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3 years ago

11

Write the position of metals, non - metals and metalloids in the periodic table.

1 answer:

Nostrana [21]3 years ago

3 0

Answer:

The metals are to the left of the line (except for hydrogen, which is a nonmetal), the nonmetals are to the right of the line, and the elements immediately adjacent to the line are the metalloids.

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(b) The mass of a single molybdenum atom is 1.59×10-22 g. Suppose enough Mo atoms were lined up like beads on a string to span a

serious [3.7K]

Answer:N=1151724138 atoms

Explanation:The atomic radius of molybdenum atom is 145 picometers so lining up molybdenum atoms is like lining up the beads of radius 145 picometers

so one bead would cover the length of 145*2=290 picometers

hence the atom's diameter =D=290 picometers

D=290 *10^(-12)

$D=290/1000000000000$

also

Distance span= 33.4 cm

Distance span= 0.334m

As the Distance span adds up with no of atoms the resulting formula is

Distance span by atoms= No of atoms * atom's diameter

so

$N=\frac{Distance Spaned}{Atom's Diameter } \\N=0.334/(0.000000000290)\\N=1151724138 atoms$

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3 years ago

¿Cómo se clasifican las disoluciones acuosas?

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2 years ago

Calculate the mass percent composition for Al(HCO3)3

ZanzabumX [31]

Answer:

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2 years ago

What property of these items do cm or mm measure: length, volume, mass or temperature?

UNO [17]

Answer:

length

Explanation:

cm measures length. Think of a ruler.

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2 years ago

How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with

konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = $a=31 cm^2$

Thickness of the gold plating = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

$V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3$

Density of the gold = $d=19.3 g/cm^3$

$m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g$

Moles of gold = $\frac{29.915 g}{197 g/mol}=0.152 mol$

$Au^{3+}+3e^-\rightarrow Au$

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

$\frac{3}{1}\times 0.152 mol=0.456 mol$ of electrons

Number of electrons = N = $0.456\times \times 6.022\times 10^{23}$

Charge on single electron = $q=1.6\times 10^{-19} C$

Total charge required = Q

$Q=N\times q$

Amount of current passes = I = 8 Ampere

Duration of time = T

$I=\frac{Q}{T}$

$T=\frac{N\times q}{I}$

$=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s$

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

7 0

2 years ago