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VARVARA [1.3K]
3 years ago
9

Can anybody help me with the wet lab on edgenuity? T-T

Chemistry
2 answers:
Jet001 [13]3 years ago
7 0
Is it the dry lab/wet lab week 1 or ?
Arte-miy333 [17]3 years ago
6 0

Answer:

I dont know if this is the right subject but i got an A+ on it. It is Life science edgenuity     This is the process part of it

Explanation:

First, I observed all of the arthropods that I would be using in my dichotomous key, Fire Ant, Bee, Butterfly, Dragonfly, Ladybug, Scorpion, Spider, and Wasp. I recorded my observations on a characteristics table that included Legs, Wings, Stinger, Antennae, and Claws.  

Next, I used my observations to create my first question, “Does the arthropod have a stinger?).  

     Then, I Looked over my notes and found the arthropds that have stingers, Fire Ant, Bee, Scorpion, and        Wasp, and sorted them into a Yes group. Also I found the arthropods in my notes that didn’t have stingers, Butterfly, Dragonfly, Ladybug, and Spider, and put them into a No group.  

     After That, I Chose a new question coming off of the Yes group, “Does the arthropod have 6 legs?”. I sorted those the arthropods that did have 6 legs into a Yes group, Fire Ant, Bee, and Wasp.  

     Then, I did the same for The arthropods who didn’t have 6 legs and the only one was Scorpion, so I wrote scorpion in the No group and that Branch was done.  

     Next, I did another question for my very first No group, “Does the arthropod have antennae?” and I sorted the arthropods into the yes and no branches. No group included, Dragonfly and Spider. Yes group included, Butterfly and Ladybug.

     Later, I did a question for the yes group coming off of the 6 legs question, “Does the arthropod have wings?”. The yes group included, Bee and Wasp. The no group included, the Fire Ant only so I wrote down Fire Ant and that branch was concluded.  

     Then, I did a question for the no group coming off of the antennae question, “Does the arthropod hve 8 legs?” and my yes group included spider only so that branch was done. My no group included Dragonfly so that branch was concluded aswell.  

     After That, I wrote a question for the yes group coming off of the wings question, “Does the arthropod have 2 wings?”. My Yes group concluded wit only the Wasp left so I had finished that branch. I finished my no group with a Bee.

     Lastly, I did my last question of the key coming off of the Yes group of the antennae question, “Does the arthropod have 4 wings?”. My no group included a Ladybug and my yes group included a Butterfly.

     In conclusion, My dichotomous key was finished with 8 athropods in each branch alone and now you can use the dichotomous to observe a Fire Ant, Bee, Wasp, Butterfly, Dragonfly, Ladybug, Spider, and Scorpion.  

                                               Please mark me brainliest :) :) :) :) :) :)

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You need to prepare .200M solution of hydrochloric acid (HCl). If you took out .830ml of the 12.0M stock solution. How much wate
Alenkinab [10]

Answer:

You must add 48.97 mL of water to make the 0.200 M diluted solution.

Explanation:

In chemistry, dilution is the reduction in concentration of a chemical in a solution. In other words, it is the process of reducing the concentration of solute in solution, simply adding more solvent to the solution.

In a dilution, the quantity or mass of the solute is not changed but only that of the solvent. As only solvent is being added, by not increasing the amount of solute the concentration of the solute decreases.

The expression for the dilution calculations is:

Cinitial* Vinitial = Cfinal* Vfinal

In this case:

  • Cinitial= 12 M
  • Vinitial= 0.830 mL
  • Cfinal= 0.200 M
  • Vfinal= ?

Replacing:

12 M*0.830 mL= 0.200 M*Vfinal

Solving:

Vfinal=\frac{12 M*0.830 mL}{0.200 M}

Vfinal= 49.8 mL

Since 0.830 mL is the volume you initially have of HCl, the amount of water you must add is:

49.8 mL - 0.830 mL= 48.97 mL

<u><em>You must add 48.97 mL of water to make the 0.200 M diluted solution.</em></u>

7 0
3 years ago
Answer and work for this problem
MArishka [77]
We can write the balanced equation for the synthesis reaction as 
     H2(g) + Cl2(g) → 2HCl(g)

We use the molar masses of hydrogen chloride gas HCl and hydrogen gas H2 to calculate for the mass of hydrogen gas H2 needed:
     mass of H2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol H2 / 2 mol HCl) * 
                           (2.02 g H2 / 1 mol H2)                        
                        = 4.056 g H2

We also use the molar masses of hydrogen chloride gas HCl and chlorine gas CL2 to calculate for the mass of hydrogen gas H2:
     mass of CL2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol Cl2 / 2 mol HCl) *
                             (70.91 g Cl2 / 1 mol Cl2)
                          = 142.4 g Cl2 

Therefore, we need 4.056 grams of hydrogen gas and 142.4 grams of chlorine gas to produce 146.4 grams of hydrogen chloride gas.
6 0
3 years ago
Help me please omg I don’t know
Anna35 [415]

Answer:

5 1 2 4and 3 this is correct way

8 0
2 years ago
Given the following information: benzoic acid = C6H5COOH hydrocyanic acid = HCN C6H5COOH is a stronger acid than HCN (1) Write t
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Answer:

The net ionic equation is

C6H5COOH+ CN-= C6H5COO- + HCN

Explanation:

From the ionic equation

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Only sodium is the spectator ion, so it cancels out, since C6H5COOH and HCN do not ionize completely they are left undissociated

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