All categories

3 years ago

How many kilograms of a fertilizer are made of pure P2O5 would be required to supply 1.69 kilogram of phosphorus to the soil?

Chemistry

1 answer:

Alekssandra [29.7K]3 years ago

3 0

Answer:

3.89 kg P2O5 must be used to supply 1.69 kg Phosphorus to the soil.

Explanation:

The molecular mass of P2O5 is

P2 = 2* 31 = 62

O5 = 5 * 16 = 80

Molecular Mass = 142

Set up a Proportion

142 grams P2O5 supplies 62 grams of phosphorus

x kg P2O5 supplies 1.69 kg of phosphorus

Though this might be a bit anti intuitive, you don't have to convert the units for this question. The ratio is all that is important.

142/x = 62/1.69 Cross multiply

142 * 1.69 = 62x combine the left

239.98 = 62x Divide by 62

239.98/62 = x

3.89 kg of P2O5 must be used.

You might be interested in

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

Answer: The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

Explanation:

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

Reduction half reaction: $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

Net reaction: $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

Would a thermos without a lid with water in it be an open or closed system? I'm so confused... also please explain your answer,

Dmitriy789 [7]

Answer:

. A closed system allows only energy transfer but no transfer of mass. Example: a cup of coffee with a lid on it, or a simple water bottle. ... In reality, a perfectly isolated system does not exist, for instance hot water in a thermos flask cannot remain hot forever.

8 0

3 years ago

Why does potassium permanganate need to be standardized right before a titration?.

Ierofanga [76]

To determine the strength of potassium permanganate with a standard solution of oxalic acid.

6 0

2 years ago

What happens to electrons in the photoelectric effect?

11111nata11111 [884]

The photoelectric effect occurs when light shines on a metal. ... Light of any frequency will cause electrons to be emitted.

5 0

2 years ago

Read 2 more answers

What Is scientific theory

sveticcg [70]

Answer:

is an explanation of an aspect of the natural world that can be repeatedly tested and verified in accordance with the scientific method, using accepted protocols of observation, measurement, and evaluation of results

8 0

3 years ago