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2 years ago

How many moles are found in 133 g of Mg3 N2

Chemistry

2 answers:

konstantin123 [22]2 years ago

7 0

Magnesium nitride weighs 100.95 g/mol

133/100.95 = 1.32 mols

scoundrel [369]2 years ago

7 0

Magneisum nitride weighs 100.95 g/mol
133/100.95 = 1.32 mols

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Explain and describe how the photoelectric effect occurs on an atomic level in terms of protons, neutrons, and electrons

grin007 [14]

Answer:

The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material. Electrons emitted in this manner are called photoelectrons.

Based on the wave model of light, physicists predicted that increasing light amplitude would increase the kinetic energy of emitted photoelectrons, while increasing the frequency would increase measured current.

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2 years ago

The chemical formula for glucose (sugar) is C6H12O6. What are the individual numbers called in a chemical formula?

givi [52]

The individual numbers which are present in chemical formula is subscript.

<h3>What is chemical formula?</h3>

Chemical formula of any compound of chemistry consist more than one atoms in it and gives idea about the number of atoms present in that compound.

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Hence, individual number are subscript.

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1 year ago

Which chemical class of hormones is not matched correctly?a) nitric oxide - lipid solubleb) steroids - lipid solublec) antidiure

stealth61 [152]

d) T3 and T4. T3 and T4 are poorly soluble in water, and more than 99% of the T3 and T4 circulating in blood is bound to carrier proteins. The main carrier of thyroid hormones is thyroxine-binding globulin, a glycoprotein synthesized in the liver.

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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

Nimfa-mama [501]

Answer:

lattice parameter = 5.3355x10^-8 cm

atomic radius = 2.3103x10^-8 cm

Explanation:

known data:

p=0.855 g/cm^3

atomic mass = 39.09 g/mol

atoms/cell = 2 atoms

Avogadro number = 6.02x10^23 atom/mol

a) the lattice parameter:

Since potassium has a cubic structure, its volume is equal to:

v = [(atoms/cell)x(atomic mass)/(p)x(Avogadro number)]

substituting values:

v =[(2)x(39.09)/(0.855x6.02x10^23)]=1.5189x10^-22 cm^3

but as the cell volume is

a^3 =v

$a=\sqrt[3]{v}=\sqrt[3]{1.5189x10^{-22} } = 5.3355x10^-8$ cm

for a BCC structure, the atomic radius is equal to

$r=\frac{ax\sqrt{3} }{4}=\frac{5.3355x10^{-8}x\sqrt{3} }{4}=2.3103x10^{-8}cm$

7 0

2 years ago

Use the balanced equation from number four above to determine the limiting reactant if we had a mixture of 5.0 moles of Fe and 4

Tom [10]

Answer:

2Fe + 3O₂ → 2Fe₂O₃

Limiting reactant: O₂

2Fe + O₂ → 2FeO

Limiting reactant: Fe

Explanation:

2Fe + 3O₂ → 2Fe₂O₃

2Fe + O₂ → 2FeO

These are the possible reactions:

In first case, 2 moles of Fe need 3 mol of oyxgen to react

If I have 5 moles of Fe, I will need (5 .3)/2 = 7.5 moles of O₂

Then, the oxygen is my limiting ( I only have 4 moles)

3 moles of O₂ need 2 moles of Fe

If I have 4 moles of O₂, I will need (4 .2)/3 = 2.66 moles (I have 5)

Fe, is the reactant in excess.

For second case, 2 moles of Fe need 1 mol of O₂, to react.

If I have 5 moles of Fe, I will need ( 5 .1) / 2 =2.5 moles of O₂

I have 4 moles of oxygen, so now it is my excess.

1 mol of O₂ need 2 moles of Fe, to react

If I have 4 moles of O₂, I will need the double of amount, 8 moles of Fe.

I have 5 moles, then the Fe is my limtiing reactant.

8 0

2 years ago