COO)$_{3}$Al. All $\textbf{125 g}$ will react.}$

5 0

3 years ago