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3 years ago

I need help with both parts, ill give brainliest and 30 points

Chemistry

2 answers:

Paul [167]3 years ago

7 0

Answer:

part a=A. A physical change, then a chemical change

part b:evidence

without physical change any sorts of chemical change do not take place.

for example: when sugar is heated on spoon the physical change takes place in sugar .sugar starts to form liquid from solid.

sweet-ann [11.9K]3 years ago

6 0

Answer:

Part A. C would be your answer I think

Sugar is made of carbon, hydrogen and oxygen atoms. When heated over a candle, these elements react with the fire to turn into a liquid. The heat causes the sugar's atoms to combine with the oxygen in the air, forming new groups of atoms. Energy is released in this chemical reaction in the form of smoke and black soot.

Part B. I don't know I just think of the answer... Meaning I'm good and science, Chemistry and all that good stuff ;-;

Explanation:

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Im stuck on these two problems

Vikki [24]

1st figure

answer

c)12

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c)28

#brianliest

3 0

4 years ago

Compare the amount of copper predicted in part A with the amount measured in part B. What percentage of the predicted amount was

timofeeve [1]

Answer:

Explanation:

Because the experiment has an different independent and dependent variable and each part has a different result of what type of copper or what type of control you are using for you experiment.

PLS MARK BRAINLIEST IF CORRECT I WORKED HARD ON THIS ANSWER,HAVE A NICE DAY/NIGHT <3

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3 years ago

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The octet rule pertains to:

Vinil7 [7]

I think the answer to c

4 0

3 years ago

If 11.7 g of aluminum reacts with 37.2 g of copper (II) sulfate according to the following reaction, how many grams of aluminum

zhannawk [14.2K]

Answer:

There is 26.59 grams of aluminium sulfate produced

Explanation:

Step 1: Data given

Mass of aluminium = 11.7 grams

Mass of copper (II) sulfate = 37.2 grams

Molar mass of Aluminium = 26.98 g/mol

Molar mass of CuSO4 = 159.61 g/mol

Molar mass of Al2(SO4)3 = 342.15 g/mol

Step 2: The balanced equation

2Al + 3CuSO4 → Al2(SO4)3 + 3Cu

Step 3: Calculate moles of Aluminium

Moles Al = mass Al / molar mass Al

Moles Al = 11.7 grams / 26.98 g/mol

Moles Al = 0.434 mol

Step 4: Calculate moles of CuSO4

Moles CuSO4 = 37.2 grams / 159.61 g/mol

Moles CuSO4 = 0.233 moles

Step 5: Calculate limiting reactant

For 2 moles of Al we need 3 moles of CuSO4

CuSO4 is the limiting reactant. It will be completely consumed (0.233 moles).

Al is in excess. There will be consumed 0.233 *(2/3) = 0.1553 moles

There will remain 0.434 - 0.1553 = 0.2787 moles

Step 6: Calculate moles of Al2(SO4)3

For 2 moles of Al we need 3 moles of CuSO4, to produce 1 mole of Al2(SO4)3 and 3 moles of Cu

For 0.233 moles CuSO4 we produce 0.233/3 = 0.0777 moles of Al2(SO4)3

Step 7: Calculate mass of Al2(SO4)3

Mass of Al2(SO4)3 = moles Al2(SO4)3 * molar mass Al2(SO4)3

Mass of Al2(SO4)3 = 0.0777 moles * 342.15g/mol

Mass of Al2(SO4)3 = 26.59 grams

There is 26.59 grams of aluminium sulfate produced

4 0

4 years ago

Consider the following reversible reaction.

dmitriy555 [2]

Answer:

Keq = [CO₂]/[O₂]

Explanation:

Step 1: Write the balanced equation for the reaction at equilibrium

C(s) + O₂(g) ⇄ CO₂(g)

Step 2: Write the expression for the equilibrium constant (Keq)

The equilibrium constant is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species. The equilibrium constant for the given system is:

Keq = [CO₂]/[O₂]

6 0

3 years ago

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