These compounds 'share' electrons to become more stable.

Is adding zinc to hydrogen chloride a chemical or physical change

BlackZzzverrR [31]

That is a chemical change.. Hope I helped!

8 0

3 years ago

A chemist makes of magnesium fluoride working solution by adding distilled water to of a stock solution of magnesium fluoride in

mojhsa [17]

The question is incomplete, here is the complete question:

A chemist makes 600. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a stock solution of 0.00154 mol/L magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.

Answer: The concentration of chemist's working solution is $5.90\times 10^{-4}M$

Explanation:

To calculate the molarity of the diluted solution (chemist's working solution), we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the stock magnesium fluoride solution

$M_2\text{ and }V_2$ are the molarity and volume of chemist's magnesium fluoride solution

We are given:

$M_1=0.00154M\\V_1=230mL\\M_2=?M\\V_2=600mL$

Putting values in above equation, we get:

$0.00154\times 230=M_2\times 600\\\\M_2=\frac{0.00154\times 230}{600}=5.90\times 10^{-4}M$

Hence, the concentration of chemist's working solution is $5.90\times 10^{-4}M$

8 0

3 years ago

How many moles of ba(oh)2 are present in 275 ml of 0.200 m ba(oh)2?

DENIUS [597]

You have to put your attention to the unit of concentration. It is expressed in terms of molarity, which is represented in M. It is the number of moles solute per liter solution. So, you simply have to multiply the molarity with the volume in liters.

Volume = 275 mL * 1 L/1000 mL = 0.275 L
Moles Ba(OH)₂ = (0.200 M)(0.275 L) = 0.055 mol

6 0

4 years ago

A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

madreJ [45]

Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

Relation between work function, wavelength, and kinetic energy is as follows.

K.E = $\frac{hc}{\lambda} - \phi$

where, h = Plank's constant = $6.63 \times 10^{-34}$ J.s

c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

From first source,

$K.E_{1}$ = $\frac{hc}{\lambda} - \phi$

1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$

$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

$K.E_{2}$ = $\frac{hc}{\lambda} - \phi$

$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

$6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

3 0

4 years ago

NO GO*GLE ANSWERS PLEASE!

dmitriy555 [2]

Answer:

Tyrel is learning about a certain kind of metal used to make satellites. He learns that infrared light is absorbed by the metal. X-ray light is transmitted through the metal. ... Yes, light can cause the metal to get warm because light carries energy.

4 0

3 years ago