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Marta_Voda [28]
3 years ago
8

Types OF SALTStwo examples​

Chemistry
1 answer:
meriva3 years ago
8 0
Sea salt and rock salt, rock salt is gathered from salt deposits found underground, sea salt is obviously from the sea :/

Hope this helps you
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The work function of a silver surface is 4.73 eV. What would happen if the silver is hit with a photon with 2.5 eV of energy?
DerKrebs [107]

Answer:Nothing, the photon just bounces off the surface.

Explanation:

According to Albert Einstein, a photoelectron can only be emitted from a metal surface when the energy of the incident photon is greater than the work function of the metal.

In the scenario described in the question, the work function of the metal is greater than the energy of the photon. Hence, the photon just bounces off the metal surface without emitting any electron.

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4 years ago
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Nonmetal halogen family atmoic mass 35
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3 years ago
A 250 ml solution of 2.0 M NaOH is diluted to 1.0 liter. What is the final concentration of the solution?
Alex73 [517]

Answer:

Diluted concentration is 0.5M

Explanation:

Let's solve this with rules of three, although there is a formula to see it easier

In 1000 mL (1L), we have 2 moles of NaOH

In 250 mL we must have (250 . 2) / 1000 = 0.5 moles of NaOH

These moles will be also in 1 L of the final volume of the diluted solution

More easy:

1 L of solution has 0.5 moles of NaOH

Then, molarity is 0.5 M

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3 0
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Picture attached, $3
Luba_88 [7]

Answer:

its -000.5

Explanation:

6 0
3 years ago
2. A quantity of 1.922g of methanol (CH3OH) was burned in a constant-volume
Cerrena [4.2K]
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
5 0
3 years ago
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