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3 years ago

Which structure could a scientist look for in a plant that would identify it as a club moss rather than a liverwort?

Chemistry

2 answers:

Marianna [84]3 years ago

7 0

<h3>Answer;</h3>

Phloem

<h3>Explanation;</h3>

Club moss plant belongs to the the family Lycopodiaceae, Lycophyte includes any spore-bearing vascular plant.
Liverworts on the other hand are bryophytes which belongs to the division bryophyta. Bryophytes are small, non-vascular plants which includes mosses, hornworts and liverworts.
Vascular plants contain vascular tissues which play an important role of transportation in plants. The major vascular tissues are phloem and xylem. Non-vascular plants on the other hand lacks the vascular tissues for transportation of substances.

Brums [2.3K]3 years ago

6 0

Answer:

Phloem

Explanation:

Took test got right

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Several students want to conduct a descriptive investigation on igneous rocks. Which process is most likely to be part of their

iris [78.8K]

Answer:

Explanation:

They will collect samples from different sites and write their observations

8 0

3 years ago

A reaction occurs in a calorimeter, resulting in the starting and final temperatures shown below. What can you say about the rea

ELEN [110]

"The reaction is exothermic and ΔH is negative" can be understood about the reaction and the enthalpy change (ΔH) during the reaction.

Option: D

Explanation:

When the reaction is positive, the process becomes endothermic, i.e. heat appears to be consumed by the system because the reaction products are more enthalpic than the reactants. When the reaction is negative, on the other hand, the process is exothermic, which is the total decrease in enthalpy is caused by heat production. Here the initial temperature is 21.0 C but increase in final temperature to 38.8 C, because if some processes require heat, others must give off heat when they take place.

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3 years ago

A flashbulb of volume 2.70 mL contains O2(g) at a pressure of 2.30 atm and a temperature of 30.0 °C. How many grams of O2(g) doe

Sholpan [36]

P = 2.30 atm

Volume in liter = 2.70 mL / 1000 => 0.0027 L

Temperature in K = 30.0 + 273 => 303 K

R = 0.082 atm

molar mass O2 = 31.9988 g/mol

number of moles O2 :

P * V = n * R* T

2.30 * 0.0027 = n * 0.082 * 303

0.00621 = n * 24.846

n = 0.00621 / 24.846

n = 0.0002499 moles of O2

Mass of O2:

n = m / mm

0.0002499 = m / 31.9988

m = 0.0002499 * 31.9988

m = 0.008 g

3 0

4 years ago

According to the law of conservation of energy, the energy released by the system must be transferred to and absorbed by the sur

Andrew [12]

True!
Energy released by system is absorbed by surroundings.

5 0

3 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

Answer: The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago