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RoseWind [281]
1 year 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]1 year 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]1 year 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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strojnjashka [21]
SThe  missing   coefficient  for  the  skeleton   equation  below  is  as  follows

skeleton   equation

Cr(s)  +  Fe(No3)2(aq)  ------> Fe (s)   +  Cr(NO3)3  (aq)
the  missing  coefficient  are  is   as  follows

 2 Cr(s)   +  3  Fe(NO3)2  ---> 3 Fe (s)  +  2 Cr(NO3)3

This  is  obtained   by  making  sure  all  the   molecules  are  balanced  in  both  sides
8 0
2 years ago
A teacher cut an apple into three wedges of the same size. She dipped one wedge in lemon juice, dipped another in water, and lef
aleksandrvk [35]

Answer:

The responding variable of this experement is the outcome and that would be  that the one in lemon juice responded and the one  in water didn't (the other one is the control). Thus the responding varible is the one in lemon juice.

Explanation:

4 0
2 years ago
Read 2 more answers
A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?
Tanzania [10]
<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume =  0.50 cm × 10.0 cm × 2.0 cm

             = 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

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Thus, the density of the gold bar is 19.3 g/cm³

3 0
1 year ago
When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3,a yellow precipitate forms and Ag+ becomes negligibly small. Whi
jok3333 [9.3K]

Answer:

Option A

Explanation:

Number of millimoles of Na3PO4 = 1 × 100 = 100

Number of millimoles of AgNO3 = 1 × 100 = 100

When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion

When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-

As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible

Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-

And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-

∴ Increasing order of concentration will be  PO43- < NO3- < Na+

4 0
2 years ago
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V - volume - 5.29 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 230 K
substituting these values in the equation 
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8 0
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