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3 years ago

How many moles of NH3 can you make from 6.20 moles of N2?

Chemistry

2 answers:

Ainat [17]3 years ago

5 0

The answer is 4 loll

german3 years ago

4 0

4 I think

$ hope it welp

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Electron transport is the process by which ________. Multiple Choice fatty acids break apart to form acetyl-CoA NADH + H+ and FA

Serggg [28]

Answer:

Electron transport is the process by NADH + H+ and FADH2 are converted to NAD+ and FAD, donating electrons and hydrogen ions to oxygen

Explanation:

3 0

3 years ago

How many mold are in 4.25x10^-2 atoms of lead

Shalnov [3]

I’m sorry I have no clue

7 0

3 years ago

Determine the molecular weight of H20. o 18 g O 18.0 g O 18.01 g O 18.016 g

Sonbull [250]

Answer:

The correct answer is D

18.016 g

Explanation:

Molecular Weight : It is the sum of atomic weights of each atoms present in the compound.

The molecular weught is measured in atomic mass.unit( amu) or simply"u"

This is calculated by using :

Molecular weight = number of atom x atomic mass of the atom

For H2O

Number of H atoms = 2

Number of O atom = 1

Molecular weight of H2O = 2(mass of H atom) + 1(mass of O atom)

Molecular weight = 2(1.00784) + 15.999

= 18.01558 u

= 18.016 u

8 0

3 years ago

Why are there 2 chloride ions for every 1 calcium in calcium chloride

tensa zangetsu [6.8K]

Answer:

See explanation

Explanation:

Calcium is divalent. This means that it donates two electrons during ionic bond formation. Since chlorine atom can only accept one electron during ionic bond formation, two chlorine atoms must accept the two electrons donated by calcium.

For this purpose, each time CaCl2 is formed, there must be two chlorine atoms for each calcium atom.

5 0

3 years ago

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

Delicious77 [7]

Answer:

For a: The edge length of the crystal is 533.5 pm

For b: The atomic radius of potassium is 231.01 pm

Explanation:

For a:

To calculate the lattice parameter or edge length of the crystal, we use the equation:

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

where,

$\rho$ = density = $0.855g/cm^3$

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

M = atomic mass of metal = 39.09 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:

$0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm$

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

Hence, the edge length of the crystal is 533.5 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 = 533.5 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm$

Hence, the atomic radius of potassium is 231.01 pm

4 0

3 years ago