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sladkih [1.3K]
3 years ago
12

Question 8 (9 points) Match the lab equipment with its purpose

Chemistry
1 answer:
Deffense [45]3 years ago
7 0

Answer:

\sf \boxed{4} \mapsto Pipet

\sf\boxed{7}\mapsto Test \:tube \: rack

\sf\boxed{3}\mapsto Test\: table

\sf\boxed{5}\mapsto Scoopula

\sf\boxed{1}\mapsto Graduated\: cylinder

\sf\boxed{9}\mapsto Bunsen \:burner

\sf\boxed{2}\mapsto Beaker

\sf \boxed{8}\mapsto Spot\: plate

\sf\boxed{6}\mapsto Goggles

Explanation:

Pipet is used to dispense a very small amount of liquid.

Test tube rack is used to hold multiple test tubes at the same time.

Test Table is used to view chemical reactions or hold or heat small amounts of substance.

Scoopula is used to dispense chemicals from a larger container.

Graduated cylinder is used to measure volume very precisely.

Bunsen burner is used to heat objects.

Beaker is used to transport heat or store substance.

Spot plate is used to observe the color changes of small quantities of a reacting mixture.

Goggles are used to protect the eyes from flying objects or chemical splashes.

_____________________________________________________

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What volume of 1.5M NaOH is needed to provide 0.75 mol of NaOH?
lutik1710 [3]
1.5M NaOH so we've 1.5 moles of NaOH in 1L of solution

1L = 1000 ml

1.5 moles of NaOH ------------in------------- 1000 ml
0.75 moles of NaOH ----------in---------------x ml
x = 500 ml

<em><u>answer: C</u></em>
5 0
3 years ago
Choose the covalent compounds from the following choices Br2 MgS SO2 KF
Novosadov [1.4K]

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Covalent Bond

                

                Between 0.4 and 1.7 then it is Polar Covalent Bond

            

                Greater than 1.7 then it is Ionic

 

For Br₂;

                    E.N of Bromine      =   2.96

                    E.N of Bromine      =   2.96

                                                   ________

                    E.N Difference             0.00          (Non Polar Covalent Bond)


For MgS;

                    E.N of Sulfur               =   2.58

                    E.N of Magnesium      =   1.31

                                                   ________

                    E.N Difference                  1.27          (Ionic Bond)


For SO₂;

                    E.N of Oxygen      =   3.44

                    E.N of Sulfur          =   2.58

                                                   ________

                    E.N Difference             0.86          (Polar Covalent Bond)


For KF;

                    E.N of Fluorine          =   3.98

                    E.N of Potassium      =   0.82

                                                   ________

                    E.N Difference                3.16          (Ionic Bond)

Result: The Bonds in Br₂ and SO₂ are Covalent in Nature.

3 0
3 years ago
What is a substance that cannot be broken down further by chemical means
IgorC [24]
Anything can be broken down, as long as it is not as small as an atom
8 0
3 years ago
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Mazyrski [523]

The rate law for this reaction is [A]².

Balanced chemical reaction used in this experiment: A + B → P

The reaction rate is the speed at which reactants are converted into products.

Comparing first and second experiment, there is no change in initial rate. The concentration of reactant B is increased by double. Initial rate does not depands on concentration of reactant B.

Comparing first and third experiment, initial rate is nine times greater, while concentration of reactant A is three times greater. Conclusion is that concentration of reactant A is squared and the rate is [A]².

More info about rate law: brainly.com/question/16981791

#SPJ4

8 0
1 year ago
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Flura [38]

Answer:

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Explanation:

This is what I think that you meant by the question listed. When balancing a chemical equation, you want to make sure that there are equal amounts of each element on each side.

Originally, the equation's elements looked like this: 1 C on left & 1 C on right; 2 H on left & 2 H on right; 2 O on left and 3 O on right. Because these are not balanced, you need to add coefficients.

When adding coefficients, you need to make sure that all of the elements stay balanced, not just one that you are trying to fix. I know that some equations are really difficult to balance, and when that is the case, there are equation balancing websites that can help out.

However, what always helps me is making a chart and continuing to keep up with the changes I am making. It is a trial and error process.

6 0
3 years ago
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