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4 years ago

Spores are a form of asexual reproduction formed in ___.

Chemistry

2 answers:

antoniya [11.8K]4 years ago

6 0

Answer & Explanation:

Spores are usually haploid and unicellular and are produced by meiosis in the sporangium of a diploid sporophyte. Under favourable conditions the spore can develop into a new organism using mitotic division, producing a multicellular gametophyte, which eventually goes on to produce gametes.

olga nikolaevna [1]4 years ago

6 0

Answer:

Spores are a form of asexual reproduction in bacteria, fungi, algae and plants.

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Water boiling and turning into steam is a chemical or physical change?

olga2289 [7]

Answer:

physical change

Explanation:

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3 years ago

Read 2 more answers

If kinetic energy increases does pressure decrease?

enyata [817]

If molecules are in a closed container then we expect the pressure to increase as the kinetic energy increases. This is because the atoms of an element collide with the walls of the container and increase the pressure.

If we use the formula $PV=nRT$ , where P is the pressure, V is the volume, n is the number of moles, R the ideal gas constant and T is the temperature. According to the formula, P is directly proportional to temperature. An increase in temperature leads to an increase in pressure.

Since we know that temperature is the average kinetic energy of molecules present. It means as we increase the temperature we increase the kinetic energy of the molecules which in turn leads to an increase in the pressure.

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3 years ago

5. A beam of photons with a minimum energy of 222 kJ/mol can eject electrons from a potassium surface. Estimate the range of wav

torisob [31]

Answer: The range of wavelengths of light that can be used to cause given phenomenon is $8.953 \times 10^{21} m$ .

Explanation:

Given: 222 kJ/mol (1 kJ = 1000 J) = 222000 J

Formula used is as follows.

$E = \frac{hc}{\lambda}$

where,

E = energy

h = Planck's constant = $6.625 \times 10^{-25} Js$

c = speed of light = $3 \times 10^{8} m/s$

Substitute the values into above formula as follows.

$E = \frac{hc}{\lambda}\\222000 J = \frac{6.625 \times 10^{-34}Js \times 3 \times 10^{8} m/s}{\lambda}\\\lambda = 8.953 \times 10^{21} m$

Thus, we can conclude that the range of wavelengths of light that can be used to cause given phenomenon is $8.953 \times 10^{21} m$ .

7 0

3 years ago

6. What parts of a nuclear reactor keep the reaction from

Shalnov [3]

Answer:

B. control rods and moderators

5 0

3 years ago

Read 2 more answers

Propane (C3H8) can be burned to produce heat for homes. The products of the reaction are CO2 and H2O. For complete combustion to

Natalija [7]

Answer:

1- 3 Moles of CO2

2- 132 g of CO2

3- 105,6 g of CO2

4- Limiting Reagent O2

Products form based on limiting reagent (384g O2) :

CO2: 316,8 g

H2O: 172,8 g

Products form based on C3H8 (132,33 g):

CO2: 396,99 g

H2O: 216,54 g

Explanation:

Atomic Masses:

C: 12

H: 1

O: 16

Molecular weights:

C3H8: 44 g

O2: 32 g

H2O: 18 g

CO2: 44 g

C3H8 + 5 O2⇒ 3 CO2 + 4 H2O

C3H8 (44g)+ O2 (160 g) ⇒ CO2 (132 g) + H2O (72 g)

In 5 moles of O2 are produced 3 moles of CO2, equivalent to 132 g

For 160g of O2 are produced 132 g of CO2, so 128 g of O2

160 g O2 ⇒ 132 g CO2

128 g O2⇒ × = 105,6 g CO2

(128×132÷160= 105,6)

The limiting reagent is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, because the reaction cannot continue without it.

If I have 132,33 g of C3H8 and 384 g O2 we can calculate:

For 44 g of CH3H8 ⇒160 g of O2

With 132,33 g of CH3H8 ⇒ ×= 481,2 g of O2

(132,33×160÷44=481,2)

As this amount exceeds the quantity of O2 that we have, we can assume that the 384 g O2 will be totally consumed.

Calculations of the products formed in base of quantity of O2 (limiting reagent):

160 g O2 ⇒ 132 g CO2

384 g O2⇒ × = 316,8 g CO2

(384×132÷160= 316,8)

160 g O2 ⇒ 72 g H2O

384 g O2⇒ × = 172,8 g H2O

(384×72÷160= 172,8)

Calculations of the products formed in base of quantity of C3H8 (excess reagent):

44 g C3H8 ⇒ 132 g CO2

132,33 g C3H8 ⇒ × = 396,99 g CO2

(132,33×132÷44=396,99)

44 g C3H8 ⇒ 72 g H2O

132,33 g C3H8⇒ × = 216,54 g H2O

(132,33×72÷44= 216,54)

3 0

3 years ago