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timofeeve [1]
3 years ago
7

What is the volume of a 3.00M(mol/L) solution made with 122g of LiF?

Chemistry
1 answer:
In-s [12.5K]3 years ago
4 0
Moles is equal to mass over molar mass

122g/25.94 (25.94 is the molar mass of LiF)

3 = (4.703/L)

L = 1.567
You might be interested in
How many photons are produced in a laser pulse of 0.862 J at 691 nm?
Inga [223]
To calculate how many photons are in a certain amount of energy (joules) we need to know how much energy is in one photon.
Start by using two equations:
Energy of a photon = Frequency * Planck's constant (6.626 * 10^(-34) J-s)
Speed of light (constant 3 * 10^8 m/s) = Frequency * Wavelength
Which means:
frequency = Speed of Light / Wavelength
So energy of a photon = (Speed of light * Planck's constant)/(Wavelength)
You may have seen this equation as E = hc/<span>λ</span>
We have a wavelength of 691 nm or 691 * 10^-9 meters
So we can plug in all of our knowns:
E = (6.626 * 10^(-34) J-s) * (3.00 * 10^8 m/s) / (691 * 10^-9 m) = 
2.88 * 10^(-19) joules per photon
Now we have joules per photon, and the total number of joules (0.862 joules)
,so divide joules by joules per photon, and we have the number of photons:
0.862 J/ (2.88 * 10^(-19) J/photon) = 3.00 * 10^18 photons.

4 0
3 years ago
\The specific heat of aluminum is 0.21 cal g°C . How much heat is released when a 10 gram piece of aluminum foil is taken out of
atroni [7]

<u>Answer: </u>The amount of heat released is 84 calories.

<u>Explanation: </u>

The equation used to calculate the amount of heat released or absorbed, we use the equation:

Q= m\times c\times \Delta T

where,

Q = heat gained  or released = ? Cal

m = mass of the substance = 10g

c = specific heat of aluminium = 0.21 Cal/g ° C

Putting values in above equation, we get:

\Delta T={\text{Change in temperature}}=(10-50)^oC=-40^oC  

Q=10g\times 0.21Cal/g^oC\times-40^oC

Q = -84 Calories

Hence, the amount of heat released is 84 calories.

8 0
3 years ago
What are acids? List few substances containing acid.
Gelneren [198K]
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  • <em><u>vinegar</u></em><em><u> </u></em>
  • <em><u>baking</u></em><em><u> </u></em><em><u>soda</u></em><em><u> </u></em>
  • <em><u>soap</u></em><em><u> </u></em>
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3 0
2 years ago
Read 2 more answers
What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2
atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}

Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

Mole of AlCl_3 = Mole of Al * Mole ratio

Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = 133.34 \frac{g}{mol}

Mass of AlCl3 can be calculated using mole formula as:

1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}

Multiplying both side by 133.34 \frac{g}{mol}

1.32 mole  * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34  \frac{g}{mol}

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

Mole of AlCl_3 = mole of Cl_2 * mole ratio

Mole of AlCl_3 = 0.55  mole  * \frac{2}{3}

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}

Multiplying both side by

133.34 \frac{g}{mol}

0.367 mol of AlCl_3 * 133.34 \frac{g}{mol}  = \frac{mass(g)}{133.34\frac{g}{mol}}   * 133.34 \frac{g}{mol}

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0
3 years ago
Read 2 more answers
How many liters are there in 28g of B?
Juliette [100K]
Hi
The volume is 58.1
The moler mass is 10.8
Molar mass details
10.81B (1*10.81)

Hope this helps
5 0
2 years ago
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