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xxMikexx [17]
3 years ago
10

Why is it important to have a control?

Chemistry
2 answers:
Alborosie3 years ago
7 0
Its important to have control so you can predict or cause what happens to what you’re doing
cricket20 [7]3 years ago
3 0

Answer:

A control is important for an experiment because it allows the experiment to minimize the changes in all other variables except the one being tested.

Explanation:

may i please have brainliest

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Describe a group of family on the periodic table
lana66690 [7]
Metals are all in the center of the periodic table and are all made up of metals
6 0
3 years ago
Read 2 more answers
An unknown triprotic acid (H3A) is titrated with NaOH. After the titration Ka1 is determined to be 0.0013 and Ka2 is determined
e-lub [12.9K]

Answer:

6.68 X 10^-11

Explanation:

From the second Ka, you can calculate pKa = -log (Ka2) = 6.187

The pH at the second equivalence point (8.181) will be the average of pKa2  and pKa3. So,

8.181 = (6.187 + pKa3) / 2

Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11

7 0
3 years ago
Which statement correctly describes extreme weather?
sveta [45]

Answer: A

Explanation:

Extreme weather events follow normal climate patterns.

5 0
3 years ago
How many moles of propane<br> react when 294 g of CO2 form?<br><br> C3H8 +502 → 3CO₂ + 4H₂O
tatiyna

2.23 moles of propane react when 294 g of CO₂ is formed .

<h3>What is moles ?</h3>

Moles is a unit which is equal to the molar mass of an element.

A reaction is given

C₃H₈ +50₂ → 3CO₂ + 4H₂O

Grams of CO₂ formed = 294 gm

In moles = 294 /44 = 6.68 moles.

Let x be the moles of C₃H₈ is x

Mole ratio of CO₂ to C₃H₈ = 3 : 1

so

6.68 /x = 3/1

x = 6.68 /3 = 2.23 moles

Therefore 2.23 moles of propane react when 294 g of CO₂ is formed .

To know more about Moles

brainly.com/question/26416088

#SPJ1

7 0
2 years ago
Determine how many millilitres of a 4.25 M HCl solution are needed to react completely with 8.75 g CaCO3?
Helen [10]

Answer:

41 mL

Explanation:

Given data:

Milliliter of HCl required = ?

Molarity of HCl solution = 4.25 M

Mass of CaCO₃ = 8.75 g

Solution:

Chemical equation:

2HCl + CaCO₃      →    CaCl₂ + CO₂ + H₂O

Number of moles of CaCO₃:

Number of moles = mass/molar mass

Number of moles = 8.75 g / 100.1 g/mol

Number of moles = 0.087 g /mol

Now we will compare the moles of  CaCO₃ with HCl.

                      CaCO₃         :          HCl

                          1               :            2

                      0.087           :         2/1×0.087 = 0.174 mol

Volume of HCl:

Molarity = number of moles / volume in L

4.25 M = 0.174 mol / volume in L

Volume in L = 0.174 mol /4.25 M

Volume in L = 0.041 L

Volume in mL:

0.041 L×1000 mL/ 1L

41 mL

8 0
3 years ago
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