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timurjin [86]
3 years ago
14

Some organisms are not able to live in an environment where there is oxygen; these types of organisms are called obligate anaero

bes. Which explanation is most plausible for how they survive without oxygen?
Chemistry
1 answer:
BabaBlast [244]3 years ago
7 0

Answer:

Microorganisms which are killed in the presence of normal concentrations of atmospheric oxygen are called obligate anaerobes.Some species are capable of surviving in upto 8% while other losing viability in 0.5%.They survive without oxygen as they use other molecules as electron acceptor instead of oxygen.

Explanation:In aerobic organisms oxygen is used in cellular respiration as an electron acceptor.oxygen accepts electrons and turns into water H2O or CO2.While obligate anaerobes other molecules as their primary electron acceptors.common electron acceptors used by them are Nitrate NO3-,  Sulphate SO42- or sulphur S,Iron,mercury and manganese.So in this way they undergo respiration and survive without oxygen.

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The rock was white and black speckled with a density of 2.6 g/mL. Physical or chemical property?
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Chemical

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Astatine, At, is below iodine in Group 17 of the Periodic Table. Which statement is most likely to be correct?
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On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?
enot [183]

Answer:

The answer is "2%"

Explanation:

Equation:

HNO_2\ (aq) \leftrightharpoons  H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\  K_a = 4.0\times \ 10^{-4}

H^{+}=?

Formula:

Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}

Let

[H^{+}] = [NO_2^{-}] = x at equilibrium

x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \  M\\\\

therefore,

[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M

Calculating the % ionization:

= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\

6 0
3 years ago
During an experiment, the percent yield of calcium chloride from a reaction was 82.38%. Theoretically, the expected amount shoul
gregori [183]

Answer:

Actual yield = 86.5g

Explanation:

Percent yield = 82.38%

Theoretical yield = 105g

Actual yield = x

Equation of reaction,

CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

Percentage yield = (actual yield / theoretical yield) * 100

82.38% = actual yield / theoretical yield

82.38 / 100 = x / 105

Cross multiply and make x the subject of formula

X = (105 * 82.38) / 100

X = 86.499g

X = 86.5g

Actual yield of CaCl₂ is 86.5g

7 0
3 years ago
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