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3 years ago

The orbit of a planet is an ellipse with the sunat one of the foci.. This is Kepler's........ second law third law first law

Chemistry

1 answer:

Svetradugi [14.3K]3 years ago

6 0

Answer:

its keplers first law

Explanation:

answer from gauth math

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Which of these does not cause a change in the reaction rate?

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A. density
Reaction rate increases with concentration.

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3 years ago

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What is the molarity of 3.0 L solution containing 250 g of Nal

IgorC [24]

Answer: C. 0.57 M

Explanation:

Molarity= Moles/Vol

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3 years ago

please help!!!!!! What is the mass of 1 mole of magnesium (Mg)? Express your answer to four significant figures. The mass of 1 m

Semmy [17]

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3 years ago

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Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

4 0

3 years ago

Nitrogen has an average atomic mass of 14.00674. it has two isotopes, n-14 and n-15, with atomic masses of 14.0031 amu and 15.00

masya89 [10]

In this case, we are going to assume that there are 100 atoms to make things easier.

Let R% be the abundance of n-15. With this in mind, we calculate the abundance of n-14 to be 100%-R%

14.0031*(100-R)% + 15.001 * R%= 14.00674
In this case, we can delete or ignore the % sign since we do not want to carry it around, however, we need to keep in mind that the final answer is in %

14.0031*(100-R) + 15.001 * R= 14.00674
1400.31-14.0031R+15.001R=1400.674
0.9979R=0.364
R=0.3648

Then, the abundance of n-15 is 0.3648%

5 0

3 years ago

The orbit of a planet is an ellipse with the sunat one of the foci.. This is Kepler's........ second law third law first law​

The orbit of a planet is an ellipse with the sunat one of the foci.. This is Kepler's........ second law third law first law