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Zanzabum
3 years ago
6

Why are group 17 elements the most reactive nonmetals?

Chemistry
1 answer:
jonny [76]3 years ago
5 0
Because as u go across groups the more reactive it is due to the ability of electrons to be given or taken away
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NiS2(s) + O2(g) --> NiO(s) + SO2(g) When 11.2 g of NiS2 react with 5.43 g of O2, 4.86 g of NiO are obtained. The theoretical
makkiz [27]

Answer:

1. The theoretical yield of NiO is 5.09g.

2. O2 is the limiting reactant.

3. The percentage yield of NiO is 95.5%

Explanation:

Step 1:

The balanced equation for the reaction is given below:

2NiS2(s) + 5O2(g) —> 2NiO(s) + 4SO2(g)

Step 2:

Determination of the masses of NiS2 and O2 that reacted and the mass of NiO produced from the balanced equation. This is illustrated below below:

Molar mass of NiS2 = 59 + (32x2) = 123g/mol

Mass of NiS2 from the balanced equation = 2 x 123 = 246g

Molar mass of o3= 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

Molar mass of NiO = 59 + 16 = 75g/mol

Mass of NiO from the balanced equation = 2 x 75 = 150g

Summary:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2 to produce 150g of NiO

Step 3:

Determination of the limiting reactant. This can be obtain as follow:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2.

Therefore, 11.2g of NiS2 will react with = (11.2 x 160)/246 = 7.28g of O2.

From the above calculation, we can see that it will take a higher mass of O2 i.e 7.28g than what was given i.e 5.43g to react completely with 11.2g of NiS2.

Therefore, O2 is the limiting reactant and NiS2 is the excess reactant.

1. Determination of the theoretical yield of NiO.

In this case, the limiting reactant will be used as all of it is consumed in the reaction. The limiting reactant is O2.

From the balanced equation above, 160g of O2 reacted to produce 150g of NiO.

Therefore, 5.43g of O2 will react to produce = (5.43 x 150)/160 = 5.09g of NiO.

Therefore, the theoretical yield of NiO is 5.09g.

2. The limiting reactant is O2. Please review step 3 above for explanation.

3. Determination of the percentage yield of NiO. This is illustrated below:

Actual yield of NiO = 4.86g

Theoretical yield of NiO = 5.09g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 4.86/5.09 x 100

Percentage yield of NiO = 95.5%

3 0
3 years ago
1) 2C2H6 + 7O2 --> 4CO2 + 6H2O
Lena [83]

Answer:

a).

2 \: moles \: of \: ethane \: react \: with \: 7 \: moles \: of \: oxygen \\ 24 \: moles \: react \: with \: ( \frac{24 \times 7}{2} ) \: moles \\  = 84 \: moles

b).

2 \: moles \: react \: with \: 7 \: moles \\ ( \frac{12 \times 2}{7} ) \: moles \: react \: with \: 12 \: moles \\  = 3 .43 \: moles \\ 1 \: mole \: weighs \: 30 \: g \\ 3.43 \: moles \: weigh \: (3.43 \times 30) \\  = 102.9 \: g

c).

746.7 g

7 0
3 years ago
An atom of ordinary hydrogen consists of an electron and a
Bess [88]

Answer: hydrogen atom is an atom of the chemical element hydrogen. The electrically neutral atom contains a single positively charged proton and a single negatively charged electron bound to the nucleus by the Coulomb force.

Explanation:

3 0
3 years ago
A sealed piston contains 400 mL of air at standard ambient pressure. The piston is
nydimaria [60]

Answer:

324.24 kPa

Explanation:

Given that;

Initial pressure P1 = 101325 Pa

V1= 400 ml

P2 = ?

V2= 125mL

From Boyle's law;

P1V1 = P2V2

P2 = P1V1/V2

P2= 101325 × 400/125

P2= 324.24 kPa

5 0
3 years ago
When you standardized the Na2S2O3, what molarity of Na2S2O3 did you obtain?
nikdorinn [45]

Answer:

0.46M NaS₂O₃ (Assuming KIO₃ solution with a concentration of 1.0M)

Explanation:

Based on the reaction:

6 Na₂S₂O₃ + KIO₃ + 5 KI + 3 H₂SO₄ → 3 Na₂S₄O₆ + 3 H₂O + 3 K₂SO₄ + 6 NaI

<em>6 moles of  Na₂S₂O₃ react per mole of KIO₃</em>

Assuming the molarity of the KIO₃ solution is 0,1M:

Moles of KIO₃: = 5.0x10⁻³L ₓ (0.1 mol / L) = <em>5.0x10⁻⁴ moles</em>

As 6 moles of thiosulfate reacted per mole of iodate:

5.0x10⁻⁴ moles KIO₃ ₓ (6 moles Na₂S₂O₃ /  1 mole KIO₃) =

<em>3.0x10⁻³ moles of Na₂S₂O₃. </em>In 6.5mL (6.5x10⁻³L):

3.0x10⁻³moles Na₂S₂O₃ / 6.5x10⁻³ L = 0.46M NaS₂O₃

6 0
3 years ago
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