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xxMikexx [17]
2 years ago
8

Explain it to you or more sentences using your own words what you learned about biomes organisms and populations

Chemistry
1 answer:
andrezito [222]2 years ago
5 0
A biome is a large area characterized by its vegetation, soil, climate, and wildlife. ... Marine biomes cover close to three-quarters of Earth's surface. Marine biomes include the ocean, coral reefs, and estuaries. Grasslands are open regions that are dominated by grass and have a warm, dry climate.
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An aluminum block has a density of 2.70 g/mL. If the mass of the block is 24.60 g, find the volume of the substance.
harina [27]

Volume of a substance can be determined by dividing mass of the substance by its density.

That can be mathematical shown as:

Density=Mass/Volume

So, Volume=Mass/Density

Here mass of the substance given as 24.60 g

Whereas density of the substance is 2.70 g/mL

So,

Volume=Mass/Density

=24.6/2.7

=9.1 mL

So volume of the substance is 9.1 mL.

8 0
3 years ago
A geochemist in the field takes a 46.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.
inysia [295]

Answer:

The solubility of the mineral compound X in the water sample is 0.0189 g/mL.

Explanation:

Step 1: Given data

The volume of water sample = 46.0 mL.

The weight of the mineral compound X after evaporation, drying, and washing = 0.87 g.

Step 2: Calculate the solubility of X in water

46.00 mL of water sample contains 0.87 g of the mineral compound X.

To calulate how many grams of the mineral compound  1.0 mL  of water sample contains:

0.87 g/46.0 mL = 0.0189 g.

This means the solubility of the mineral compound X in the water sample is 0.0189 g/mL.

3 0
3 years ago
Which ion in the ground state has the same electron configuration as an atom of
Sergeu [11.5K]

Answer:

Ca^+2

Explanation:

Hence, the correct option is A.

4 0
3 years ago
A 25.0 mL sample of 0.100 M lactic acid (HC3H503, Ka = 1.4 E-4) is titrated with 0.100 M NaOH solution. Calculate the pH of the
aleksklad [387]
I’m a bit confused too
4 0
3 years ago
One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 C, and the temperature af
Licemer1 [7]

Answer:

fH = - 3,255.7 kJ/mol

Explanation:

Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

4 0
4 years ago
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