Pl help it’s for a grade

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

frez [133]

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.

3 0

3 years ago

Read 2 more answers

This green solution of chromium(III) can further be reduced by zinc metal to a blue solution of chromium(II) ions. Write the bal

Contact [7]

Answer:

Half-reactions:

Cr³⁺ + 1e⁻ → Cr²⁺; Zn → Zn²⁺ + 2e⁻

Net ionic equation:

2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺

Explanation:

The Cr³⁺ is reduced to Cr²⁺:

<h3>Cr³⁺ + 1e⁻ → Cr²⁺ -Half-reaction 1-</h3>

Zn is oxidized to Zn²⁺:

<h3>Zn → Zn²⁺ + 2e⁻ -Half-reaction 2-</h3>

Twice the reduction of Cr:

2Cr³⁺ + 2e⁻ → 2Cr²⁺

Now this reaction + Oxidation of Zn:

2Cr³⁺ + 2e⁻ + Zn → 2Cr²⁺ + Zn²⁺ + 2e⁻

<h3>2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ - Net ionic equation</h3>

6 0

3 years ago

What substance is present in a solution consisting solely of electrolytes?

Debora [2.8K]

Electrolytes are substances which are capable of producing a solution that has the ability to conduct electricity. These electrolytes are formed from the dissociation of salts into ions. The positive ions are named cations while the negative ions are termed anions.

Hence, the answer to these item is SALT. Salt are substances produced by the reaction of cations and anions.

4 0

3 years ago

Read the descriptions below of two substances and an experiment on each. Decide whether the result of the experiment tells you t

inessss [21]

Answer:

Sample A - Mixture

Sample B - (can't decide)

Explanation:

We know a mixture as a sample that is made up of two or more substances. Based on the results from the experiment conducted on sample A, the sample is a mixture. Each colour that appeared on the paper represents one of the components of the mixture.

For Sample B, at a particular sharp temperature, the crystals begin to appear. That temperature at which the first crystal appears is actually the melting point of the solid. We were also told that only half of the clear liquid was crystallized meaning that other substances may still be contained in the remaining liquid. Crystallization is a separation technique that depends on differences in melting points of substances. We can't decide if the sample is pure because we have no further information about what happened to the remaining liquid. That would have told us if the liquid remaining was just the solvent used to dissolve B which could have also been evaporated to leave only the pure sample.

4 0

3 years ago

Lead all chlorate is mixed with hydrolylic acid. Each solution is 0.85 molar. Write balanced, molecular, ionic, and net equation

d1i1m1o1n [39]

Answer:

Here's what I get

Explanation:

Solubility rules

Salts containing halides are generally soluble. Important exceptions to this rule are halides of silver, mercury, and lead(II).
All acetates, chlorates, and perchlorates are soluble

So, PbCl₂ is insoluble, and Pb(ClO₃)₂ is soluble.

1. "Molecular" equation

$\rm Pb(ClO_{3})_{2}(aq) + 2HCl(aq) \longrightarrow \, PbCl_{2}(s) + 2HClO_{3}(aq)$

2. Ionic equation

Convert the soluble salts to their hydrated ions.

HCl and HClO₃ are strong acids. Convert them to their ions.

$\rm Pb^{2+}(aq) + 2ClO_{3}^{-}(aq)+ 2H^{+}(aq) + 2Cl^{-}(aq) \longrightarrow \, PbCl_{2}(s) + 2H^{+}(aq) + 2ClO_{3}^{-}(aq)$

3. Net ionic equation

Cancel all ions that appear on both sides of the reaction arrow (in boldface).

$\rm {Pb}^{2+}(aq) + \textbf{2ClO}_{3}^{-}(aq)+ \textbf{2H}^{+}(aq) + 2Cl^{-}(aq) \longrightarrow \, PbCl_{2}(s) + \textbf{2H}^{+}(aq) + \textbf{2ClO}_{3}^{-}(aq)$

The net ionic equation is

$\rm {Pb}^{2+}(aq) + 2Cl^{-}(aq) \longrightarrow \, PbCl_{2}(s)$

4. Theoretical yield

We have the volumes and concentrations of two reactants, so this is a limiting reactant problem.

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

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

M_r: 278.11

Pb(ClO₃)₂ + 2HCl ⟶ PbCl₂ + 2HClO₃

Volume/mL: 125 95

c/mol·L⁻¹: 0.85 0.85

(ii) Calculate the moles of each reactant

$\text{Moles of Pb(ClO$_{3}$)}_{2} = \text{0.125 L} \times \dfrac{\text{0.85 mol}}{\text{1 L }} = \text{0.1062 mol}\\\text{Moles of HCl} = \text{0.095 L} \times \dfrac{\text{0.85 mol}}{\text{1 L }} = \text{0.08075 mol}$

(iii) Identify the limiting reactant

Calculate the moles of PbCl₂ we can obtain from each reactant.

From Pb(ClO₃)₂:

The molar ratio of PbCl₂:Pb(ClO₃)₂ is 2:2

Moles of PbCl₂ = 0.1062 × 2/2 =0.1062 mol PbCl₂

From HCl :

The molar ratio of PbCl₂:HCl is 1 mol PbCl₂:2 mol HCl.

Moles of PbCl₂ = 0.08075 × 1/2 = 0.04038 mol PbCl₂

The limiting reactant is HCl because it gives the smaller amount of PbCl₂.

(iv) Calculate the theoretical yield of PbCl₂.

$\text{Theor. yield of PbCl}_{2} = \text{0.0438 mol} \times \dfrac{\text{278.11 g}}{\text{ 1 mol}} = \textbf{11.2 g}$

5. Calculate the actual yield of PbCl₂

$\text{Actual yield} = \text{11.2 g theor.} \times \dfrac{\text{ 68 g actual}}{\text{100 g theor,}} = \textbf{7.6 g}$

6. Calculate [ClO₃⁻]

Original concentration of Pb(ClO₃)₂ = 0.85 mol·L⁻¹

Original concentration of ClO₃ = 2 × 0.85 = 1.70 mol·L⁻¹

The solution was diluted by the addition of HCl.

Total volume = 125 + 95 =220 mL

c₁V₁ = c₂V₂

1.70 mol·L⁻¹ × 125 mL = c₂ × 220 mL

212.5 mol·L⁻¹ = 200 c₂

c₂ = (212.5 mL)/200 = 1.06 mol·L⁻¹

7. Calculate [Pb²⁺].

Moles of Pb²⁺ originally present = 0.1062 mol

Moles of Pb²⁺removed = 0.04038 mol

Moles of Pb²⁺ remaining = 0.0659 mol

c = 0.0659 mol/0.220 L = 0.299 mol·L⁻¹

8 0

4 years ago