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ella [17]
3 years ago
14

Which of the following questions about the flu is a scientific question?

Chemistry
2 answers:
qwelly [4]3 years ago
5 0
D. Is a fly shot effective in preventing the flu?

Answer A touches on the topics of economics and industrial feasibility of the flu shot.
B is more of an ethical / situational question.
C is a social question.
D is the only question which focuses on the science of microbiology and medicine.
OLEGan [10]3 years ago
5 0

Answer:

the answer d is correct!

Explanation:

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What is the total number of pairs of electrons shared between the carbon atom and the oxygen atom in a molecule of methanal?
valentinak56 [21]
Answer is: between the carbon atom and the oxygen atom in a molecule of methanal there are two (2) <span>pairs of electrons.

Methanal or formaldehyde (H</span>₂C=O)<span> </span><span>is the simplest of the </span><span>aldehydes.
In methanal, bond between carbon and oxygen is double covalent bond (one sigma and one pi bond), which means there are four shared electrons or two pair of electrons.</span>
8 0
3 years ago
Both pyrrole and pyridine are aromatic compounds, and undergo electrophilic aromatic substitution (EAS). Using a resonance argum
soldi70 [24.7K]

Answer:

See attached picture for both electrophillic substitution in pyrole and in pyridine.

Explanation:

5 0
3 years ago
A mixture of methanol and methyl acetate contains 15.0 wt% methanol.
Usimov [2.4K]

The mixture flow rate in lbm/h = 117.65 lbm/h

<h3>Further explanation</h3>

Given

15.0 wt% methanol

The flow rate of the methyl acetate :100 lbm/h

Required

the mixture flow rate in lbm/h

Solution

mass of methanol(CH₃OH, Mw= 32 kg/kmol) in mixture :

\tt 15\%\times 200~kg=30~kg\\\\mol=\dfrac{mass}{MW}=\dfrac{30~kg}{32~kg/kmol}=0.9375~kmol

mass of the methyl acetate(C₃H₆O₂,MW=74 kg/kmol,85% wt) in 200 kg :

\tt 85\%\times 200=170~kg\\\\mol=\dfrac{170}{74}=2.297~kmol

Flow rate of the methyl acetate in the mixture is to be 100 lbm/h.

1 kg mixture = 0.85 .methyl acetate

So flow rate for mixture :

\tt \dfrac{1~kg~mixture}{0.85~methyl~acetat}\times 100~lbm/h=117.65~lbm/h

5 0
3 years ago
The solid compound, K2SO4, contains?
givi [52]
The solid compound, K2SO4 contains a cation called K+ and an anion called SO42-. In this case, there are 2 atoms of potassium, 1 atom of sulfur and 4 moles of oxygen. The compound also contains ionic bonds because of the composing non-metals and metal. 
3 0
3 years ago
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e
notsponge [240]

<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-   E^o_{Zn^{2+}/Zn}=-0.76V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)   E^o_{Sn^{2+}/Sn}=-0.136V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.136-(-0.76)=0.624V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}

where,

E_{cell} = electrode potential of the cell = 0.660 V

E^o_{cell} = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

[Zn^{2+}]=2.5\times 10^{-3}M

[Sn^{2+}] = ?

Putting values in above equation, we get:

0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

[Sn^{2+}]=9.0\times 10^{-3}M

Hence, the concentration of Sn^{2+} ions is 9.0\times 10^{-3}M

3 0
3 years ago
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