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2 years ago

What would the % p, on the npk ratio, be reported as for a 100g sample containing 7.5g of phosphorous?

2 answers:

6 0

When the concentration is expressed in percentage, you simply have to divide the amount of substance to the total amount of the mixture, then multiply it by 100. In this case, when you want to find the percentage of P in the sample, the solution is as follows:

%P = 7.5 g/100 g * 100 = 7.5%

pshichka [43]2 years ago

6 0

Ans: 7.50% of P

Given:

Mass of NPK sample = 100 g

Mass of P in sample = 7.5 g

To determine:

The % P in the given sample

Explanation:

The percent of a particular substance (say,X) in a given total amount (M) is generally expressed as:

% X = [Mass of X/Total mass]*100

In this case:

%P = [mass of P/mass of NPK sample]*100

= [7.5/100]*100 = 7.5%

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Water treatment plants commonly use chlorination to destroy bacteria. a byproduct is chloroform (chcl3), a suspected carcinogen

antiseptic1488 [7]

100. ppb of chcl3 in drinking water means 100 g of CHCl3 in 1,000,0000,000 g of water

Molarity, M

M = number of moles of solute / volume of solution in liters

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molar mass of CHCl3 = 119.37 g/mol

number of moles of solute = 100 g / 119.37 g/mol = 0.838 mol

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M = 0.838 / 1,000,000 = 8.38 * 10^ - 7 M <----- answer

Molality, m

m = number of moles of solute / kg of solvent

number of moles of solute = 0.838

kg of solvent = kg of water = 1,000,000 kg

m = 0.838 moles / 1,000,000 kg = 8.38 * 10^ - 7 m <----- answer

mole fraction of solute, X solute

X solute = number of moles of solute / number of moles of solution

number of moles of solute = 0.838

number of moles of solution = number of moles of solute + number of moles of solvent

number of moles of solvent = mass of water / molar mass of water = 1,000,000,000 g / 18.01528 g/mol = 55,508,435 moles

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7 0

2 years ago

Read 2 more answers

What is the 2s in 42,256

MissTica

42,256 = 2,000

42,256 = 200

together they'd be 2,200 (if that's what you needed as well)

5 0

2 years ago

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If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the

jeyben [28]

25 g of NH₃ will produce 47.8 g of (NH₄)₂S

Explanation:

2 NH₃ + H₂S ----> (NH₄)₂S

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S

1 g of NH₃ will produce $\frac{68}{34}$ g of (NH₄)₂S

25g of NH₃ will produce $\frac{65}{34} X 25 g$ of (NH₄)₂S

= 47.8 g of (NH₄)₂S

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S

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2 years ago

Consider the solutions, 0.04 m urea [(NH2)2C=O)], 0.04 m AgNO3 and 0.04 m CaCl2. Which has (i) the highest osmotic pressure, (ii

Lana71 [14]

Answer:

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

Explanation:

The colligative properties depend upon the number of solute particles in a solution.

The following four are the colligative properties:

a) osmotic pressure : more the concentration of the solute, more the osmotic pressure

b) vapor pressure: more the concentration of the solute, lesser the vapor pressure.

c) elevation in boiling point: more the concentration of the solute, more the boiling point.

d) depression in freezing point: more the concentration of the solute, lesser the freezing point.

the number of particle produced by urea = 1

the number of particle produced by AgNO3 = 2

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As concentrations are same, CaCl2 will have more number of solute particles and urea will have least

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

5 0

2 years ago