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charle [14.2K]
3 years ago
10

Please help! It’s due today. I’ll give brainliest

Chemistry
2 answers:
MArishka [77]3 years ago
5 0

Answer:

look at it is the correct answer because if a new gas has formed bubbles will how or change of colour so look at it

Explanation:

zavuch27 [327]3 years ago
3 0
Iiiiii you guys have to have the kids tomorrow so you
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6th grade science i mark as brainliest​
kiruha [24]

Answer:

it would be 5 i think

Explanation:

cuz 50 divided by 10 !

7 0
3 years ago
Read 2 more answers
A + B = AB
Ahat [919]

The greatest amount of AB would be produced if the equilibrium constant of the reaction is equal to 1.0 \;X \;10^5. Hence, option D is correct.

<h3>What is an equilibrium constant?</h3>

A number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

The equilibrium constant expression is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants.

If the value of K is greater than 1, the products in the reaction are favoured. If the value of K is less than 1, the reactants in the reaction are favoured.

Hence, option D is correct.

Learn more about the equilibrium constant here:

brainly.com/question/10038290

#SPJ1

7 0
2 years ago
All of the deer, grass, maple trees, and owls in a forest would be considered a
Alexxx [7]
A habitat because it has animals and trees
4 0
3 years ago
Read 2 more answers
How many molecules are in
Olin [163]
It’s B
Because 6.02
5 0
3 years ago
30 POINTS! ----- How many moles of oxygen gas are needed to completely react with 145 grams of aluminum? Report your answer with
gladu [14]
First a balanced reaction equation must be established:
4Al _{(s)}    +   3 O_{2}  _{(g)}     →    2 Al_{2} O_{3}

Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
                                                                                    =  4.83 mols

Now the mole ratio of Al : O₂ based on the equation is  4 : 3  
                                                                [4Al  + 3 O₂ → 2 Al₂O₃]

∴ if moles of Al = 4.83 moles
  then moles of O₂ = (4.83 mol ÷ 4) × 3
                              =  3.63 mol   (to  2 sig. fig.) 

Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum. 


     
4 0
4 years ago
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