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2 years ago

When the equation below is completely balanced, what will be the coefficient for Fe2O3?

Chemistry

1 answer:

Mrac [35]2 years ago

5 0

What’s the equation?

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What are the two types of pure substances? Give definitions for each.

BabaBlast [244]

Water and dimond are the 2 pure substances

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2 years ago

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Ammonium dihydrogen phosphate, formed from the reaction of phosphoric acid with ammonia, is used as a crop fertilizer as well as

Tema [17]

Answer:

a)The mass percentages of nitrogen and phosphorus in the compound is 24.34% and 26.95% respectively.

b) 14.78 grams ammonia is incorporated into 100. g of compound.

Explanation:

a) Ammonium dihydrogen phosphate that is $NH_6PO_4$ .

Molecular mass of ammonium dihydrogen phosphate = M

M = 115 g/mol

Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100$

Percentage of nitrogen:

$\frac{1\times 28 g/mol}{115 g/mol}\times 100=24.34\%$

Percentage of phosphorus:

$\frac{1\times 31g/mol}{115 g/mol}\times 100=26.95\%$

b) Percentage of ammonia in 1 molecule of ammonium dihydrogen phosphate.

Molar mass of ammonia = 17 g/mol

$\%=\frac{17 g/mol}{115 g/mol}\times 100=14.78\%$

Amount of ammonia in 100 grams of compound:

14.78% of 100 g of ammonium dihydrogen phosphate:

$\frac{14.78}{100}\times 100 g=14.78 g$

14.78 grams ammonia is incorporated into 100. g of compound.

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3 years ago

PLEASE HELP!! ASAP!! WILL MARK BRAINLIEST!!!

sdas [7]

Answer:

Alkali metals are soft and have low melting points.

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2 years ago

Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L

const2013 [10]

Answer:

The answer is 1.15m.

Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.

We can find the number of H2SO4 moles by using its molarity

C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288

Since water has a density of 1.00kgL, the mass of solvent is

m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg

Therefore, molality is

m=nmass.solvent=0.288moles0.250kg=1.15m

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2 years ago

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Hydrogen bonds are

ella [17]

Answer:

Explanation:

hydrogen bonds are initially weak and collectively strong because they are partially positive ions

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