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Tanya [424]
3 years ago
12

3. Potassium chlorate, KC103, decomposes when heated to produce potassium

Chemistry
1 answer:
FromTheMoon [43]3 years ago
5 0

Answer: oxygen=2.547g

Explanation:

Based on the question,it was observed that the reaction is reversible

2 moles of KClO3 gives 2 moles of KCl and three moles of O2

Molar mass for KClO3 is 245 g/mol

Molar mass for O2 is 96 g/mol

We are to find the mass of O2 and we Are given the mass KCLO3 is 6.50g

245g of KClO3 gives 96g of O2

6.50g of KClO3 gives xg of O2

Cross multiply

245x=624

X=624/245

X=2.547g

Therefore the gram of oxygen is 2.547g

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Water supplies are often treated with chlorine as one of the processing steps in treating wastewater. Estimate the liquid diffus
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Answer:

⇒D_AB= 1.21×10^(-9)

Explanation:

Wike chang  equation is given as:

D_{AB}= \frac{117.3\times10^{-18}\times\(\phi\times M_B)^{0.5}\times T}{\mu\times\nu^{0.6}}

Where

D_AB= diffusivity of chlorine in water

Φ= 2.26 for water as solvent

ν= 0.0484 for chlorine as solute

M_B = Molecular weight of water

τ= temperature=289 K

μ= viscosity = 1.1×10^{-3}

Now putting these values in the above equation we get

D_{AB}= \frac{117.3\times10^{-18}\times\(\2.26\times18)^{0.5}\times289}{\1.1\times10^{-3}\times\0.0484^{0.6}}

⇒D_AB= 1.21×10^(-9)

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3 years ago
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Based on the table, which atom has a charge of –1?<br> 1<br> 2<br> 3<br> 4
Shkiper50 [21]

An electron

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3 years ago
When doing medical research with human subjects, which four limitations are unavoidable?
Fofino [41]

Answer:

Answer

The limitations are-

1. Privacy of the individuals involved in the research process.

2. Physical and psychological risks should be minimized

3. The subjects should be chosen equitably

4. Only reasonable exposure to risks is admissible.

Explanation

The privacy of the individuals involved in the research must be taken into consideration. This will ensure safety of patient data and information. The risk of physical and physiological well being of the person must be taken into consideration in such a research. In addition to that, the subject must be made to understand every procedure and the risks involved before testing. Moreover, only minimal exposure to risks is allowed and must have been previously tested in animals to avoid deaths.

8 0
2 years ago
A sample of o2 gas (2.0 mmol) effused through a pinhole in 5.0 s. it will take __________ s for the same amount of co2 to effuse
s344n2d4d5 [400]
Answer is: 5,9.
Effusion is leakage of gas through a small hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. R<span>elative rates of effusion is related to the molecular mass.
</span>M(O₂) = 32g/mol.
M(CO₂) = 44g/mol.
t₁ = 5s.
t₂ = ?
t₂/t₁ = √(M(CO₂)/M(O₂)).
t₂ = 1,17·5s = 5,9s.

8 0
3 years ago
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