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Nostrana [21]
3 years ago
11

Which of the following is an example of maintaining homeostasis?

Chemistry
2 answers:
rosijanka [135]3 years ago
4 0
I think that it is drinking water, it maintains a balance in hydration in the body.
stich3 [128]3 years ago
3 0
The answer to this is Drinking Waer
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How will the rate at which a solid solute dissolves change if the solution is stirred?
MakcuM [25]

Answer:

The rate at which the solute dissolves will increase.

Explanation:

If a solution is stirred, the rate at which a solute dissolves would increase substantially provided the solution is not yet saturated.

Stiring would cause more of the solution to come in contact with every part of the solute. It will increase the surface area of contact for the solution to act which will shoot up the rate of reaction. Stiring helps to bring solutes in solutions into a more close contact with the molecules or compounds of the medium.

6 0
3 years ago
Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if
mrs_skeptik [129]

The final volume of the methane gas in the container is 6.67 L.

The given parameters;

  • <em>initial volume of gas in the container, V₁ = 2.65 L</em>
  • <em>initial number of moles of gas, n₁ = 0.12 mol</em>
  • <em>additional concentration, n = 0.182 mol</em>

The total number of moles of gas in the container is calculated as follows;

n_t = 0.12 + 0.182 = 0.302 \ mol

The final volume of gas in the container is calculated as follows;

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L

Thus, the final volume of the methane gas in the container is 6.67 L.

Learn more here:brainly.com/question/21912477

5 0
3 years ago
Flourine is found to undergo 10% radioactivity decay in 366 minutes determine its halflife​
yuradex [85]

Answer:

\boxed{\text{2408 min}}

Explanation:

The integrated rate law for radioactive decay is

\ln\dfrac{N_{0}}{N_{t}} = kt

1. Calculate the decay constant

\begin{array}{rcl}\ln \dfrac{100}{90} & = & k \times 366\\\\1.054 & = & 366k\\\\k & = & \dfrac{1.054 }{366}\\\\k & = & 2.879 \times 10^{-4} \text{ min}^{-1}\\\end{array}\\\\

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k}\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{2.879 \times 10^{-4} \text{ min}^{-1}} = \text{2408 min}\\\\\text{The half-life for decay is } \boxed{\textbf{2408 min}}

8 0
3 years ago
When the paramagnetic [co(cn)6] 4– ion is oxidized to [co(cn)6] 3– , the ion becomes diamagnetic. however, when the paramagnetic
S_A_V [24]
Answer:  
Basically, paramagnetic and diamagnetic refer to the way a chemical species interacts with a magnetic field. More specifically, it refers to whether or not a chemical species has any unpaired electrons or not. 
 A diamagnetic species has no unpaired electrons, while a paramagnetic species has one or more unpaired electrons. 
 Now, I won't go into too much detail about crystal field theory in general, since I assume that you're familiar with it. 
 So, you're dealing with the hexafluorocobaltate(III) ion, [CoF6]3â’, and the hexacyanocobaltate(III) ion, [Co(CN)6]3â’. 
 You know that [CoF6]3â’ is paramagnetic and that [Co(CN)6]3â’ is diamagnetic, which means that you're going to have to determine why the former ion has unpaired electrons and the latter does not. 
 Both complex ions contain the cobalt(III) cation, Co3+, which has the following electron configuration 
  Co3+:1s22s22p63s23p63d6 
 For an isolated cobalt(III) cation, all these five 3d-orbitals are degenerate. The thing to remember now is that the position of the ligand on the spectrochemical series will determine how these d-orbtals will split. 
 More specifically, you can say that 
  a strong field ligand will produce a more significant splitting energy, Δ  a weak field ligand will produce a less significant splitting energy, Δ 
 Now, the spectrochemical series looks like this 
 http://chemedu.pu.edu.tw/genchem/delement/9.htmhttp://chemedu.pu.edu.tw/genchem/delement/9.htm 
 Notice that the cyanide ion, CNâ’, is higher on the spectrochemical series than the fluoride ion, Fâ’. This means that the cyanide ion ligands will cause a more significant energy gap between the eg and t2g orbitals when compared with the fluoride ion ligands. 
 http://wps.prenhall.com/wps/media/objects/3313/3393071/blb2405.htmlhttp://wps.prenhall.com/wps/media... 
 In the case of the hexafluorocobaltate(III) ion, the splitting energy is smaller than the electron pairing energy, and so it is energetically favorable to promote two electrons from the t2g orbitals to the eg orbitals → a high spin complex will be formed. 
 This will ensure that the hexafluorocobaltate(III) ion will have unpaired electrons, and thus be paramagnetic. 
 On the other hand, in the case of the hexacyanocobaltate(III) ion, the splitting energy is higher than the electron pairing energy, and so it is energetically favorable to pair up those four electrons in the t2g orbitals → a low spin complex is formed. 
 Since it has no unpaired electrons, the hexacyanocobaltate(III) ion will be diamagnetic.
6 0
3 years ago
Which element has a grater atomic radius ?phosphorus or bismuth?
LuckyWell [14K]

According to periodic trends, bismuth has the greatest atomic radius.

6 0
3 years ago
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