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Alekssandra [29.7K]
4 years ago
13

The silkworm feeds on what​

Chemistry
2 answers:
nydimaria [60]4 years ago
7 0

Answer:

The various species raised today are distinguished by the quality of the silk they produce. Silkworms feed on the leaves of the mulberries (genus Morus) and sometimes on the Osage orange (Maclura pomifera). Bombyx Mori will not bite, making it an ideal worm for feeding most reptiles, amphibians and other animals.

Explanation:

AfilCa [17]4 years ago
4 0
Silkworms only eat mulberry leaves and/or artificial silkworm diet (Silkworm Chow). If you are going to feed the newly hatched silkworms mulberry leaves instead of Silkworm Chow be sure and only feed them the new growth (tiny 1/2 to 1 inch leaves) or they will not be able to eat them because their jaws are too weak.
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The substances are placed in separate containers at room temperature, and each container is gradually cooled. Which of these sub
algol13

Answer:

Benzene

Explanation:

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3 0
3 years ago
The molar solubilities of the following compounds (in mol/L) are:
IceJOKER [234]

<u>Answer:</u> The decreasing order of K_{sp} is AgSCN>AgBr>AgCN

<u>Explanation:</u>

  • <u>For AgBr:</u>

The balanced equilibrium reaction for the ionization of silver bromide follows:

AgBr\rightleftharpoons Ag^{+}+Br^-

                 s      s

The expression for solubility constant for this reaction will be:

K_{sp}=[Ag^{+}][Br^-]

We are given:

s=7.3\times 10^{-7}mol/L

Putting values in above equation, we get:

K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.3\times 10{-7})^2=5.33\times 10^{-13}

Solubility product of AgBr = 5.33\times 10^{-13}

  • <u>For AgCN:</u>

The balanced equilibrium reaction for the ionization of silver cyanide follows:

AgCN\rightleftharpoons Ag^{+}+CN^-

                    s      s

The expression for solubility constant for this reaction will be:

K_{sp}=[Ag^{+}][CN^-]

We are given:

s=7.7\times 10^{-9}mol/L

Putting values in above equation, we get:

K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.7\times 10{-9})^2=5.93\times 10^{-17}

Solubility product of AgCN = 5.33\times 10^{-17}

  • <u>For AgSCN:</u>

The balanced equilibrium reaction for the ionization of silver thiocyanate follows:

AgSCN\rightleftharpoons Ag^{+}+SCN^-

                     s      s

The expression for solubility constant for this reaction will be:

K_{sp}=[Ag^{+}][SCN^-]

We are given:

s=1.0\times 10^{-6}mol/L

Putting values in above equation, we get:

K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(1.0\times 10{-6})^2=1.0\times 10^{-12}

Solubility product of AgSCN = 1.0\times 10^{-12}

The decreasing order of K_{sp} follows:

AgSCN>AgBr>AgCN

4 0
4 years ago
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Explanation:

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Let's consider a scenario in which the resting membrane potential changes from −70 mV to +70 mV, but the concentrations of all i
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