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3 years ago

Which is the empirical formula for a compound that contains 64.75g nitrogen and 185.25 oxygen

Chemistry

1 answer:

Svetach [21]3 years ago

5 0

Answer:

What is the formula for a compound that contains 64.75 g nitrogen and 185.25 g oxygen? D. N2O5

The name of a hydrate is calcium chloride dihydrate. What is its formula? B. CaCl2 x 2H20

Explanation:

BRAINLIEST PLZZZZ

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Imagine that you traveled to the Moon for a vacation and discovered that your weight is different there. How would your weight b

dmitriy555 [2]

Answer

No you did not lose weight.

Explanation:

Weight is the force acting on an object, Gravity is a force on earth which does not apply to the moon so your weight may change depending in where you are but your mass will stay the same because mass measures the matter of an object ITSELF! Have a nice day....

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2 years ago

Why won't a fat–soluble vitamin dissolve in water?

Nataliya [291]

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3 years ago

How many atoms of carbon are represented in 3CH3CH20

pishuonlain [190]

6 carbon atoms

          H H
         | |
3 x   H - C - C - O - H
   | |
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2 years ago

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

You are provided with a stock solution with a concentration of 1.0x10-5 M. You will be using this to make two standard solutions

artcher [175]

Answer:

1. V₁ = 2.0 mL

2. V₁ = 2.5 mL

Explanation:

You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.

To calculate the volume required (V₁) in each dilution we will use the dilution rule.

C₁ . V₁ = C₂ . V₂

where,

C are the concentrations

V are the volumes

1 refers to the initial state

2 refers to the final state

1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.

C₁ . V₁ = C₂ . V₂

(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL

V₁ = 2.0 mL

2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.

C₁ . V₁ = C₂ . V₂

(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL

V₁ = 2.5 mL

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3 years ago