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SashulF [63]
3 years ago
8

why was the mass lost when the reaction was done in the normal setup but stayed the same when it was done in the gas collection

setup
Chemistry
1 answer:
german3 years ago
6 0
<h2>Answer : Law of conservation of mass</h2><h3>Explanation :</h3>

The law of conservation of mass states that in any reaction mass is neither created nor lost it has to remain constant in a system.

In this case, when the reaction setup was done in normal way the mass was  lost in surrounding was not considered nor being calculated; whereas when the reaction was studied in a closed system where the gas was collected after the reaction the mass changes was noted down which helped to prove the point of law of conservation of mass and energy.

One can consider an example of soda can where the carbonated drink contains pressurized carbon dioxide gas. when opened the gas bubbles gets lost into the surroundings and we don't measure the mass changes. Instead if the soda can was opened in such a way where the gas evolved was measured then the mass changed would remain the same.

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The solubility of Z is 60 g/ 100 g water at 20 °C. How many grams of solution are produced when a saturated solution is prepared
V125BC [204]

Answer:

Saturated solution = 180 gram

Explanation:

Given:

Solubility of Z = 60 g / 100 g water

Given temperature =  20°C

Amount of water = 300 grams

Find:

Saturated solution

Computation:

Saturated solution = [Solubility of Z] × Amount of water

Saturated solution = [60 g / 100 g] × 300 grams

Saturated solution = [0.6] × 300 grams

Saturated solution = 180 gram

3 0
3 years ago
Does ecstasy drug effect how long you live?
const2013 [10]

Answer:

yes it gives some bad effect

Explanation:

6 0
2 years ago
What explanation can you give for why the sodium-potassium pump does not run out of ions to move in or out of the cell
rewona [7]

The sodium-potassium pump does not run out of ions since ion exchange is essential for the action potential to take place and to maintain homeostasis.

The cell has variable concentrations of different substances compared to the environment that surrounds it, with significant differences with sodium and potassium.

  • The main function of the sodium-potassium pump is to maintain homeostasis of the intracellular medium, controlling the concentrations of these two ions.

  • In order to carry out the adequate exchange of sodium and potassium ions in the extra and intracellular medium, the cells need an active transport process that is carried out thanks to the sodium potassium pump.

  • This process is needed for the maintenance and functioning of cells, and it is essential for the action potential to be executed, necessary for the transmission of electrical impulses from neuron to neuron.

Therefore, we can conclude that the sodium potassium pump produces an exchange of potassium ions for sodium ions which keeps the cellular system functioning properly.

Learn more here: brainly.com/question/24336764

6 0
2 years ago
an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem
OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

P\cdot V = n\cdot R\cdot T,

\displaystyle V = \frac{n\cdot R\cdot T}{P}.

where

  • P is the pressure of the gas,
  • V is the volume of the gas,
  • n is the number of moles of particles in this gas,
  • R is the ideal gas constant, and
  • T is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of R will be 8.314 if P, V, and T are in SI units. Convert these values to SI units:

  • P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa;
  • V shall be in cubic meters, \rm m^{3};
  • T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K.

Apply the ideal gas law:

\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}.

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