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Brrunno [24]
2 years ago
14

Find the % composition for each element in Zn(CIO3)2

Chemistry
1 answer:
bezimeni [28]2 years ago
7 0

Zn = 28.15%

Cl = 30.53%

O = 41.32%

<h3>Further explanation</h3>

Given

Zn(CIO3)2 compound

Required

The % composition

Solution

Ar Zn = 65.38

Ar Cl = 35,453

Ar O = 15,999

MW Zn(CIO3)2 = 232.3

Zn = 65,38/232.3 x 100% = 28.15%

Cl = (2 x 35.453) / 232.3 x 100% = 30.53%

O = (6 x 15.999) / 232.3 x 100% = 41.32%

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lisabon 2012 [21]

Answer: c. Some health risks are increased by heredity, which manifest under certain environmental conditions.

Explanation:

Many abiotic factors (non-living factors) such as radiations, hazardous substances present in the environment such as air, water and soil may originate from the industries, mining practices, fossil fuels and landfills. Some of the substances are carcinogenic and mutagenic in nature. These are capable of affecting the genetic make up of the organism. The genetic variations or mutations occurs may transmit from the parent to the offsprings.

Therefore, on the basis of the above information, c. Some health risks are increased by heredity, which manifest under certain environmental conditions. is the correct option.

3 0
3 years ago
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If the starting volume of a hot air balloon is 55,500 m3and the initial temperature is 21 °C, what is the temperature inside the
Dmitry_Shevchenko [17]

Answer:

T₂ = 392 K

Explanation:

Given that,

Initial volume of the hot air balloon, V₁ = 55500 m³

Initial temperature, T₁ = 21°C = 294 K

Final volume, V₂ = 74000 m³

We need to find the final temperature inside the balloon. The relation between the temperature and volume is given by charles law i.e.

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

Where

T₂ is the final temperature

So,

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}\\\\T_2=\dfrac{294\times 74000 }{55500 }\\\\T_2=392\ K

So, the new temperature is 392 K.

8 0
2 years ago
When lithium reacts with bromine to form the compound LiBr, each lithium atom:
yKpoI14uk [10]
(3) loses one electron and becomes positively charged 
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A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
Georgia [21]

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NaOH + CH3COOH → CH3COONa + H2O 

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH 

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH 

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L 

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pH using Henderson - Hasselbalch equation: 

pH = pKa + log ([salt]/[acid]) 

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7 0
3 years ago
What is the purpose of finding oxidation states in the half-reaction method for balancing equations?
vazorg [7]

Answer:

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