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3 years ago

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

1 answer:

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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Whick ions profuce similar colors in the flame tests?

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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:

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

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

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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