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3 years ago

The amount of dissolved oxygen in water may decrease because of the

2 answers:

8 0

<span>The amount of dissolved oxygen in water may decrease because of the increase in organic matter in the water. <span>Aquatic organisms breathe and use oxygen. Large amounts of oxygen are consumed by the decomposition of bacteria (when there are large amounts of dead matter to decompose, there will be a significant number of bacteria). Examples: dead organic matter (algae), wastewater, garden waste, oils and fats, all this results in a decrease in dissolved oxygen in the water.</span></span>

Crazy boy [7]3 years ago

4 0

Answer:

Answer would be d

Explanation:

Increase in organic matter in the water

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T/F: Stars die.<br> True<br> False

Yuki888 [10]

Answer:

The answer is true

Explanation:

could you brainliest again they said I plagarized when I didn't

8 0

2 years ago

Given the isotope 2Fes, which has an actual mass of 55.934939 u: a) b) Determine the mass defect of the nucleus in atomic mass u

SSSSS [86.1K]

Answer:

Mass defect of each iron-56 nuclei:

The binding energy per nucleon of Iron-56 is approximately 8.6 MeV.

Explanation:

According to the physics constants table on Chemistry Libretexts:

Proton rest mass: $\rm 1.0072765\;amu$ ;
Neutron rest mass: $\rm 1.0086649\; amu$ .
Speed of light in vacuum: $\rm 2.99792458\times 10^{8}\;m\cdot s^{-1}$ .
Charge on an electron: $\rm 1.6021765\times 10^{-19}\;C$ .

The mass defect of a nucleus is equal to the sum of the mass of its parts (protons and, in most cases, neutrons) minus the mass of the nucleus.

The atomic number of iron is 26. There are 26 protons in each iron-56 nucleus. The mass number 56 indicates that there are 56 nucleons (neutrons and protons) in each iron-56 nucleus. The other $56 - 26 = 30$ particles are neutrons.

The mass of protons and neutrons in each iron-56 nucleus will be:

$\rm 26 \times 1.0072765 + 30 \times 1.0086649 = 56.464736\;amu$ .

According to this question, the mass of an iron-56 nucleus is equal to 55.934939 amu. The mass defect will be

$\rm 56.464736 - 55.934939 = 0.514197\;amu$ .

By the mass-energy equivalence,

$E = m\cdot c^{2}$ .

Refer to this equation, the speed of light in vacuum $c^{2}$ is the conversion factor between mass $m$ and energy $E$ . The value of $c$ is usually given only in SI units $\rm m\cdot s^{-1}$ . Accordingly, the value of $c^{2}$ will be in the SI unit $\rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}$ .

Convert million electron-volts to joules.

One electron-volt is equal to the electrical work done moving an electron across a potential difference of one volt.

$\begin{aligned}\rm 1 MeV&= \rm 10^{6}\; eV\\ &= \rm (10^{6}\times 1.6021765\times 10^{-19}\;C)\times 1\; V\\&=\rm 1.6021765\times 10^{-19}\;J\end{aligned}$ .

Convert the unit of $c^{2}$ from $\rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}$ to the desired $\rm MeV \cdot amu^{-1}$ :

$\begin{aligned}c^{2} &= \rm {\left(2.99792458\times 10^{8}\;m\cdot s^{-1}\right)}^{2}\\&=\rm 8.987551787\times 10^{16}\; m^{2}\cdot s^{-2}\\ &= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\\&= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\times \frac{1\;MeV}{1.6021765\times 10^{-13}\;J}\times \frac{1\times 10^{-3}\;kg}{6.022142\times 10^{23}\;amu}\\&\approx \rm 931.602164\;MeV\cdot amu^{-1}\end{aligned}$ .

Total binding energy in each iron-56 nucleus:

$\begin{aligned}E &= m\cdot c^{2}\\&= \rm 0.514197\;amu \times 9.31602164\;MeV\cdot amu^{-1} \\&=\rm 479.027038\; MeV \end{aligned}$ .

Again, the mass number 56 indicates that there are 56 nucleons in each iron-56 nucleus. The binding energy per nucleon of iron-56 $\mathrm{^{56}Fe}$ will be:

$\displaystyle \rm \frac{479.027038\; MeV}{56} \approx 8.6\; MeV$ .

6 0

3 years ago

A car traveled 112.0m in 24.500s, what was a car’s average speed?

kodGreya [7K]

Average speed of the car is 4.57 m/s

Explanation:

Speed is calculated by the rate of change of displacement.

It is given by the formula, Speed = Distance/Time

Here, distance = 112 m and time = 24.5 s

Speed of the car = 112/24.5 = 4.57 m/s

3 0

3 years ago

The mass of a string is 7.7 × 10-3 kg, and it is stretched so that the tension in it is 190 N. A transverse wave traveling on th

Scilla [17]

Length of the strings = 2.33 m

3 0

2 years ago

How are size and mass of the moon different from that of the earth?

Elodia [21]

<span>The diameter of the Moon is 3,474 km. Now, let's compare this to the Earth. The diameter of the Earth is 12,742 km. This means that the Moon is approximately 27% the size of the Earth.</span>

3 0

2 years ago