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zloy xaker [14]
3 years ago
6

Which components can be found in both comparative and experimental investigations, but are never found in descriptive investigat

ions? Check all that apply.
- question
- hypothesis
- variables
- control group
- prediction
- conclusion
Chemistry
2 answers:
Reptile [31]3 years ago
6 0

Answer:

HYPOTHESIS, VARIABLES, AND CONTROL GROUP

Explanation:

YOUR WELCOME :)

alex41 [277]3 years ago
4 0

Answer:

Hypothesis and Variables

Explanation:

I just took the quiz

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You have a graduated cylinder with 10 mL of water in it.
Law Incorporation [45]

Answer:

<h3>The answer is 10 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 300 g

volume = final volume of water - initial volume of water

volume = 40 - 10 = 30 mL

We have

density =  \frac{300}{30}  =    \frac{30}{3}  \\

We have the final answer as

<h3>10 g/mL</h3>

Hope this helps you

5 0
3 years ago
James exclaims to his chemistry teacher that a new element has been discovered that fits between Nickel and Copper on the period
myrzilka [38]

Aye you have the same class as me bruh I need help on some chemistry qustions

8 0
3 years ago
Read 2 more answers
Write an overall equation for the acid-base reaction that would be required to produce the following salt. MgCl2
harina [27]
Mg+Cl2--> MgCl2

Magnesium plus chlorine equals magnesium chloride
7 0
3 years ago
Read 2 more answers
For the reaction 2kclo3(s)→2kcl(s)+3o2(g) calculate how many grams of oxygen form when each quantity of reactant completely reac
barxatty [35]
First, we need to get the molar mass of:

KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol

KCl =39.1 + 35.5 = 74.6 g/mol

O2 = 16*2 = 32 g/mol

From the given equation we can see that:

every 2 moles of KClO3 gives 3 moles of O2

when mass = moles * molar mass

∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g

and the mass of O2 then = 3 mol * 32g/mol = 96 g

so, 245.2 g of KClO3 gives 96 g of O2

A) 2.72 g of KClO3: 

when 245.2 KClO3 gives → 96 g  O2

   2.72 g KClO3 gives →  X

X = 2.72 g KClO3 * 96 g O2/245.2 KClO3

    = 1.06 g of O2

B) 0.361 g KClO3:

when 245.2 g KClO3 gives → 96 g O2

     0.361 g KClO3 gives → X

∴ X = 0.361g KClO3 * 96 g / 245.2 g

       = 0.141 g of O2

C) 83.6 Kg KClO3:

when 245.2 g KClO3 gives → 96 g O2

       83.6 Kg KClO3 gives  →  X

∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3

     = 32.7 Kg of O2

D) 22.4 mg of KClO3:

when 245.2 g KClO3 gives → 96 g O2

        22.4 mg KClO3 gives → X

∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3

      = 8.8 mg of O2

     

 


7 0
3 years ago
The solubility of oxygen gas in water at 40 ∘c is 1.0 mmol/l of solution. What is this concentration in units of mole fraction?
juin [17]

The formula for mole fraction is:

mole fraction of solute = \frac{number of moles of solute}{total number of moles of solution}    -(1)

The solubility of oxygen gas = 1.0 mmol/L  (given)

1.0 mmol/L means 1.0 mmol are present in 1 L.

Converting mmol to mol:

1.00 mmol\times \frac{1 mol}{1000 mmol} = 0.001 mol

So, moles of oxygen = 0.001 mol

For moles of water:

1 L of water = 1000 mL of water

Since, the density of water is 1.0 g/mL.

Density = \frac{mass}{volume}

Mass = 1.0 g/ml\times 1000 mL = 1000 g

So, the mass of water is 1000 g.

Molar mass of water = 18 g/mol.

Number of moles of water = \frac{1000 g}{18 g/mol} = 55.55 mol

Substituting the values in formula (1):

mole fraction = \frac{0.001}{55.55+0.001}

mole fraction = 1.8\times 10^{-5}

Hence, the mole fraction is 1.8\times 10^{-5}.

7 0
3 years ago
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