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Alona [7]
3 years ago
13

If a precipitation reaction occurs, what will be the products of the unbalanced

Chemistry
1 answer:
levacccp [35]3 years ago
7 0

Answer:

{ \sf{B.  \: AgCl _{(s)}  + NaNO _{3(aq)}}}

Explanation:

Silver chloride precipitates out because it has a low value of Ksp.

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How does the motion of objects with similar mass compare to objects<br> of different mass?
emmasim [6.3K]

Explanation:

<em><u>in fact , we can use newtons second law of motion (see the SPT: Force topic) to calculate the acceleration in each of these cases</u></em>

<em><u>in fact , we can use newtons second law of motion (see the SPT: Force topic) to calculate the acceleration in each of these caseshope it helps you like me plz</u></em>

7 0
2 years ago
What is the atomic number for an element whose mass number is 78,
Ann [662]

Answer:

protons = atomic numbermass number = protons + neutronsatomic number = mass number - neutrons = 78-38 = 40

Explanation:

5 0
2 years ago
Read 2 more answers
What was one main point of Dalton's atomic theory?
sweet-ann [11.9K]

Answer:

C. That atoms made up the smallest form of matter

Explanation:

The crux of the Dalton's atomic theory is that atoms are the smallest form of matter. He propositioned that atoms is an indivisible particle and beyond an atom, no form of matter exists.

Series of discoveries through time have greatly shaped the Dalton's atomic theory. The discovery of cathode rays by J.J Thomson in 1897 opened up the atom. Atoms were now seen to be made up of some negatively charged particles. Ernest Rutherford through his gold foil experiment proposed the nuclear model of the atom.

4 0
2 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
3 years ago
How well did tossing the pennies simulate half lives?
SSSSS [86.1K]
Its a 50% chance that approx 1/2 of the pennies will land on tails. The next toss will result the same. and so on and so on. showing how a reaction would slowly eliminate 1/2 of the remaining lives per reaction, until nothing is left. I think it is a good stimulation.
5 0
3 years ago
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