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PilotLPTM [1.2K]
3 years ago
6

You are given an unknown type of clothing dye. how could you use the procedures in this lab to see if this dye is a mixture?

Chemistry
2 answers:
Arte-miy333 [17]3 years ago
8 0
One separation technique to be used is the paper chromatography. This works by separating the components of the mixture through the difference of their concentrations. There is a stationary phase and the mobile phase, which flows through the stationary phase. The components travel at different rates and is usually signified by the colors. If more than one color would appear, that means that the dye is a mixture.
Ket [755]3 years ago
8 0

Use the paper chromatography method and see how each color ink separates in different solvents.

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Which of the following contains a nonpolar covalent bond? O2 CO NaCI HF
jasenka [17]

Answer: The nonpolar covalent bond happens in an O₂ molecule.

Explanation: Nonpolar bonds happen where there's no difference between the polarity of the atoms involved. This usually happens when the same atoms bond to each other, which explains why the O₂ molecule is the one who has this bond.

5 0
3 years ago
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andreyandreev [35.5K]
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8 0
3 years ago
The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO
Irina-Kira [14]

Answer:

Amount of excess Carbon (ii) oxide left over = 23.75 g

Explanation:

Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂

Molar mass of Fe₂O₃ = 160 g/mol;

Molar mass of Carbon (ii) oxide = 28 g/mol

From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g)  of carbon (ii) oxide

450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide

Therefore the excess reactant is carbon (ii) oxide.

Amount of excess Carbon (ii) oxide left over = 260 - 236.25

Amount of excess Carbon (ii) oxide left over = 23.75 g

5 0
3 years ago
Ozone has a molecular formula of O3. If 7.92*1024 atoms of oxygen react to form ozone, how many
prisoha [69]

337.92 moles of Ozone will be produced

1 Oxygen atom is 8 g

1 mole of ozone, O3 = 8 * 3 = 24 g

7.92 * 1024 = 8110.08 g

1 mole = 24 g

? moles = 8110.08 g

? = 337.92 moles


Read more on moles here:

brainly.com/question/15356425

Hope it helps

7 0
2 years ago
A sample of a compound contains 32.0 g C and 8.0 g H. Its molar mass is 30.0 g/mol. What is the compound’s molecular formula
Grace [21]
The compound's molecular formula is C2H6. This is obtained by:
  
            mass             moles               divided by smallest moles
C          32g         32/12 = 2.67                                 1
H           8g           8/1.01 = 7.92                        approx. 3

Next, divide both terms by the smallest number of moles, 2.67. This gives 1 and 3. So the empirical formula is CH3 which has a molar mass of 15g/mol. Given the molar mass of the molecular formula as 30g/mole, we can calculate the factor by which to multiply the subscripts of CH3.

X = molar mass of molecular formula / molar mass of empirical formula = 30/15
X=2

So (CH3)2 is C2H6.
3 0
3 years ago
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