MgO(s) + H2O(l) → Mg(OH)2(aq)

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

german

<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.

6 0

3 years ago

What is one property that is different between water and oil? *

Arlecino [84]

Oil is less dense than water, so the difference would be its density. Water is a good solvent, which means It can dissolve other substances.

3 0

2 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

A gas is contained in a cylinder with a volume of 2.9 L at a temperature of 32.7oC and a pressure of 645.3 torr. The gas is then

stealth61 [152]

Answer: 41 atm

Explanation:

Given that:

Original Volume of gas V1 = 2.9L

Temperature T1 = 32.7°C

Convert Celsius to Kelvin

(32.7°C + 273 = 305.7K)

Pressure P1 = 645.3 torr

New Volume V2 = 0.23 L

New temperature T2 = 894.7°C

Convert Celsius to Kelvin

(894.7°C + 273 = 1167.7K)

New pressure = ?

Then, apply the combined gas equation

(P1V1)/T1 = (P2V2)/T2

(645.3 torr x 2.9L)/305.7K = (P2 x 0.23L)/1167.7K

1871.37 / 305.7 = 0.23P2 / 1167.7

To get P2, Cross multiply

1871.37 x 1167.7 = 305.7 x 0.23P2

2185198.749 = 70.311P2

Divide both sides by 70.311

2185198.749/70.311 = 70.311P2/70.311

31079.045 torr = P2

Now, convert pressure in torr to atmosphere

Since 760 torr = 1 atm

31079.045 torr = Z

cross multiply

760 torr x Z = 31079.045 torr x 1 atm

Z = 31079.045 torr / 760 torr

Z = 40.89 atm (Round to the nearest whole number as 41 atm)

Thus, new pressure of gas is 41 atm

3 0

3 years ago

Can someone help please !!!!

Naya [18.7K]

Answer:

x = 231/90 Th

Explanation:

3 0

3 years ago