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3 years ago

What is the five spheres?❤️❤️

1 answer:

Taya2010 [7]3 years ago

5 0

The Lithosphere ("rock sphere") is the ground you are standing on and the whole inside of Earth.

The Hydrosphere ("water sphere") includes all of the rivers, lakes and oceans of Earth.

The Cryosphere ("icy cold sphere") is the frozen part of Earth: the glaciers, icebergs at sea, and the huge icecaps in Greenland and Antarctica.

The Biosphere ("Life sphere") includes all living things: the trees in the park, the birds in the air, the fly on your wall, the viruses that make you sick, your pets, and even you and all your friends!

The Atmosphere ("Air Sphere") is the envelope of air that surrounds the whole Earth.

The Exo- or Celestial Sphere ("Outside or heavenly sphere") includes the whole universe beyond the top of the atmosphere--the Sun, Moon, and stars, as well as the asteroids and the little bits of dust that make meteors when they hit the atmosphere.

You might be interested in

Which of the following represents the law of conservation of mass (a balanced chemical equation)?

lilavasa [31]

The reaction represents the law of conservation of mass (a balanced chemical equation :

2NaN₃ → 2Na + 3N₂

<h3>Further explanation </h3>

Conservation of mass applies to a closed system, where the masses before and after the reaction are the same, so balanced chemical equations show that mass is conserved in chemical reactions.

So we just have to look at the amount of each type of element present in the reactants and products must be the same

1. Hg + O₂ → HgO

not balanced, because element O not equal(2 in left, 1 in right)

2. 2NaN₃ → 2Na + 3N₂

balanced, because element in reactant and product are the same

Na, left = 2, right=2

N, left =2x3=6, right=3x2=6

3. C+O → CO₂

not balanced, because element O not equal(1 in left, 2 in right)

4. CH₄ + O₂ → CO₂ +H₂O

not balanced, because element O and H not equal

8 0

3 years ago

A solution contains 10.20 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL

AveGali [126]

The question is incomplete, here is the complete question:

A solution contains 10.20 g of unknown compound dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.21°C. The mass percent composition of the compound is 60.98% C , 11.94% H , and the rest is O.

What is the molecular formula of the compound?

Answer: The molecular formula for the given organic compound is $C_6H_{14}O_2$

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 50.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{50.0mL}\\\\\text{Mass of water}=(1g/mL\times 50.0mL)=50g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and the freezing point of solution

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=i\times K_f\times m$

Or,

$\text{Freezing point of pure solution}-\text{freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution (water) = 0°C

Freezing point of solution = -3.21°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal boiling point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute = 10.20 g

$M_{solute}$ = Molar mass of solute = ?

$W_{solvent}$ = Mass of solvent (water) = 50.0 g

Putting values in above equation, we get:

$(0-(-3.21))^oC=1\times 1.86^oC/m\times \frac{10.20\times 1000}{M_{solute}\times 50}\\\\M_{solute}=\frac{1\times 1.86\times 10.20\times 1000}{3.21\times 50}=118.2g$

Calculating the molecular formula:

We are given:

Percentage of C = 60.98 %

Percentage of H = 11.94 %

Percentage of O = (100 - 60.98 - 11.94) % = 27.08 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 60.98 g

Mass of H = 11.94 g

Mass of O = 27.08 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{60.98g}{12g/mole}=5.082moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{11.94g}{1g/mole}=11.94moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{27.08g}{16g/mole}=1.69moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.69 moles.

For Carbon = $\frac{5.082}{1.69}=3$

For Hydrogen = $\frac{11.94}{1.69}=7.06\approx 7$

For Oxygen = $\frac{1.69}{1.69}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 7 : 1

The empirical formula for the given compound is $C_3H_7O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 118.2 g/mol

Mass of empirical formula = 59 g/mol

Putting values in above equation, we get:

$n=\frac{118.2g/mol}{59g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(7\times 2)}O_{(1\times 2)}=C_6H_{14}O_2$

Hence, the molecular formula for the given organic compound is $C_6H_{14}O_2$

8 0

3 years ago

A testable question is one that _____.

cestrela7 [59]

Answer:

One that “Can be answered by conducting an experiment”

Explanation:

8 0

3 years ago

Read 2 more answers

A teapot with a surface area of 785 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic

lina2011 [118]

Answer:

Given info: The surface area of teapot plated with silver is 700 cm2700 cm2, the cell is powered by 12.0-V12.0-V, the resistance of the cell is 1.80 Ω1.80 Ω, thickness of silver layer is 0.133-mm0.133-mm and density of silver is 10.5×103 kg/m310.5×103 kg/m3.

Write the expression for the mass of the silver.

m=ρAdm=ρAd (1)

Here,

ρρ is the density of silver.

AA is the surface area.

dd is the thickness of the silver layer.

Substitute 10.5×103 kg/m310.5×103 kg/m3 for ρρ, 700 cm2700 cm2 for AA and 0.133-mm0.133-mm for dd in equation (1) to find mass of the silver.

m=(10.5×103 kg/m3)(700 cm2(10−21 cm)2)(0.133-mm(10−3 m1 mm))=0.0978 kg(103 g1 kg)=97.8 gm=(10.5×103 kg/m3)(700 cm2(10−21 cm)2)(0.133-mm(10−3 m1 mm))=0.0978 kg(103 g1 kg)=97.8 g

Thus, the mass of silver is 97.8 g97.8 g.

Write the expression for number of moles of silver.

n=mWan=mWa (2)

Here,

mm is the mass of silver.

WaWa is the atomic weight.

The atomic weight of silver is 107.8 g/mole107.8 g/mole

Substitute 97.8 g97.8 g for mm, and 107.8 g/mole107.8 g/mole for WaWa in equation (2) to find number of moles of silver.

n=97.8 g107.8 g/mole=0.907 moln=97.8 g107.8 g/mole=0.907 mol

Thus, the number of moles of silver is 0.907 mol0.907 mol.

7 0

4 years ago

How is matter measured

Maksim231197 [3]

Answer:

Explanation:Matter can be described by certain measurements. They include length, area, and volume. Length tells you the number of units that fit along one edge on an object. In the metric system, units of length are based on the meter.

3 0

3 years ago