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neonofarm [45]
2 years ago
14

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

m 1.40 kg of foil?
Chemistry
1 answer:
horrorfan [7]2 years ago
5 0

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)

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if you mixed 72.9g hydrochloric acid(aq) with 150g silver acetate(aq), what would be the limiting reagent?​
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Answer:

                      Silver Acetate would be the Limiting Reagent.

Explanation:

                    The balance chemical equation for the given double displacement reaction is as;

                            HCl + AgC₂H₃O₂ → AgCl + HC₂H₃O₂

Step 1: <u>Calculate Moles of Starting Materials:</u>

Moles of HCl:

                      Moles  =  Mass / M.Mass

                      Moles  =  72.9 g / 36.46

                      Moles =  1.99 moles

Moles of AgC₂H₃O₂:

                      Moles  =  150 g / 166.91 g/mol

                      Moles  =  0.898 moles

Step 2: <u>Find out Limiting reagent as:</u>

According to balance chemical equation.

              1 mole of HCl reacts with  =  1 mole of AgC₂H₃O₂

So,

         1.99 moles of HCl will react with  =  X moles of AgC₂H₃O₂

Solving for X,

                     X =  1.99 mol × 1 mol / 1 mol

                     X =  1.99 mol of AgC₂H₃O₂

Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.

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3 years ago
How many molecules are in 1 mol of the chemical equation shown above
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5 0
3 years ago
Read 2 more answers
(Please help, ASAP) How many grams are in 3.45x10^23 atoms of P?
Daniel [21]

one mole of P weights about 31 grams

in one mole there are 6.022*10^23 atoms

we use the rule of threes

6.022*10^23atoms......weight..........31 grams

3.45*10^23 atoms.........weight...........x grams

x=(3.45*10^23*31)/6.022*10^23

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2 years ago
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