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belka [17]
3 years ago
15

Bread is made with which fungus? morel yeast mold chytrid

Chemistry
2 answers:
Troyanec [42]3 years ago
8 0
The answer is Yeast!!
marin [14]3 years ago
3 0

Answer:

yeast

Explanation:

sorry if incorrect

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Draw the products obtained from the reaction of 1 equivalent of HBr with 1 equivalent of 2,5-dimethyl-1,3,5-hexatriene. Draw the
fgiga [73]

Answer:

Here's what I get.

Explanation:

According to Markovnikov's rule, the H will add to a terminal carbon, generating three resonance stabilized carbocations.

The Br⁻ ion will add to any of the three carbocations.

There are three possible products:

  1. 5-bromo-2,5-dimethylhexa-1,3-triene (1)
  2. 3-bromo-2,5-dimethylhexa-1,4-triene (<em>2</em>)
  3. 1-bromo-2,5-dimethylhexa-2,4-triene (3)

6 0
3 years ago
Calculate the moles of 336 L of carbon dioxide gas (CO2) at STP.
fenix001 [56]

Answer:

14.99 mols of CO2

Explanation:

STP= 273 k , 1 atm , .0821 gas constant

N=PV/RT

N= 1x336/.0821x273

14.99 mols of CO2

659.6 grams of CO2

8 0
3 years ago
Please help I need help:(((​
Elza [17]

Answer:

I am not understand bro a question

7 0
3 years ago
A zinc rod is placed in 0.1 MZnSO4 solution at 298 K. Write the electrode reaction and calculate the potential of the electrode.
yanalaym [24]

<u>Answer:</u> The potential of electrode is -0.79 V

<u>Explanation:</u>

When zinc is dipped in zinc sulfate solution, the electrode formed is Zn^{2+}(aq.)/Zn(s)

Reduction reaction follows:  Zn^{2+}(0.1M)+2e^-\rightarrow Zn(s);(E^o_{Zn^{2+}/Zn}=-0.76V)

To calculate the potential of electrode, we use the equation given by Nernst equation:

E_{(Zn^{2+}/Zn)}=E^o_{(Zn^{2+}/Zn)}-\frac{0.059}{n}\log \frac{[Zn]}{[Zn^{2+}]}

where,

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

E^o_{cell} = standard electrode potential of the cell = -0.76 V

n = number of electrons exchanged = 2

[Zn]=1M    (concentration of pure solids are taken as 1)

[Zn^{2+}]=0.1M

Putting values in above equation, we get:

E_{cell}=-0.76-\frac{0.059}{2}\times \log(\frac{1}{0.1})\\\\E_{cell}=-0.79V

Hence, the potential of electrode is -0.79 V

4 0
3 years ago
A solution contains the ions Ag , Pb2 , and Ni2 . Dilute solutions of NaCl, Na2SO4, and Na2S are available to separate the posit
Delicious77 [7]

Answer:

The solutions should be added in this order NaCl > Na2SO4 > Na2S

Explanation:

Silver is insoluble as a chloride, so the silver ions get precipitated on addition of chloride ion as silver chloride.  This means Ag+ would be removed the first.

So we will add NaCl in the first step.

The following reaction will occur.

Ag+ + Cl- → AgCl(s)

Both, Pb2 and Ni are soluble as chlorides. (lead chloride is soluble as a hot solution but will ppt when colder).

When we add Na2SO4, Pb2+ will get precipitated (because it's insoluble) as PbSO4 and Ni will remain soluble as NiSO4 is soluble in water.

The reaction that will occur is:

Pb^2+ + SO4^2- → PbSO4(s)

Nickel is insoluble as a sulfide. So when we will add Na2S, nickel will be precipitated as sulfide and be able to separate and be collected.

The solutions should be added in this order NaCl > Na2SO4 > Na2S

6 0
4 years ago
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