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2 years ago

hii what the procedures of making organic fertilizer and its precautions. can u answer it i kinda need it now

Chemistry

2 answers:

frez [133]2 years ago

3 0

Answer:

Go through your kitchen waste. Vegetables and fruit peelings are the number one food remnants you should keep aside. ...

Add other organic materials to the compost. ...

Collect some garden waste. ...

Create the compost. ...

Apply the fertilizer.

rosijanka [135]2 years ago

3 0

Answer:

<h3>Abstract </h3>

Organic fertilizers are an essential source for plant nutrients and a soil conditioner in agriculture. Due to its sources and the composition of the organic inputs as well as the type, functionality and failures of the applied treatment process, the organic fertilizer may contain various amounts of infectious agents and toxic chemicals, especially the antibiotics that can be introduced to the subsequent food chain. A range of human and animal pathogens of bacterial, viral and parasitic origin have been the cause of food-borne epidemics due to unintended contamination from organic fertilizers. The use of antibiotics by humans and in animal feeds will also end up in the organic fertilizers. These antibiotics and other chemicals, depending on the sources of the organics, will enhance the likelihood of occurrence of resistant and multi-resistant strains of microorganisms in society and have been reported to cause ecotoxicological environmental effects and disruption of the ecological balance. Exposure of microorganisms to sublethal concentration of antibiotics in the organic products induces antibiotic resistance. WHO guidelines for the reuse of excreta and other organic matters identify the risk for the exposed groups to the reuse of the

Explanation:

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Balance the following chemical equation by providing the correct coefficients fe+h2so4 fe(so4)3 + h2

Dmitriy789 [7]

Answer:

2Fe + 3H2SO4 + Fe2(SO4)3+ 3H2

Explanation:

1. Fe (SO4) 3 is an incorrectly written formula because iron is trivalent as we can see by this three ahead of SO4. SO4 is divalent always.

2. since (SO4) is 3, this three shows us that there must be 3 in the reactants as well.

so now there is 3H2SO4

3. Since we have added 3 to one hydrogen we must add another. So now it's 3H2

4. and finally iron. In Fe2 (SO4) 3 we see this 2 in front of Fe which means it goes 2Fe.

3 0

2 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

2 years ago

If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did

user100 [1]

Answer:

4.5 L water we have in litres (L).

Explanation:

$Q=m\times c \times \Delta T$

where

$\Delta T$ = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0 cal/(g℃))

m is the mass of water

Plugging in the values

$\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$$

So,

Volume of water = mass/density

$\\\\=\frac{4500 \mathrm{g}}{\frac{1.09}{\mathrm{mL}}}=4500 \mathrm{mL}$$$

=4.5 L (Answer)

6 0

3 years ago

Explain the process of radioactive decay. What happens during the radioactive decay? What is the result at the end of radioactiv

nignag [31]

1) Radioactive decay is the spontaneous decomposition of the unstable nucleus of an atom.

2) The emission of a particle or a photon.

For example, alpha decay is radioactive decay in which an atomic nucleus emits an alpha particle (helium nucleus).

3) The result is usually more stable element with smaller atomic number.

For example, in alpha decay atom transforms into an atom with an atomic number that is reduced by two and mass number that is reduced by four.

For example nuclear fission is radioactive decay process in which the nucleus of an atom splits into smaller parts and huge amount of energy is released.

7 0

2 years ago

Read 2 more answers

C) A battery of emf, E drives a current of 0.25A when connected to a 5.5ohms resistor. When

boyakko [2]

Answer:

your ans is battery emf v+1r

5 0

3 years ago