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RoseWind [281]
2 years ago
13

You are trying to make balloon sculptures. You twist the balloons gently, but they keep popping. Besides trimming your nails, ho

w could you you prevent that? Why will your strategy work?
Chemistry
2 answers:
Maurinko [17]2 years ago
7 0

Answer:

              By decreasing pressure.

Explanation:

                      In order to prevent balloons from popping while making sculptures, it is suggested to decrease the pressure of air in parent  balloon. Decreasing the pressure of parent balloon will allow it to twist easily and make designs.

This strategy will work according to Boyle's Law which states that, "Pressure and Volume are inversely proportional to each  other at constant temperature".

Mathematically,

                                       P ∝ 1/V

Or,

                                       P = k/V

Or,

                                       PV = k

Hence, as the new designs made after twisting are of less volume, therefore it is good to decrease the pressure in advance otherwise the resulting less volume will increase the pressure of daughter small balloons  and will explode them.

blondinia [14]2 years ago
6 0
I would simply remove some air from the balloons because with less air pressure the balloons will be less tight and more twistable without popping so by doing this it will be possible to sculpt the balloons into varied shapes successfully.
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N a football game, two players tackle each other so hard that they both fly in opposite directions after they hit each other. �W
evablogger [386]

Answer:

Newton's Third Law

Explanation:

Newton's Third Law stipulates that for every action there is an equal and opposite reaction.

So when the two players are tackling they exert a force on each other.

If player 1 tackles (exerts a force) player 2,  player 2 will exert an equal and opposite reaction on player 1 as stated in Newton's Third Law.

Therefore when they tackle each other so hard they both experience reaction forces so powerful that they fly in opposite directions.

Thus this is an example of the Newton's Third Law.

4 0
2 years ago
An experiment requires that each student use an 8.5 cm length of magnesium ribbon. How many students can do the experiment if th
melisa1 [442]
570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students
7 0
2 years ago
Read 2 more answers
If the actual yield of a reaction is 37.6 g while the theoretical yield is 112.8 g what is the percent yield
Zigmanuir [339]
<h2>Hello!</h2>

The answer is:

The percent yield of the reaction is 32.45%

<h2>Why?</h2>

To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.

We are given that:

ActualYield=37.6g\\TheoreticalYield=112.8g

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.

So, calculating we have:

PercentYield=\frac{ActualYield}{TheoreticalYield}*100\\\\PercentYield=\frac{37.6g}{112.8g}*100=0.3245*100=32.45(percent)

Hence, we have that the percent yield of the reaction is 32.45%.

Have a nice day!

8 0
2 years ago
Read 2 more answers
Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r
Scilla [17]

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.

6 0
2 years ago
How many grams of the excess reagent are left over when 6.00g of CS2 gas react with 10.0g of Cl2 gas in the following reaction:
WARRIOR [948]

Answer:

CS₂ = 2.43 g

Explanation:

Data Given:

CS₂ gas = 6.00g

Cl₂ = 10.0g

Reaction Given:

                   CS₂(g) + 3Cl₂(g) --------> CCl₄(l) + S₂Cl₂(l)

Solution

Limiting Reagent :

The reactant which is less in amount control the amount of product and know as limiting reagent.

Excess Reagent:

The amount of reactant which are in excess and leftover at the end of reaction and product formed.

Now we have to find the reactant that is in excess

For this we will look at the Reaction

                       CS₂   +    3Cl₂  -------->   CCl₄   +  S₂Cl₂

                       1 mol     3 mol               1 mol      1 mol

we come to know from the above reaction that

1 mole of CS₂ react with 3 mole of Cl₂ to produce 1 mole of CCl₄ and 1 mole of  S₂Cl₂

Now to convert mole to mass we required molar masses

molar mass of CS₂ = 12 +  2(32)

molar mass of CS₂ = 76 g/mol

molar mass of Cl₂ = 2(35.5)

molar mass of Cl₂ = 71 g/mol

if we represent mole in grams then

             CS₂              +    3Cl₂             -------->   CCl₄   +  S₂Cl₂

      1 mol (76 g/mol)     3 mol (71 g/mol)

                CS₂   +    3Cl₂   -------->   CCl₄   +  S₂Cl₂

                 76 g       213 g

So,

we come to know that 76 g of CS₂ will combine with 213 g of Cl₂ to form product.

So now look for the ratio of both reactant

                   CS₂    :    Cl₂

                   76 g   :    213 g

                    1  g     :      2.8 g

So, the for every one gram of CS₂ required 2.8g Cl₂

So from this details

we apply unity formula

                 2.8 g of Cl₂ ≅  1 g of CS₂

                 10 g of Cl₂ ≅  ? g of CS₂

by doing cross multiplication

                    g of CS₂ = 10 x 1 / 2.8

                    g of CS₂ = 3.6 g

So,

10.0g of Cl₂ used 3.57 g of CS₂

It showed that 3.57 g of CS₂ used and the remaining amount of CS₂ is

               Remaining amount of CS₂ = 6 - 3.57 = 2.43 g

 

8 0
2 years ago
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