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kumpel [21]
3 years ago
7

Why are chemical sanitizers ineffective in killing microbes present in cracks and crevices?

Chemistry
1 answer:
Strike441 [17]3 years ago
6 0
Microbes are so small that they are able to "hide" in the cracks and crevices in your hand even with the help of sanitizer it only kills 99% not 100% so font dump that soap yet because you'll be wanting to kill that 1% that is causing you disease and fermintation
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6) Determine la concentración de una solución que ha sido preparada añadiendo 80cc de agua a 320cc de solución alcohólica al 20%
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Answer:

16% v/v es la nueva concentración de alcohol en la solución

Explanation:

El porcentaje volumen/volumen (% v/v) es definido como 100 veces la relación entre el volumen de soluto (Alcohol en este caso) y el volumen total de la solución (Agua + Alcohol). Para resolver esta pregunta necesitamos hallar el volumen de alcohol y el de agua:

<em>Volumen alcohol:</em>

320cc * (20cc etanol / 100cc) = 64cc etanol

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80cc + (320cc-64cc) = 336cc agua

<em>% v/v:</em>

64cc / (336cc + 64cc) * 100

= 16% v/v es la nueva concentración de alcohol en la solución

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3 years ago
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Why is plastic not classified as a composite?
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During a synthesis reaction, 1.8 grams of magnesium reacted with 6.0 grams of oxygen. What is the maximum amount of magnesium ox
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Answer:

2.9 grams.

Explanation:

  • From the balanced reaction:

<em>Mg + 1/2O₂ → MgO,</em>

1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.

  • We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:

no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.

no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.

<em>So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1.0 mole of Mg produce → 1.0 mol of MgO.

∴ 0.074 mol of Mg produce → 0.074 mol of MgO.

<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.074 mol)(40.3 g/mol) = <em>2.98 g.</em>

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