All categories

1 year ago

How do chemists model the valence electrons in a metal

Chemistry

1 answer:

Bumek [7]1 year ago

5 0

Answer:

the valence electrons of atoms in a pure metal can be modeled as a sea of electrons

You might be interested in

Which state of matter had the greatest distance the individual particles?

mylen [45]

Particles in gaseous state have the greatest distance between them. This is because the gas particles have a greater kinetic energy because of which they move far apart from each other. If we take water, in steam (water vapor) the particles are far apart from each other when compared to liquid water and solid ice.

6 0

2 years ago

Balance the following equations:

neonofarm [45]

Hope this helps! If you dont understand balancing equations in general, say so in the comments, I’m happy to help

7 0

3 years ago

How many moles of sodium (Na) are there in a<br> sample of 5.87 x 1024 atoms of sodium?

Leno4ka [110]

9.74x 2351 that's the answer

6 0

3 years ago

If you want to use a serial dilution to make a 1/50 dilution. The first dilution you make is a 1/5 dilution with a total volume

uranmaximum [27]

Answer:

0.2 mL stock solution, 0.8 solvent, 0.1 mL first solution and 0.9 solvent

Explanation:

The final volume for fist solution is 1 mL and concentration must will be 1/5, then 1 mL/5=0.2 mL. For complete the 1 mL add the missing solvent volume 1 mL-0.2 mL=0.8 mL. For second solution, assuming final volume is 1 mL, and concentration 1/10, then we have 1 mL /10=0.1 mL solution 1/5. Completing volume, 1 mL-0.1 mL= 0.9 mL solvent.

7 0

2 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

7 0

2 years ago