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3 years ago

15

The plant in the photo is called an Asparagus Fern. This fern does not make seeds for reproduction. How does a fern reproduce? Q

uestion 7 options: A fern is a cone-bearing plant. A fern is a spore-producing plant. A fern is a flowering plant. A fern is a monocot and dicot plant.

1 answer:

Pavlova-9 [17]3 years ago

3 0

Ferns use both sexual and asexual reproduction methods. In sexual reproduction, a haploid spore grows into a haploid gametophyte. ... The sporophyte produces spores, completing the life cycle. Asexual methods of reproduction include apogamy, poliferous frond tips, and rhizome spreading.

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Symbol for the magnesium oxide reactants<br>

Stella [2.4K]

Answer:

MgO

Explanation:

Magnesium oxide is formed up of the ions Mg2+ and O2

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3 years ago

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What is the best explanation for why a particle is striking point X?

const2013 [10]

Answer:

What is the best explanation for why a particle is striking point X? When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.

Explanation:

thank me later

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2 years ago

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Identify the acid and conjugate base pair in the following equation:

Serjik [45]

Answer:

hola no ablo inalgun moderador puede acabar a este otro este me borró respuesta que no debió borrar

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2 years ago

Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate

Anvisha [2.4K]

Answer:

$-3272$ kJ/mol

Explanation:

Given and known facts

Mass of Benzene $= 0.187$ grams

Mass of water $= 250$ grams

Standard heat capacity of water $= 4.18$ J/g∙°C

Change in temperature ΔT $= 7.48$ °C

Heat

$=250 * 4.18 * 7.48\\=7816.6 \\=7.82$

Heat released by benzine is - 7.82 kJ

Now, we know that

$0.187$ grams of benzene release $= -7.82$ kJ heat

So, $1$ g benzine releases

$\frac{ -7.82 }{0.187}\\= -41.8$

kJ/g

$0.187 * \frac{1}{78.108}=0.00239$ mol C6H6

Heat released

$= \frac{-7.82}{ 0.00239}$

$=-3272$ kJ/mol

4 0

2 years ago

Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the

GarryVolchara [31]

Answer : The final temperature of the metal block is, $25^oC$

Explanation :

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminum = 55 g

$m_2$ = mass of water = 0.48 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminum = $25^oC$

$T_2$ = temperature of water = $25^oC$

$c_1$ = specific heat of aluminum = $0.900J/g^oC$

$c_2$ = specific heat of water= $4.184J/g^oC$

Now put all the given values in equation (1), we get

$55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]$

$T_{final}=25^oC$

Thus, the final temperature of the metal block is, $25^oC$

6 0

3 years ago