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3 years ago

If a balloon is squeezed, what happens to the pressure of the gas inside the balloon?

Chemistry

2 answers:

Goshia [24]3 years ago

8 0

Answer: Option (a) is the correct answer.

Explanation:

Pressure is defined as the force applied on per unit area.

Mathematically, Pressure = $\frac{Force}{Area}$

When we squeeze a balloon then it means we are applying pressure on it. Also, we are decreasing the area of balloon by applying more amount of force on it.

As area is inversely proportional to pressure. So, a decrease in area will lead to an increase in pressure.

Hence, we are increasing the pressure on the balloon by squeezing it.

Thus, we can conclude that if a balloon is squeezed, then pressure of the gas inside the balloon increases.

Nuetrik [128]3 years ago

5 0

Your answer is
A. it increases

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when particles of matter are in direct contact and movement occurs in the substance due to convection, thermal energy transfers

liq [111]

Thermal energy transfers in a solid state, due to convection, in metalic substances.

This is because the covalent bonds between the atoms are being broken and reformed again while the metal is experiencing stress. covalent bonds store energy.

4 0

3 years ago

I need help solving this!

zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, $CH_{4}$ .

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol$

The given reaction equation is as follows.

$C + 2H_{2} \rightarrow CH_{4}$

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

$Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol$

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, $CH_{4}$ .

5 0

3 years ago

Magda has the two magnets shown below what will happen if magda tries to push the north poles of the two magnets toward one anot

ryzh [129]

They will repel one another because opposites attract, but like charges will repel.

8 0

3 years ago

Select the equations below that represent physical changes.

Ierofanga [76]

Physical changes- any change in the physical properties of the substance is a physical change. Generally physical properties include color, state, size , shape, odour, appearance . In any phyical change, no new substance is formed.

Chemical changes- any change in the chemical properties of the substance is a chemical change. Chemical properties changes only when something new is formed during a reaction.

∴ Part A- 2 H2O(l) → 2 H2(g) + O2(g)

In this, H2 and O2 is produced during the reaction which is different from H20. Thus, it is a chemical change.

Part B- H2O(l) → H2O(s)

In this, only the state of water changes from liquid to solid and no new product is formed. Thus, it is a physical change.

Part C-CO2(s) → CO2(g)

In this also, only the state of CO2 changes from solid to gas and no new product is formed. Thus, it is a physical change.

Part D-H2(g) → 2 H(g)

In this reaction, H2 molecule is dissociated into 2 hydrogen atom leading to formation of new products. Thus, it is a chemical change.

Finally, equations that represent physical changes are - B and C

B. H2O(l) → H2O(s)

C. CO2(s) → CO2(g)

6 0

3 years ago

A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat

Lady_Fox [76]

Explanation:

(a) The given data is as follows.

Load applied (P) = 1000 kg

Indentation produced (d) = 2.50 mm

BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$

Now, putting the given values into the above formula as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$ = $\frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}$

= $\frac{2000}{9.98}$

= 200

Therefore, the Brinell HArdness is 200.

(b) The given data is as follows.

Brinell Hardness = 300

Load (P) = 500 kg

BHI diameter (D) = 10 mm

Indentation produced (d) = ?

d = $\sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}$

= $\sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}$

= 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0

3 years ago