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Tanzania [10]
3 years ago
5

What class of organism is vital in the recycling of nutrients in the ecosystem?

Chemistry
1 answer:
asambeis [7]3 years ago
6 0
I believe that would be a decomposer
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Your sick puppy is diagnosed with roundworms. They grow inside of the puppy. What type of symbiotic relationship do the puppy an
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D. Parasitism , the roundworm benefits from the relationship, but the puppy is harmed
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In a chemical equation, (aq) after one of the substances means that it is which of these? a solid formed from two ionic substanc
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The answer is dissolved in water, because in chemistry, (aq) is shorthand for aqueous solution.
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What can create sediment
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Read 2 more answers
Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%, respectively)
Reika [66]

Answer:

Average atomic mass = 79.9034 amu

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

<u>For first isotope: </u>

% = 50.69 %

Mass = 78.9183 amu

<u>For second isotope: </u>

% = 49.31 %

Mass = 80.9163 amu

Thus,  

Average\ atomic\ mass=\frac{50.69}{100}\times {78.9183}+\frac{49.31}{100}\times {80.9163}\ amu

Average\ atomic\ mass=40.0036+39.8998\ amu

<u>Average atomic mass = 79.9034 amu</u>

4 0
3 years ago
In going from room temperature (25 C) to 10 C above room temperature, the rate of reaction doubles. Calculate the activation ene
Sauron [17]

Answer:

Ea=5.29 × 10⁴ J/mol

Explanation:

In going from 25 °C (298 K) to 35 °C (308 K), the rate of the reaction doubles. Since the rate of the reaction depends on the rate constant (k), this implies that the rate constant doubles. We can find the activation energy (Ea) using the two-point form of the Arrhenius equation.

ln\frac{k_{2}}{k_{1}} =\frac{-Ea}{R} .(\frac{1}{T_{2}}-\frac{1}{T_{1}})\\ln\frac{2k_{1}}{k_{1}}=\frac{-Ea}{8.314J/K.mol}.(\frac{1}{308K}-\frac{1}{298K} )\\Ea=5.29 \times 10^{4} J/mol

8 0
3 years ago
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