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notsponge [240]
3 years ago
13

Is plastic wrap anf mixture or substance

Chemistry
2 answers:
defon3 years ago
4 0
Plastic wrap is a substance
maxonik [38]3 years ago
4 0
Its a mixture not a substance i promise

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A certain weak acid, ha, has a ka value of 6.0×10−7. calculate the percent ionization of ha in a 0.10 m solution.
joja [24]
For the purpose, we will use the equation for determining the dissociation constant from concentration and <span>percent of ionization:

Kd = c </span>× α²

α = √(Kd/c) × 100%

Kd = 6.0×10⁻⁷

c(HA) = 0.1M

α = √(6.0×10⁻⁷/0.1)  × 100% =  0.23%

So, in the solution, the acid <span>percent of ionization will be just 0.23%.</span>

5 0
3 years ago
Main job is to sort and package proteins and other substances in a plant cell
melomori [17]

Answer:

Main job of golgi bodies is to sort and package proteins and other substances in a plant cell.

Explanation:

Golgi bodies are also called post office of the cell because it modify and distribute proteins for the cell. First, proteins are made in the organelle of the cell i. e. endoplasmic reticulum. From here, it is send to the Golgi apparatus for modification. Golgi bodies add some special structures with the protein and this protein leaves golgi bodies which is used by the cell where it is needed.

8 0
3 years ago
Ammonia, nh3, is a weak base with a kb = 1.8 x 10-5. in a 0.8m solution of ammonia, which has a higher concentration: ammonia (n
insens350 [35]
Answer is: concentration ammonia is higher than concentration of ammonium ion.
Chemical reaction of ammonia in water: NH₃ + H₂O → NH₄⁺ + OH⁻.
Kb(NH₃) = 1,8·10⁻⁵.
c₀(NH₃) = 0,8 mol/L.
c(NH₄⁺) = c(OH⁻) = x.
c(NH₃) = 0,8 mol/L - x.
Kb = c(NH₄⁺) · c(OH⁻) / c(NH₃).
0,000018 = x² /  0,8 mol/L - x.
solve quadratic equation: x = c(NH₄⁺) = 3,79·10⁻³ mol/L.
7 0
3 years ago
Hydrofluoric acid and Water react to form fluoride anion and hydronium cation, like this HF(aq) + H_2O(l) rightarrow F(aq) + H_3
maksim [4K]

Answer:

Kc = 1.09x10⁻⁴

Explanation:

<em>HF = 1.62g</em>

<em>H₂O = 516g</em>

<em>F⁻ = 0.163g</em>

<em>H₃O⁺ = 0.110g</em>

<em />

To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:

Kc = [H₃O⁺] [F⁻] / [HF]

<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>

<em />

[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M

[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M

[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M

Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]

<h3>Kc = 1.09x10⁻⁴</h3>
7 0
3 years ago
When the following equation is balanced, what is the coefficient of oxygen?
Gwar [14]

Answer:

The answer is 3

C2H5OH + O2 CO2 +H2O (unbalanced)

C2H5OH +3O2(g). 2CO2(g)+3H2O(balanced)

4 0
3 years ago
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