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german
2 years ago
14

What is the final volume of a gas (in liters) when the initial volume is 14.00 L at

Chemistry
1 answer:
8_murik_8 [283]2 years ago
4 0

Answer:

V_2=1344L

Explanation:

Hello there!

In this case, since we have a problem about volume-pressure relationship, the idea here is to use the Boyle's law to calculate the final volume as shown below:

P_2V_2=P_1V_1\\\\V_2=\frac{P_2V_2}{P_1}\\

Then, we plug in the initial and final pressures and the initial volume to obtain:

V_2=\frac{14.00L*0.9600atm}{0.01000atm}\\\\V_2=1344L

Regards!

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Which process releases energy that eventually produces lightning in the thunderstorm?
Ira Lisetskai [31]

Answer:

Answer: B. Water condenses to form clouds.

Explanation:

When the moisture condenses, this results in the release of energy. The energy causes the air to be warm and results in the rise of air in the upper atmosphere. This process results in the instability in the atmosphere and cumulonimbus clouds are formed. These clouds support lightening during a thunderstorm.

4 0
3 years ago
Americans spend up to $100 billion annually for bottled water (41 billion gallons). The only beverages with higher sales are car
grigory [225]

<u>Answer:</u> C) be hypertonic to Tank B.

<u>Explanation: </u>

<u> The ability of an extracellular solution to move water in or out of a cell by osmosis</u> is known as its tonicity. Additionally, the tonicity of a solution is related to its osmolarity, which is the <u>total concentration of all the solutes in the solution. </u>

Three terms (hypothonic, isotonic and hypertonic) are used <u>to compare the osmolarity of a solution with respect to the osmolarity of the liquid that is found after the membrane</u>.  When we use these terms, we only take into account solutes that can not cross the membrane, which in this case are minerals.

  1. If the liquid in tank A has a lower osmolarity (<u>lower concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypotonic with respect to the latter.
  2. If the liquid in tank A has a greater osmolarity (<u>higher concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypertonic with respect to the latter.
  3. If the liquid in tank A has the same osmolarity (<u>equal concentration of solute</u>) as the liquid in tank B, the liquid in tank A would be isotonic with respect to the latter.

In the case of the problem, option A is impossible because the minerals can not cross the membrane, since it is permeable to water only. There is no way that the concentration of minerals decreases in tank A, so <u>the solution in this tank can not be hypotonic with respect to the one in Tank B. </u>

Equally, both solutions can not be isotonic and neither we can say that the solution in tank A has more minerals that the one in tank B because the liquid present in tank B is purified water that should not have minerals. Therefore, <u>options B and D are also not correct.</u>

Finally, the correct option is C, since in the purification procedure the water is extracted from the solution in tank A to obtain a greater quantity of purified water in tank B. In this way, the solution in Tank A would be hypertonic to Tank B.

5 0
3 years ago
The smallest unit of an element that can exist either alone or in combination with other such particles of the same or different
Mekhanik [1.2K]

Answer

im not quite sure but I think the answer is <em>D atom</em><em> </em>

Explaination

5 0
2 years ago
Pls help me on this it’s past due ty
mihalych1998 [28]

Answer:

sorry

Explanation:

6 0
2 years ago
When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.
Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2

And the resulting percent yield:

Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%

Regards!

4 0
3 years ago
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