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3 years ago

Name two different ecosystems.Name one living part of each ecosystem.Name one non living part of each ecosystem.

2 answers:

7 0

Ecosystems: Coral Reef, Savanah
Living: In coral reef; clown fish In Savanah: Emu
Nonliving: In Coral reef: Water In Savanah: sun

ss7ja [257]3 years ago

5 0

There is a freshwater ecosystem one living thing is fishes a non living thing is lots of concentrations of dissolved oxygen. Another ecosystem would be a forest ecosystem one living thing would be insects one non living would be water

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In a single displacement reaction between Sodium Phosphate and Barium, how much of each product (in grams) will be formed from 1

mixer [17]

Answer:

14.6 g of barium phosphate

3.35 g of sodium metal

Explanation:

2Na3PO4(aq) + 3Ba(s) -------> Ba3(PO4)2(aq) + 6Na(s)

The first step in any such reaction is to but down the balanced reaction equation according to the stoichiometry of the reaction.

The two products formed are barium phosphate and sodium metal.

Number of moles of barium corresponding to 10.0g of barium = mass of barium/ molar mass of barium

Molar mass of barium = 137.327 g

Number of moles of barium = 10/137.327

Number of moles of barium = 0.0728 moles

For barium phosphate;

3 moles of barium yields 1 mole of barium phosphate

0.0728 moles yields 0.0728 moles × 1/3 = 0.0243 moles of barium phosphate

Molar mass of barium phosphate = 601.93 g/mol

Therefore mass of barium phosphate = 0.0243 moles × 601.93 g/mol = 14.6 g of barium phosphate

For sodium metal

3 moles of barium yields 6 moles of sodium metal

0.0728 moles of barium yields 0.0728 × 6 / 3 = 0.1456 moles of sodium

Molar mass of sodium metal= 23 gmol-1

Mass of sodium metal= 0.1456g × 23 gmol-1 = 3.35 g of sodium metal

4 0

3 years ago

Determine the empirical formula of the compound formed when 1.2g of magnesium reacts with 3.55g of chlorine.Take the molar mass

masha68 [24]

Explanation:

For Mg, (1.2 g Mg/24 g Mg) = 0.05 mol Mg.

For Cl, (3.55 g Cl/35.5 g Cl) = 0.1 mol Cl

So the ratio now is

Mg:Cl = 0.05 : 0.1 = 1:2

I got the 1:2 ratio by dividing both by the smallest number, which is 0.05 mol. Therefore, the empirical for formula of the substance is $MgCl_2$

5 0

3 years ago

A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is −0.3 kJ/K

Karolina [17]

Answer:

W = -120 KJ

Explanation:

Since the piston–cylinder assembly undergoes an isothermal process, then the temperature is constant.

Thus; T1 = T2 = 400K

change in entropy; ΔS = −0.3 kJ/K

Formula for change in entropy is written as;

ΔS = Q/T

Where Q is amount of heat transferred.

Thus;

Q = ΔS × T

Q = -0.3 × 400

Q = -120 KJ

From the first law of thermodynamics, we can find the workdone from;

Q = ΔU + W

Where;

ΔU is Change in the internal energy

W = Work done

Now, since it's an ideal gas model, the change in internal energy is expressed as;

ΔU = m•C_v•ΔT

Where;

m is mass

C_v is heat capacity at constant volume

ΔT is change in temperature

Now, since it's an isothermal process where temperature is constant, then;

ΔT = T2 - T1 = 0

Thus;

ΔU = m•C_v•ΔT = 0

ΔU = 0

From earlier;

Q = ΔU + W

Thus;

-120 = 0+ W

W = -120 KJ

8 0

3 years ago

Which two factors that influence weather are caused by the uneven heating of

just olya [345]

B, sorry if i am wrong.

8 0

3 years ago

Read 2 more answers

I need help with this plz help

schepotkina [342]

Answer:

$\rm ^{103}_{\phantom{1}40}Zr$ , zirconium-103.

Explanation:

In a nuclear reaction, both the mass number and atomic number will conserve.

Let $^{A}_{Z}\mathrm{X}$ represent the unknown particle.

The mass number of a particle is the number on the upper-left corner. The atomic number of a particle is the number on its lower-left corner under the mass number. For example, for the particle $^{A}_{Z}\mathrm{X}$ , $A$ is the mass number while $Z$ while $Z$ is the atomic number.

Sum of mass numbers on the left-hand side of the equation:

$\underbrace{239}_{^{239}_{\phantom{2}94}\mathrm{Pu}} + \underbrace{1}_{^{1}_{0}\mathrm{n}} = 240$ .

Note that there are three neutrons on the right-hand side of the equation. Sum of mass numbers on the right-hand side:

$\underbrace{A}_{^{A}_{Z}\mathrm{X}} + \underbrace{134}_{^{134}_{\phantom{2}54}\mathrm{Xe}} + \underbrace{3\times 1}_{3\;^{1}_{0}\mathrm{n}} = A + 137$ .

Mass number conserves. As a result,

$A + 137 = 240$ .

Solve this equation for $A$ :

$A = 103$ .

Among the five choices, the only particle with a mass number of 103 is $\rm ^{103}_{\phantom{1}40}Zr$ . Make sure that atomic number also conserves.

3 0

3 years ago