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gogolik [260]
3 years ago
14

How many grams of fluorine must be reacted with excess lithium iodide to produce 10.0 grams of lithium fluoride?

Chemistry
1 answer:
rewona [7]3 years ago
8 0
Answer:
             7.32 g of F₂

Solution:
              The equation is as follow,

                                   2 LiI  +  F₂    →    2 LiF  +  I₂

According to equation,

           51.88 g (2 mole) of LiF is produced from  =  37.99 g (1 mole) F₂
So,
                          10 g of LiF will be produced by  =  X g of F₂

Solving for X,
                      X  =  (10 g × 37.99 g) ÷ 51.88 g

                      X  =  7.32 g of F₂
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Pls help i beg
marusya05 [52]

Answer:

The increasing order of conductivity is O< Ge< Mn.

Explanation:

Electrical conductivity is defined as the measure of the ability of a material to conduct electrical current through it. The conductivity depends on the atomic and molecular structure of the material.

Metals are good conductors because they have a structure with many electrons with weak bonds, and this allows their movement instead non-metals have between four and eight valence electrons, which lack this tendency.

The conductivity increases in the periodic table from top to bottom and from right to left.

oxygen is a nonmetal therefore it is a bad conductor.

Germanium is a metalloid whose conductivity is greater than a nonmetal and worst than a metal.

Manganese is a metal,in this case, it is a good conductor.

4 0
2 years ago
Is a Big Mac an element, compound, or mixture?
Lunna [17]
It is a mixture. it can easily be separated. 
4 0
3 years ago
A student prepared a stock solution by dissolving 10.0 g of KOH in enough water to make 150. mL of solution. She then took 15.0
yanalaym [24]

Answer: The concentration of KOH for the final solution is 0.275 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

 V_s = volume of solution in ml = 150 ml

moles of solute =\frac{\text {given mass}}{\text {molar mass}}=\frac{10.0g}{56g/mol}=0.178moles

Now put all the given values in the formula of molality, we get

Molality=\frac{0.178\times 1000}{150}=1.19M

According to the dilution law,

C_1V_1=C_2V_2

where,

C_1 = molarity of stock solution = 1.19 M

V_1 = volume of stock solution = 15.0 ml

C_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 65.0 ml

Putting in the values we get:

1.19\times 15.0=M_2\times 65.0

M_1=0.275M

Therefore, the concentration of KOH for the final solution is 0.275 M

5 0
3 years ago
Rolls of foil are 300mm wide and 2.020mm thick. (The density of foil is 2.7 g/cm^3). What maximum length of foil can be made fro
horrorfan [7]

Answer:  8556 mm, or 855.6 cm (8560 mm to 3 sig figs)

Explanation:  Convert mm to cm by dividing by 10 (1cm/10mm)

Find the area of the foil face in cm^2 (30cm*0.2020cm) = 0.606 cm^2

Calculate the volume occupied by 1.40 kg of foil in cm^3.  1.40kg = 1400g

1.400g/(2.7 g/cm^3) = 518.5 cm^3 for 1.40 kg Au

   Volume = Area (of the face) * Length  

We want Length:

Length = Volume/Area

L = (518.5 cm^3/0.606 cm^2)

L = 855.6 cm (8556 mm)  Round to 3 sig figs (856 cm and 8560 mm)

5 0
2 years ago
if the theoretical yield of a reaction is 26.0 grams and you actually recovered 22.0 grams what is the precent yield
netineya [11]
Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100  

∴ if theoretical yield is 26 g, but only 22.0 is recovered from the reaction, 
then Percentage Yield = (22 g ÷ 26 g) × 100  
                                       =  84.6 %
8 0
3 years ago
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