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sweet-ann [11.9K]
3 years ago
15

Which of the following laboratory tools would be most appropriate for determining the volume of an irregular shape?

Chemistry
1 answer:
Verdich [7]3 years ago
7 0
For an irregularly shaped objects, we use a GRADUATED CYLINDER to determine its volume by water displacement method. We fill the glassware first with water then measure the volume. Next, drop the object and read off the new volume. Calculate the difference between the volumes. 
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Ba2+ and Cu2+, and Sr2+ and Li+

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3 years ago
PLEASE HELP!! GIVING BRAINLIEST
RSB [31]

Answer:

The answer is "0.00172172603".

Explanation:

Given:

Mass (M) = 1.60 \times  10^{-3} \ g\\\\Density (D) = 9.293 \times 10^{-1} \ \frac{g}{cm^3}\\\\Volume (V) =  ?

Formula:

\to \bold{V = \frac{M}{D}}

       = \frac{1.60 \times  10^{-3}}{9.293 \times  10^{-1}} \\\\  = \frac{1.60 \times  10^{1}}{9.293 \times  10^{3}} \\\\ = \frac{1.60 \times  10}{9.293 \times  1000} \\\\ = \frac{1.60 }{9.293 \times  100} \\\\ = \frac{1.60 }{929.3 } \\\\= 0.00172172603

4 0
3 years ago
8.632 nm + 8.3 nm - 30.0 nm=
Misha Larkins [42]

Answer:

13.06800 nanometers

Explanation:

i hope you had understand !

5 0
3 years ago
12) _Na+ _H2O → ___ NaOH + __H2
TEA [102]

Answer:

Explanation:

2 Na+ 2 H2O →  2 NaOH + 1 H2

4 0
3 years ago
When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II)
katen-ka-za [31]

Answer:

3.81 g Pb

Explanation:

When a lead acid car battery is recharged, the following half-reactions take place:

Cathode: PbSO₄(s) + H⁺ (aq) + 2e⁻ → Pb(s) + HSO₄⁻(aq)  

Anode: PbSO₄(s) + 2 H₂O(l) → PbO₂(s) + HSO₄⁻(aq) + 3H⁺ (aq) + 2e⁻

We can establish the following relations:

  • 1 A = 1 c/s
  • 1 mole of Pb(s) is deposited when 2 moles of e⁻ circulate.
  • The molar mass of Pb is 207.2 g/mol
  • 1 mol of e⁻ has a charge of 96468 c (Faraday's constant)

Suppose a current of 96.0A is fed into a car battery for 37.0 seconds. The mass of lead deposited is:

37.0s.\frac{96.0c}{s} .\frac{1mole^{-} }{96468c} .\frac{1molPb}{2mole^{-} } .\frac{207.2gPb}{1molPb} =3.81gPb

6 0
3 years ago
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