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kifflom [539]
2 years ago
14

Distinguish air mass and a front

Chemistry
1 answer:
Readme [11.4K]2 years ago
3 0
Air mass is how much matter is in air. Front is the side that something faces forward with.
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Which of the following leads to a higher rate of diffusion?
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A field worker is exposed to a xylene for a duration of 8 weeks at 40 hrs/wk. The concentration of xylene in the workplace is 40
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Answer:

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.

Explanation:

Number of hours worker exposed to xylene = 40 hr/week\times 8 week = 320 hours

The concentration of xylene in the workplace =40 \mu g/m^3

The worker is inhaling air at a rate of 0.9 m^3/hr.

Amount xylene inhaled by worker in an hour :

= 40\mu g/m^3\times 0.9 m^3/hr=36 \mu g/hr

Amount xylene inhaled by worker in 320 hours:

36 \mu g/hr\times 320 hr=11,520 \mu g=11,520\times 0.001 mg=11.520 mg

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Amount xylene inhaled by worker in 320 hours = 11.520 mg

1 day = 24 hours

Amount xylene inhaled by worker in 1 day:

\frac{24}{320}\times 11.520 mg=0.864 mg

Assuming 70 kg body mass, the chronic daily intake of xylene :

\frac{0.864 mg/day}{70 kg}=0.01234 mg/ kg day\approx 0.012 mg/ kg day

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.

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2 years ago
How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?
Kipish [7]

Answer:

V KOH = 41 mL

Explanation:

for neutralization:

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∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ <em>C</em> KOH = 0.0050 N = 0.0050  eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH

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