fibrous roots; taproots
Explanation:
A taproot root consists of one large main root with smaller roots that branches off into the soil, while the fibrous roots consists of several main roots that branches off to form one mass of roots.
- The root system in plants helps them to absorb water and other nutrients from the soil.
- A taproot is much more like extension of stem that penetrates into the ground.
- It tapers at the end with many other smaller roots branching and networking from it.
- The fibrous root is a series of roots directly from the stem that independent of one another.
- Roots are used by plants for anchorage into the soil.
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Answer:
P₂ = 13.9 atm (3 sig. figs.)
Explanation:
The pressure (P), Volume (V) relationship with Temperature (T) & mass (n) held constant is an inverse proportionality. That is Boyles Law ...
P ∝ 1/V => P = k/V => k = P·V
For two pressure-volume conditions, the proportionality constant (k) remains constant where k₁ = k₂ and P₁·V₁ = P₂·V₂ => P₂ = P₁·V₁/V₂
Given:
P₁ = 1.31 atm.
V₁ = 5.51 L
P₂ = ?
V₂ = 0.520 L
V₂ = (1.31 atm)(5.51L)/(0.520L) = 13.88096154 atm (calc. ans.) = 13.9 atm (3 sig. figs.)
HF and NaF - If the right concentrations of aqueous solutions are present, they can produce a buffer solution.
<h3>What are buffer solutions and how do they differ?</h3>
- The two main categories of buffers are acidic buffer solutions and alkaline buffer solutions.
- Acidic buffers are solutions that contain a weak acid and one of its salts and have a pH below 7.
- For instance, a buffer solution with a pH of roughly 4.75 is made of acetic acid and sodium acetate.
<h3>Describe buffer solution via an example.</h3>
- When a weak acid or a weak base is applied in modest amounts, buffer solutions withstand the pH shift.
- A buffer made of a weak acid and its salt is an example.
- It is a solution of acetic acid and sodium acetate CH3COOH + CH3COONa.
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Answer:
The answer to your question is C₂HO₃
Explanation:
Data
Hydrogen = 3.25%
Carbon = 19.36%
Oxygen = 77.39%
Process
1.- Write the percent as grams
Hydrogen = 3.25 g
Carbon = 19.36 g
Oxygen = 77.39 g
2.- Convert the grams to moles
1 g of H ----------------- 1 mol
3,25 g of H ------------- x
x = (3.25 x 1) / 1
x = 3.25 moles
12 g of C ---------------- 1 mol
19.36 g of C ---------- x
x = (19.36 x 1) / 12
x = 1.61 moles
16g of O --------------- 1 mol
77.39 g of O --------- x
x = (77.39 x 1)/16
x = 4.83
3.- Divide by the lowest number of moles
Carbon = 3.25/1.61 = 2
Hydrogen = 1.61/1.61 = 1
Oxygen = 4.83/1.61 = 3
4.- Write the empirical formula
C₂HO₃
