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2 years ago

Three blocks of the same mass are placed in front of you. Block A has a volume of

Chemistry

1 answer:

Ivan2 years ago

8 0

Given :

Three block of same mass name A , B and C .

Sides of block A , B and C is 3.0 cm , 5.0 cm and 10.0 cm .

To Find :

Which block has the higher density .

Solution :

We know , density $\rho$ is given by :

$\rho=\dfrac{M}{V}$ ......1 )

Here , V is volume .

Now , volume V for cube is given :

$V=a^3$ .......2 )

( Here , a is the side of cube )

Now ,form equation 1 we can see if mass remains constant then density decrease with increase in volume.

Therefore , cube with minimum side will have higher density , in this case it is 3 cm block .

Hence , this is the required solution .

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A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contain

krek1111 [17]

Answer:

- 0.0249% Sb/cm

$-1.2465 * 10^9 \frac{atoms}{cm^3.cm}$

Explanation:

Given that:

One surface contains 1 Sb atom per 10⁸ Si atoms and the other surface contains 500 Sb atoms per 10⁸ Si atoms.

The concentration gradient in atomic percent (%) Sb per cm can be calculated as follows:

The difference in concentration = $\delta_c$

The distance $\delta_x$ = 0.2-mm = 0.02 cm

Now, the concentration of silicon at one surface containing 1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per 10⁸ silicon atoms can be calculated as follows:

$\frac{\delta_c}{\delta_c} = \frac{(1/10^8 -500/10^8)}{0.02cm} *100%$

= - 0.0249% Sb/cm

b) The concentration $(c_1)$ of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:

$c_1 = \frac{(8 si atoms/unit cells)(1/10^3)}{(lattice parameter)^3/unit cell}$

Lattice parameter = 5.4307 Å; To cm ; we have

= $5.4307A^0* \frac{10^{-8}cm}{ A^0}$

$c_1 = \frac{(8 si atoms/unit cells)(1/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}$

= $0.00499*10^{17}atoms/cm^3$

The concentration $(c_2)$ of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:

$c_1 = \frac{(8 si atoms/unit cells)(500/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}$

= $\frac{4*10^{-3}}{1.601*10^{-22}}$

= $2.4938*10^{17}atoms/cm^3$

Finally, to calculate the concentration gradient

$(\frac{\delta _c}{\delta_ x}) = \frac{c_1-c_2}{\delta_x}$

$(\frac{\delta _c}{\delta_ x}) = \frac{0.00499*10^{17}-2.493*10^{17}}{0.02}$

$= -1.2465 * 10^9 \frac{atoms}{cm^3.cm}$

8 0

2 years ago

A tank holds 6.90 m3 of water. What is the volume in ft3? (3.28 ft = 1 m, Remember significant figures, but this answer does not

AleksAgata [21]

Volume of tank = $6.9 m^{3}$ (given)

Since, $1 m = 3.28 ft$

So,

$1 m^{3} = 35.287 ft^{3}$

For $6.9 m^{3}$ :

$6.9\times 35.287 ft^{3} = 243.4803 ft^{3}$

The significant rule for multiplication, states that the number of significant figures in the answer obtained by multiplication is determined by the value with the lowest number of significant digits.

Since, the minimum number of decimal places in the above multiplication operation is 1 so, the final result must be upto 1 decimal place only.

$243.4803 ft^{3} \simeq 243.5 ft^{3}$

Hence, volume in $ft^{3}$ is 243.5.

5 1

2 years ago

Read 2 more answers

The average concentration of bromde ion in seawater is 65 mg of bromide ion per kg of seawater. what is the molarity of the brom

allsm [11]

Usually concentrations are expressed as molarity, or moles of solute per liter solution. First, convert the mass of bromide ion to moles. The molar mass of bromine is 79.904 g/mol.

Moles of bromine = 65 mg * 1 g/1000 mg * 1 mol/79.904 g = 8.135×10⁻⁴ moles

Next, convert the mass of seawater to volume using the density.

Volume of seawater = 1 kg * 1 m³/ 1,025 kg * 1000 L/1 m³ = 0.976 L

Thus,
Molarity = 8.135×10⁻⁴ moles/0.976 L = 8.335×10⁻⁴ M

5 0

2 years ago

Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because

olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$