What is the lewis dot structure of phosphorus decasulfide

You are provided with a compound fertilizer, 40-15-10.Calculate the quantity of fertilizer to add to a one hectare field supply

victus00 [196]

Answer:

a. 300 kg of Fertilizer

b. 225 kg of fertilizer

c.400 Kg of fertilizer

d.600 Kg of fertilizer

Explanation:

The percentage composition ratio of Nitrogen, Phosphorus and Potassium bag of the given fertilizer is 40:15:10.

The percentages can be expressed as fractions as follows:

For nitrogen; 40/100 = 0.4

For phosphorus; 15/100 = 0.15

For potassium; 10/100 = 0.1

To find the quantity of fertilizer required to add to a hectare to supply the given amount of nutrients, the amount to be provided is divided by the percentage or fractional compostion of each nutrient.

Quantity of fertilizer required to add to a hectare to supply;

a. Nitrogen at 120 kg/ha = 120/0.4 = 300 Kg of fertilizer

b.. Nitrogen at 90 Kg/ha = 90/0.4 = 225 Kg of fertilizer

c. Phosphorus at 60 kg/ha = 60/0.15 = 400 Kg of fertilizer

d. Potassium at 60 kg/ha = 60/0.1 = 600 Kg of fertilizer

5 0

2 years ago

A volleyball contains 1.36 g of nitrogen gas. How many moles of N2 gas are in the volleyball?

Igoryamba

There is a lot of nitrogen gas in a volleyball choices?

3 0

3 years ago

Green plants use light from the Sun to drive photosynthesis, a chemical reaction in which liquid water and carbon dioxide gas fo

Luda [366]

Answer:

0.800 mol of O2

Explanation:

Calculate the moles of oxygen produced by the reaction of 0.800mol of carbon dioxide.

The balanced equation for the reaction is given as;

6CO2 + 6H2O → C6H12O6 + 6O2

From the reaction;

6 mol of CO2 produces 6 mol of O2

0.0800 mol of CO2 would produce x mol of O2

6 = 6

0.0800 = x

Solving for x;

x = 6 * 0.800 / 6

x = 0.800 mol

5 0

2 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

3 years ago

What mass of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution?

inna [77]

Answer:

459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution

Explanation:

Molarity is a measure of the concentration of a solute in a solution that indicates the amount of moles of solute that appear dissolved in one liter of the mixture. In other words, molarity is the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by the following expression:

$Molarity=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$

In this case:

Molarity: 1.56 M= 1.56 $\frac{moles}{liter}$
Number of moles of calcium chlorine= ?
Volume= 2.657 liters

Replacing:

$1.56 M=\frac{Number of moles of calcium chlorine}{2.657 liters}$

Solving:

Number of moles of calcium chlorine= 1.56 M* 2.657 liters

Number of moles of calcium chlorine= 4.14 moles

In other side, you know:

Ca: 40 g/mole
Cl: 35.45 g/mole

Then the molar mass of the calcium chloride CaCl₂ is:

CaCl₂= 40 g/mole + 2* 35.45 g/mole= 110.9 g/mole

Now it is possible to apply the following rule of three: if in 1 mole there is 110.9 g of CaCl₂, in 4.14 moles of the compound how much mass is there?

$mass=\frac{4.14 moles*110.9g}{1 mole}$

mass= 459.126 g

459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution

3 0

2 years ago