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Andrej [43]
3 years ago
14

Which of the following "spheres" is NOT involved in the water cycle?

Chemistry
2 answers:
olya-2409 [2.1K]3 years ago
5 0
B is the correct answer.
The exosphere is the outermost region of the atmosphere the gradually merges out into space. The water cycle does not occur within this region.
harina [27]3 years ago
5 0
The answer is B. exosphere
You might be interested in
Which subshells (s, p, d, f, or g) can have electrons with the indicated magnetic quantum number (ml)?
amm1812

Answer:

=3 means is 3 or greater so that would be f and g subshells

=0 means is 0 or greater so that would be s, p, d, f and g subshells

=1 means is 1 or greater so that would be p, d, f, and g subshells

=4 means is 4 or greater so that would be g only

4 0
2 years ago
Calculate the grams of solute in each of the following solution: 278 mL of 0.038 M Fe2(SO4)3
Goryan [66]

Answer:  4.22 grams of solute is there in 278 ml of 0.038 M Fe_2(SO_4)_3

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

Now put all the given values in the formula of molality, we get

0.038M=\frac{n}{0.278L}

n=0.0105mol  

mass of  Fe_2(SO_4)_3 = moles\times {\text {Molar Mass}}=0.0105\times 399.88g/mol=4.22g

Thus 4.22 grams of solute is there in 278 ml of 0.038 M Fe_2(SO_4)_3

3 0
3 years ago
An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
Tju [1.3M]

Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0
3 years ago
The following reactions take place in the ozone layer by the absorption of ultraviolet light.
____ [38]

Answer:

Option B will require a shorter wave length of light.

Explanation:

The bonding between Ozone (O3) and Oxygen (O2) can be used to explain why the breaking of oxygen into Oxygen radicals will require a shorter wave length.

  • The bond between Oxygen (O2) is a double bond while Ozone (O3) has an intermediate bond between a double bond and a single bond.
  • The bond order of Oxygen (O2) is equals 2 while that of Ozone (O3) is 1.5. Since the bond order of oxygen is higher, it will require more energy to break the bond compared to breaking the Ozone (O3) bond.
  • Recall that Energy is inversely proportional to wave length.
  • So it will require a shorter wave length to break the Oxygen (O2) into its radicals.
5 0
3 years ago
In a single displacement reaction between sodium phosphate and barium, how much of each product (in grams) will be formed from 1
weeeeeb [17]

Answer:

A. 3.36g of Na.

B. 14.62g of Ba3(PO4)2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Ba + 2Na3PO4 → 6Na + Ba3(PO4)2

Next, we shall determine the mass of Ba that reacted and the mass of Na and Ba3(PO4)2 produced from the equation.

This is illustrated below:

Molar Mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Na = 23g/mol

Mass of Na from the balanced equation = 6 x 23 = 138g

Molar mass of Ba3(PO4)2 = (3 x 137) + 2[31 + (4x16)] = 411 + 2[31 + 64] = 601g/mol

Mass of Ba3(PO4)2 from the balanced equation = 1 x 601 = 601g

Summary:

From the balanced equation above,

411g of Ba reacted to produce 138g of Na and 601g of Ba3(PO4)2.

A. Determination of the mass of Na produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 138g of Na.

Therefore, 10g of Ba will react to produce = (10 x 138)/411 = 3.36g of Na.

Therefore, 3.36g of Na is produced.

B. Determination of the mass of Ba3(PO4)2 produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 601g of Ba3(PO4)2.

Therefore, 10g of Ba will react to produce = (10 x 601)/411 = 14.62g of Ba3(PO4)2.

Therefore, 14.62g of Ba3(PO4)2 is produced.

7 0
3 years ago
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