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3 years ago

How is specific gravity or density used in identifying minerals

1 answer:

7 0

<span>In summary, density and specific gravity are properties used to help identify minerals. Density is a measure of the mass of a certain volume of the sample. Specific gravity is a unitless measure, and it is the ratio of the mass of a substance to the mass of an equal volume of water.</span>

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A point charge is placed 100 m from a 6 µC charge generating an electric field. What is the strength of the electric field on th

allochka39001 [22]

Answer:

Explanation:

The answer is 5 on edge.

6 0

4 years ago

Read 2 more answers

onsider 1000 mL of a 1.00 × 10-4 M solution of a certain acid HA that has a Ka value equal to 1.00 × 10-4. Water was added or re

solmaris [256]

Answer:

The volume of the final solution, V = 0.0305L

Explanation:

Number of moles = Concentration * volume

Concentration of HA = 1.00 * 10⁻⁴M

Volume of HA = 1000mL = 1 L

Number of moles of HA = 1.00 * 10⁻⁴ * 1

Number of moles of HA = 1.00 * 10⁻⁴ mols

Equation of reaction:

HA → H⁺ + A⁻

If 1 mol of HA produces 1 mol of H⁺ and A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of H⁺ and A⁻.

Since only 16% dissociation occurs = 0.16

Number of moles of H⁺ produced = 0.16 * 1.00 * 10⁻⁴

Number of moles of H⁺ produced = 1.6 * 10⁻⁵mols

Number of moles of A⁻ produced = 0.16 * 1.00 * 10⁻⁴

Number of moles of A⁻ produced = 1.6 * 10⁻⁵mols

Since 16% of HA dissociated into H⁺ and A⁻, 84% of HA is left

Number of mols of HA left = 0.84 * 1.00 * 10⁻⁴

Number of mols of HA left = 8.4 * 10⁻⁵mols

Concentration = num of moles/volume

Let the volume of the final solution be V

Conc of HA = 8.4 * 10⁻⁵/V

Conc of H⁺ = 1.6 * 10⁻⁵/V

Conc of A⁻ = 1.6 * 10⁻⁵/V

To calculate the dissociation constant

$k_{a} = [H^{+} ][A^{-} ]/[HA]$

$k_{a}= [1.6 * 10^{-5} /V][1.6 * 10^{-5} /V]/[8.4 * 10^{-5} /V]\\k_{a}= 3.05 * 10^{-6} /V\\k_{a} = 1.00 * 10^{-4}\\ 1.00 * 10^{-4} = 3.05 * 10^{-6} /V\\V= 3.05 * 10^{-6}/ 1.00 * 10^{-4}\\V=3.05 * 10^{-2}\\V=0.0305 L$

3 0

3 years ago

An airplane is traveling 25° west of north at 300 m/s when a wind with velocity 100 m/s directed 35° east of north begins to blo

hodyreva [135]

Answer:

The resultant velocity is 360.5 m/s and direction 79° north of east.

Explanation:

Given that,

Velocity of airplane = 300 m/s

Velocity of wind = 100 m/s

Angle θ₁ = 25°

Angle θ₂ =35°

The horizontal velocity component

Using formula of velocity

$v_{x}=v_{1}\cos\theta-v_{2}\cos\theta$

Put the value into the formula

$v_{x}=300\cos65-100\cos55$

$v_{x}=69.42\ m/s$

The vertical velocity component

Using formula of velocity

$v_{y}=v_{1}\sin\theta+v_{2}\sin\theta$

Put the value into the formula

$v_{y}=300\sin65+100\sin55$

$v_{y}=353.8\ m/s$

We need to calculate the resultant velocity

Using formula of resultant velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{69.42^2+353.8^2}$

$v=360.5\ m/s$

We need to calculate the direction of the resultant velocity

Using formula of direction

$\tan\theta=\dfrac{v_{y}}{v_{x}}$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{353.8}{69.42})$

$\theta=79^{\circ}$

Hence, The resultant velocity is 360.5 m/s and direction 79° north of east.

3 0

3 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$