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notsponge [240]

2 years ago

10

True or False: Mutually beneficial relationships, predator-prey relationships, and competition can happen in different environme

nts.

1 answer:

Masja [62]2 years ago

8 0

Answer:

Yes :)

Explanation:

Mutual relationships can include birds cleaning out aligators' teeth, which helps with their hygiene, and gives the bird a meal!

Predator-prey would be a cheetah-antelope.

Competition could be an invasive species :)

You might be interested in

Explain why the C―Ha bond is much more acidic than the C―Hb bond in pentan-2-one. Select the single best answer. 253 Ha is less

mafiozo [28]

Answer:

Ha is more acidic than Hb because loss of Ha forms a resonance-stabilized conjugate base.

Explanation:

The carbon atom that is next to the carbonyl group in pentan-2-one is known as the alpha carbon atom, this carbon atom bears the Ha, the alpha hydrogen atoms.

Ha is more acidic than Hb because, loss of Ha leads to the formation of a resonance stabilized enolate ion. This resonance stabilization of the ion formed makes loss of Ha an easier process than loss of Hb, hence the answer above.

3 0

4 years ago

In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to

labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of $NH_3$ = 100 g

Molar mass of $NH_3$ = 27 g/mole

Molar mass of $N_2$ = 28 g/mole

First we have to calculate moles of $NH_3$ .

$\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles$

The given balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the given reaction, we conclude that

2 moles of $NH_3$ produced from 1 mole of $N_2$

3.7 moles of $NH_3$ produced from $\frac{1mole}{2mole}\times 3.7mole=1.85moles$ of $N_2$

Now we have to calculate the mass of $N_2$ .

Mass of $N_2$ = Moles of $N_2$ × Molar mass of $N_2$

Mass of $N_2$ = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0

3 years ago

The smallest unit that can exist as an element and still have the properties of that element is a(n):

Yuri [45]

The smallest unit that can exist as an element and still have proprieties of that element is an Atom<span />

8 0

3 years ago

Read 2 more answers

Omg GUYS I NEED HELPPP

Ilia_Sergeevich [38]

27) Partial pressure of oxygen: 57.8 kPa

29) Final volume: 80 mL

30) Final volume: 8987 L

31) Due to property of water of being polar, ice floats on water

Explanation:

27)

In a mixture of gases, the total pressure of the mixture is the sum of the partial pressures:

$p_T = p_1 + p_2 + ... + p_N$

In this problem, the mixture contains 3 gases (helium, carbon dioxide and oxygen). We know that the total pressure is

$p_T=201.4 kPa$

We also know the partial pressures of helium and carbon dioxide:

$P_{He}=125.4 kPa\\P_{CO_2}=18.2 kPa$

The total pressure can be written as

$p_T=p_{He}+p_{CO_2}+p_{O_2}$

where $p_{O_2}$ is the partial pressure of oxygen. Therefore, we find

$p_{O_2}=p_T-p_{He}-p_{CO_2}=201.4-125.4-18.2=57.8 kPa$

29)

Assuming that the pressure of the gas is constant, we can apply Charle's law, which states that:

"For an ideal gas at constant pressure, the volume of the gas is proportional to its absolute temperature"

Mathematically,

$\frac{V}{T}=const.$

where

V is the volume of the gas

T is the Kelvin temperature

We can re-write it as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Here we have:

$V_1 = 42 mL$ (initial volume)

$T_1=-89^{\circ}C+273=184 K$ is the initial temperature

$T_2=77^{\circ}C+273=350 K$ is the final temperature

Solving for V2, we find the final volume:

$V_2=\frac{V_1 T_2}{T_1}=\frac{(42)(350)}{184}=80 mL$

30)

For this problem, we can use the equation of state for ideal gases, which can be written as

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where in this problem:

$p_1 = 102.3 kPa$ is the initial pressure

$V_1=1975 L$ is the initial volume

$T_1=25^{\circ}C+273=298 K$ is the initial temperature

$p_2=21.5 kPa$ is the final pressure

$T_2=12^{\circ}C+273=285 K$ is the final temperature

And solving for V2, we find the final volume of the balloon:

$V_2=\frac{p_1 V_1 T_2}{p_2 T_1}=\frac{(102.3)(1975)(285)}{(21.5)(298)}=8987 L$

31)

A molecule of water consists of two atoms hydrogen bond with an atom of oxygen ( $H_2 O$ ) in a covalent bond.

While the molecul of water is overall neutral, due to the higher electronegativity of the oxygen atom, electrons are slightly shifted towards the oxygen atom; as a result, there is a slightly positive charge on the hydrogen side, and a slightly negative charge on the oxygen side (so, the molecules is said to be polar).

As a consequence, molecules of water attract each other, forming the so-called "hydrogen bonds".

One direct consequence of the polarity of water is that ice floats on liquid water.

Normally, for every substance on Earth, the solid state is more dense than the liquid state. However, this is not true for water, because ice is less dense than liquid water.

This is due to the polarity of water. In fact, when the temperature of water is decreased to freezing point and water becomes ice, the hydrogen bondings "force" the molecules to arrange in a lattice structure, so that the molecules become more spaced when they turn into solid state. As a result, ice occupies more volume than water, and therefore it is less dense, being able to float on water.

Learn more about ideal gases:

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#LearnwithBrainly

4 0

3 years ago

In an electroplating process, copper (ionic charge +2e, atomic weight 63.6 g/mol) is deposited using a current of 10.0 A. What m

salantis [7]

Answer : The mass of copper deposit is, 1.98 grams

Explanation :

First we have to calculate the charge.

Formula used : $Q=I\times t$

where,

Q = charge = ?

I = current = 10 A

t = time = 10 min = 600 sec (1 min = 60 sec)

Now put all the given values in this formula, we get

$Q=10A\times 600s=6000C$

Now we have to calculate the number of atoms deposited.

As, 1 atom require charge to deposited = $2\times (1.6\times 10^{-19})$

Number of atoms deposited = $\frac{(6000)}{2\times(1.6\times 10^{-19})}=1.875\times 10^{22}$ atoms

Now we have to calculate the number of moles deposited.

Number of moles deposited = $\frac{(1.875\times 10^{22})}{(6.022\times 10^{23})}=0.03113$ moles

Now we have to calculate the mass of copper deposited.

1 mole of Copper has mass = 63.5 g

Mass of Copper Deposited = $63.5\times 0.03113 =1.98g$

Therefore, the mass of copper deposit is, 1.98 grams

5 0

4 years ago