Ask question

Ask question

All categories

2 years ago

10

A student observes that a patch of grass in a garden can reproduce without seed formation by using the structures that grow hori

zontally from the roots. Which form of vegetative propagation does the student observe?
A). vines

B). tubers

C). seeds

D). runners

1 answer:

liraira [26]2 years ago

3 0

RUNNERS , well that what i would pick

You might be interested in

What other kinds of empirical evidence can be used to show common ancestry among organisms?

kotegsom [21]

Answer:There are, however, other forms of empirical evidence that support the theory of evolution. The fossil record, anatomical and embryological homologous, and genetic similarities are also used to support the theory of evolution. Many pieces of evidence support the theory that birds and dinosaurs are related.

Explanation:hope i am helpful

4 0

3 years ago

) Do you think the pH of 1,0 M tri-methyl ammonium (CH3)3NH+, pKa = 9.80, will be higher or lower than that of 1.0 M phenol, C6H

Elanso [62]

Answer:

1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).

2. The difference in pH values is 4.95.

Explanation:

1. The pH of a compound can be found using the following equation:

$pH = -log([H_{3}O^{+}])$

First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.

<u>Trimethyl ammonium</u>:

We can calculate [H₃O⁺] using the Ka as follows:

(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺

1.0 - x x x

$Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}$

$10^{-pKa} = \frac{x*x}{1.0 - x}$

$10^{-9.80}(1.0 - x) - x^{2} = 0$

By solving the above equation for x we have:

x = 0.097 = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01$

<u>Phenol</u>:

C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺

1.0 - x x x

$Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}$

$10^{-10} = \frac{x^{2}}{1.0 - x}$

$1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0$

Solving the above equation for x we have:

x = 9.96x10⁻⁶ = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00$

Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.

2. The difference in pH values for the two acids is:

$\Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95$

Therefore, the difference in pH values is 4.95.

I hope it helps you!

7 0

3 years ago

PLEASE ANSWERRRRRRRRRRR

andrew11 [14]

Answer:

B if I remember right.

Explanation:

5 0

2 years ago

Iron is extracted from iron oxide in the Blast Furnace: Fe 2 O 3 + 3 CO → 2 Fe + 3 CO 2

arsen [322]

a. mass of iron = 69.92 g

b. percent yield = 93%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

a.

Reaction

Fe₂O₃+3CO⇒2Fe+3CO₂

MW Fe₂O₃ : 159.69 g/mol

mol Fe₂O₃

$\tt \dfrac{100}{159,69}=0.626$

mol Fe₂O₃ : mol Fe = 1 : 2

mol Fe :

$\tt \dfrac{2}{1}\times 0.626=1.252$

mass of Fe(Ar=55.845 g/mol) :

$\tt 1.252\times 55.845=69.92~g$

b.

actual yield = 65 g

theoretical yield = 69.92 g

percent yield :

$\tt =\dfrac{65}{69.92}=0.93=93\%$

8 0

2 years ago

Please provide thorough explanation

Arlecino [84]

Possible products in this reaction are the products 2 and 3.

<h3>What is the Friedel Crafts reaction?</h3>

The Friedel Crafts reaction is one in which the electrophile is created by a Lewis acid reaction between the AlCl3 and the alkylhalide reactant .

Now we know that there is the possibility of two products in this reaction due to a resonance shift as such possible products in this reaction are the products 2 and 3.

Learn more about Friedel Crafts reaction:brainly.com/question/14993566?

#SPJ1

4 0

1 year ago