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-Dominant- [34]
2 years ago
7

How is scientific observation done properly and safely using laboratory equipment?

Chemistry
1 answer:
FrozenT [24]2 years ago
6 0
Using the proper protective equipment from goggles to lab coat and gloves. Also includes ventilation fan to prevent toxic fumes and emergency eyewash station
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Water is heated to boiling and water vapor (steam) is released. Chemical change or physical change?
Vlad [161]
Physical because it is still H2O
6 0
3 years ago
The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?
vazorg [7]
C + O2= CO2
n  =  \frac{m}{mw}
n =  \frac{5.4}{12}  \\ n  = 0.45 \: mol \: of \: carbon
n =  \frac{13.6}{12 + 16 \times 2} \\ n =  \frac{13.6}{44}  \\ n = 0.31 \: mol \: of \: carbon \: dioxide
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %
4 0
3 years ago
Color is an example of
Nuetrik [128]

Answer:

a physical property

Explanation:

Color does not change state.

- Hope that helped! Please let me know if you need further explanation.

3 0
2 years ago
Lithium reacts with oxygen in a synthesis reaction. How much lithium is needed to react with 2.6 g of oxygen?
Vikentia [17]

Answer: A

Explanation:

i just did this test

5 0
2 years ago
Does anyone know the answer to these?
Rashid [163]

Answer:

The answer to your question is below

Explanation:

Data

mass of CaCO₃ = 155 g

mass of HCl = 250 g

mass of CaCl₂ = 142 g

reactants = CaCO₃ + HCl

products = CaCl₂ + CO₂ + H₂O

1.- Balanced chemical reaction

             CaCO₃ + 2HCl   ⇒    CaCl₂ + CO₂ + H₂O

2.- Limiting reactant

molar mass of CaCO₃ = 40 + 12 + 48 = 100 g

molar mass of HCl = 2[1 + 35.5 ] = 73 g

theoretical proportion CaCO₃ /HCl = 100 / 73 = 1.37

experimental proportion CaCO₃ /HCl = 155 / 250 = 0.62

As the experimental proportion was lower than the theoretical proportion the limiting reactant is CaCO₃

3.-

Calculate the molar mass of CaCl₂

CaCl₂ = 40 + 71 = 111 g

          100 g of CaCO₃ ------------------ 111 g of CaCl₂

           155 g of CaCO₃ ----------------- x

               x = (155 x 111) / 100

               x = 17205 / 100

              x = 172.05 g of CaCl₂

4.- percent yield

Percent yield = 142 / 172.05 x 100 = 82.5 %

5.- Excess reactant

    100 g of CaCO₃  -------------------- 73 g of HCl

     155 g of caCO₃ ------------------- x

           x = (155 x 73)/100

           x = 133.15 g

Mass of HCl = 250 - 133.15

                    = 136.9 g

4 0
2 years ago
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