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klio [65]
3 years ago
8

Hydrogen Cyanide, HCN, is a poisonous gas. The

Chemistry
1 answer:
Nikolay [14]3 years ago
5 0

mass of NaCN = 0.539 g

Further explanation

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

2NaCN (s) + H₂SO₄ (aq) ⇒ Na₂S0₄ (aq) + 2HCN (g)

MW HCN = 27,0253 g/mol

MW NaCN = 49,0072 g/mol

Assume there is 1 kg air, and mass of HCN = 300 mg

mol HCN :

mass 300 mg = 0.3 g

\tt mol=\dfrac{0.3}{27.0253}=0.011

NaCN : HCN = 2 : 2(mole ratio from equation), so :

mol NaCN = mol HCN = 0.011

mass NaCN :

\tt mass=mol\times MW\\\\mass=0.011\times 49.0072=0.539~g

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How many sub-energy levels does the 2nd MEL have?
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Answer:

n=  | Shell            |       Maximum Number of Electrons

1    | 1st Shell      |      2

2   | 2nd Shell      |      8

3   | 3rd Shell        |      18

4   |  4th Shell        |      32

Explanation: cause :)

5 0
3 years ago
A sample of bleach was analyzed as in this procedure. The only procedural difference is that the student weighed out the bleach
Bogdan [553]

Answer:

% = 5.69%

Explanation:

To do this, we need to write the equations taking place here. First, this is a REDOX reaction where the hypoclorite and thiosulfate solution reacts. The balanced equations are:

ClO⁻ + 2I⁻ + 2H⁺ -------> Cl⁻ +  I₂ + H₂O

I₂ + 2S₂O₃²⁻ -----------> 2I⁻ + S₄O₆²⁻

We already have the required volume and concentration of the thiosulfate solution, so we can calculate the moles of thiosulfate. With this moles, we can calculate the moles of hypochlorite, then the mass and finally the %.

The moles of thiosulfate would be:

moles S₂O₃²⁻ = V * M

moles S₂O₃²⁻ = 0.01324 * 0.0732 = 9.69x10⁻⁴ moles

Now according to the above reactions, we can see that

moles I₂ = moles ClO⁻

and

moles I₂ / moles S₂O₃²⁻ = 1/2

Therefore, let's calculate the moles of ClO⁻:

moles ClO⁻ = 9.69x10⁻⁴ / 2 = 4.845x10⁻⁴ moles

Now, we can calculate the mass of these moles, using the molar mass of sodium hypochlorite which is 74.44 g/mol:

m = 74.44 * 4.845x10⁻⁴

m = 0.036 g

Finally the % of this, in the bleach sample would be:

% = 0.036 / 0.634 * 100

<h2>% = 5.69%</h2>
6 0
3 years ago
Which of these is a chemical property?
ankoles [38]

Answer:

The second choice, or flammability.

Explanation:

The flammability of something is how easy it is for it to burn or ignite.

7 0
3 years ago
Read 2 more answers
Write the balanced standard combustion reaction for the c5h7on3s2
AlexFokin [52]

Answer:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Explanation:

When C5H7ON3S2 under go Combustion, the following are obtained as illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

Now, let us balance the equation. This is illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

There are 3 atoms of N on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of C5H7ON3S2 and 6 in front of N2 as shown below:

4C5H7ON3S2 + O2 —> CO2 + H2O + 6N2 + S2

There are 20 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 20 in front of CO2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + H2O + 6N2 + S2

There are 28 atoms on H on the left side and 2 atoms on the right side. It can be balance by putting 14 in front of H2O as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + S2

There are 8 atoms of S on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of S2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, there are a total of 54 atoms of O on the right side and 6 atoms on the left side

It can be balance by putting 25 in front of O2 as shown below:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, we can see that the equation is balanced.

4 0
3 years ago
What is an appropriate stepwise synthesis for the following synthesis that uses ethyl 3-methylbutanoate as the only source of ca
gayaneshka [121]

Explanation:

Using ethyl 3-methylbutanoate as your only source of carbon and using any other reagents necessary, propose a stepwise synthesis for the following conversion.

I need help drawing the product formed in step three.

Thanks!

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3 years ago
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