What 3 gases does the flame test check for and what should you expect to see/hear?

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

3 years ago

To calculate the enthalpy change for the reaction, 2CO (g) + O2 (g) Imported Asset 2 CO2 (g), you can use ΔHf0 values for each r

svetoff [14.1K]

Answer:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

Explanation:

Chemical equation:

CO + O₂ → CO₂

Balanced chemical equation:

2CO + O₂ → 2CO₂

The standard enthalpy for the formation of CO = -110.5 kj/mol

The standard enthalpy for the formation of O₂ = 0 kj/mol

The standard enthalpy for the formation of CO₂ = -393.5 kj/mol

Now we will put the values in equation:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol + 0]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol]

ΔH0reaction = -283 kj/mol

7 0

3 years ago

A major component of gasoline is octane (C8H18). When liquid octane is burned in air it reacts with oxygen gas to produce carbon

Paraphin [41]

Answer:

The answer to your question is 0.4 moles of Oxygen

Explanation:

Data

Octane (C₈H₈)

Oxygen (O₂)

Carbon dioxide (CO₂)

Water (H₂O)

moles of water = ?

moles of Oxygen = 1

Balanced chemical reaction

C₈H₈ +10O₂ ⇒ 8CO₂ + 4H₂O

Reactant Element Products

8 C 8

8 H 8

20 O 20

Use proportions to solve this problem

10 moles of Oxygen ----------------- 4 moles of water

1 mol of Oxygen ------------------ x

x = (4 x 1) / 10

x = 4 / 10

x = 0.4 moles of water

7 0

3 years ago

Firewood burns and ashes remain<br><br> is it a chemical or physical change?

erastova [34]

It is a physical change because you can not put it back like it was

4 0

4 years ago

Read 2 more answers

How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56

neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.0156

Thus,

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.0156=e^{-k\times t}$

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}$

$t = 6\times t_{1/2}$

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

6 0

3 years ago