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yarga [219]
3 years ago
14

A helium balloon has a pressure of 0.2 kPa and a volume of 15 L. If the pressure increases to 0.5 kPa, what volume will the ball

oon have?
1.5 L

6.0 L

0.007 L

37.5 L
Chemistry
1 answer:
Alex17521 [72]3 years ago
7 0

Answer:

6.0 L

Explanation:

For this question, we can use

P1×V1= P2 × V2

where

P1 (initial pressure)= 0.2 kPa

V1 (initial volume)= 15L

P2 (final pressure)= 0.5 kPa

V2(final volume)= ?

Since we are trying to find final volume, we can rearrange the equation to make V2 the subject.

V2= (P1 × V1)/ P2

V2= (0.2 ×15)/0.5

V2 =6 L

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murzikaleks [220]

Answer:

A

Explanation:

14=pH + pOH

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pOH= -log[OH]

[OH]= 10^-12

=1×10^-12

5 0
1 year ago
A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C・g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('
polet [3.4K]

Answer:

a=5.65cm

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

Q_{Pt}=-Q_{Deu}

We can represent the heats in terms of mass, heat capacities and temperatures:

m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})

Thus, we solve for the mass of platinum:

m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g

Next, by using the density of platinum we compute the volume:

V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3

Which computed in terms of the edge length is:

V=a^3

Therefore, the edge length turns out:

a=\sqrt[3]{180cm^3}\\ \\a=5.65cm

Best regards.

6 0
3 years ago
Nvermind i messed u. thats not what i meant to outtb ycgdgc8ydsaox agdydac bc7w
jasenka [17]

Answer:

same

Explanation:

yes, I agree 100% and you are correct *applaud*

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