why the total mass of the products would be less than the total weight of the reactant after a chemical reaction?

Chemistry

1 answer:

Lisa [10]2 years ago

4 0

The best explanation would be that Gases were released during the Chemical reaction, causing a loss of Mass.

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Which reaction would cause an increase in entropy? A. SO2Cl2(g) SO2(g) + Cl2(g) B. PCl3(g) + Cl2(g) PCl5(g) C. 2CO(g) + O2(g) 2C

IrinaVladis [17]

A. SO2Cl2(g) --> SO2(g) + Cl2(g)
1 mole of SOCl2 becomes 1 mole SO2 and 1 mole Cl2 
1 mole --> 2 moles 
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8 0

3 years ago

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A gas at 200 K occupies a volume of 350 ml. What temperature is needed to increase the volume

Rudik [331]

Answer: Temperature in constant pressure is 286 K

Explanation: If pressure remains constant, then V/T = constant.

V1 = 350 ml and T1 = 200 K and V2 = 500 ml. V1/T1 = V2/T2

and T2 = T1· V2 / V1 = 200 K · 500 ml / 300 ml = 285,7 K

5 0

2 years ago

If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio

bulgar [2K]

Answer:

$m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

$n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2$

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

$m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2$