Determine the elements present in sodium hydrogen carbonate

According to the law of conservation of mass, when sodium, hydrogen, and oxygen react to form a compound, the mass of the compou

krok68 [10]

I think the correct term to fill in the blank would be equal. According to the law of conservation of mass, when sodium, hydrogen, and oxygen react to form a compound, the mass of the compound is equal the sum of the masses of the individual elements. This particular states that the mass cannot be created nor destroyed. So, for every process, the total inflow of mass should be the same total mass for the outflow since mass should be constant all through out. So, for a reaction, the total mass of the reactants should be equal to the total mass of the products.

6 0

4 years ago

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

Maksim231197 [3]

The theoretical yield of $SnS_2$ will be 4.20 grams while the percent yield will be 7.93%

<h3>How is yield calculated?</h3>

From the equation of the reaction, the mole ratio of $SnBr_4$ to $Na_2S$ is 1:2.

Mole of 48.1 mL, 0.478 M $SnBr_4$ = 0.478 x 48.1/100 = 0.023 mols

Mole of 48.8 mL, 0.160 M $Na_2S$ = 0.160 x 48.8/1000 = 0.0078 moles

$SnBr_4$ $Na_2S$ is the limiting reactant.

Mole ratio of $SnBr_4$ and $SnS_2$ = 1:1

Equivalent mole of $SnS_2$ = 0.023 moles

Mass of 0.023 noles $SnS_2$ = 0.023 x 182.81 = 4.20 grams

With 0.0333 g of $SnS_2$ recovered, percent yield = 0.333/4.2 x 100 = 7.93%

More on yields of reactions can be found here: brainly.com/question/17042787

#SPJ1

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)

Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.

1) Calculate the theoretical yield of SnS2. ;

2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.

7 0

2 years ago

According to the following reaction, how many grams of chloric acid (HClO3) are produced in the complete reaction of 31.6 grams

gogolik [260]

Answer:

$m_{HClO_3}=12.7gHClO_3$

Explanation:

Hello,

Considering the reaction:

$3Cl_2(g)+3H_2O(l)-->5HCl+HClO_3$

The molar masses of chlorine and chloric acid are:

$M_{Cl_2}=35.45*2=70.9g/mol\\M_{HClO_3}=1+35.45+16*3=84.45g/mol$

Now, we develop the stoichiometric relationship to find the mass of chloric acid, considering the molar ratio 3:1 between chlorine and chloric acid, as follows:

$m_{HClO_3}=31.6gCl_2*\frac{1molCl_2}{70.9gCl_2} *\frac{1molHClO_3}{3mol Cl_2} *\frac{85.45g HClO_3}{1mol HClO_3} \\m_{HClO_3}=12.7gHClO_3$

Best regards.

4 0

4 years ago

Sean is moving heavy furniture by sliding it across the floor. he puts a blanket under Meath the furniture why did this help him

Neko [114]

The answer is C.....

5 0

3 years ago

How many grams of H2O are there in 6.02 x 1023 molecules of water?

tamaranim1 [39]

Answer:

By definition, there are 6.022×1023 such molecules, or NA such molecules in ONE mole of water. And thus in such a quantity there are NA oxygen atoms, and 2×NA hydrogen atoms...and the mass associated with this numerical quantity of water molecules is approx. 18⋅g ...

4 0

3 years ago

Determine the elements present in sodium hydrogen carbonate ​

Determine the elements present in sodium hydrogen carbonate