Convert 14 millimeters to it equivalent in meter

You have a ballon filled with hydrogen gas which keeps it at a constant pressure, regardless the volume. The initial volume of t

abruzzese [7]

Answer:

619°C

Explanation:

Given data:

Initial volume of gas = 736 mL

Initial temperature = 15.0°C

Final volume of gas = 2.28 L

Final temperature = ?

Solution:

Initial volume of gas = 736 mL (736mL× 1L/1000 mL = 0.736 L)

Initial temperature = 15.0°C (15+273 = 288 K)

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = T₁V₂/V₁

T₂ = 2.28 L × 288 K / 0.736 L

T₂ = 656.6 L.K / 0.736 L

T₂ = 892.2 K

K to °C:

892.2 - 273.15 = 619°C

7 0

2 years ago

Nvm i got it lol how do i delete

Lady_Fox [76]

Answer:

delete what?

Explanation:

3 0

2 years ago

In the EXPLORE section of your lesson 4.08 on Potential energy there were several animations to watch that provided a graphic il

Lunna [17]

Answer:

This is because no energy is being created or destroyed in this system

Explanation:

I think this is correct? I hope it helps.

7 0

2 years ago

Read 2 more answers

A 500-ml glass beaker is filled to the brim with ethyl alcohol at a temperature of 5. 00ºc. how much will overflow when its temp

Setler [38]

The volume of ethyl alcohol overflow will be 9.265mL.

To calculate the volume of ethyl alcohol overflow, we will use

∆V=V(á∆T)

Where,

∆V=volume expand

á=volumetric expansion Coefficients =0.00109/°C

V=initial volume =500ml

T'=final temp. =22°C

T=initial temp=5°C

Now,

∆V=500×(0.00109/°C) ×(22-5)

=500×(0.00109) (17)

=9.265ml

Thus, we find that the volume of ethyl alcohol overflow will be 9.265 mL.

learn more about ethyl alcohol:

brainly.com/question/11237501

#SPJ4

5 0

1 year ago

Find the volume of a gas STP if it’s volume is 80.0 mL at 109 kpa and-12.5c

photoshop1234 [79]

Answer:

Approximately $8.38 \times 10^8\; \rm mL$ , which is the same as $8.38 \times 10^5 \; \rm L$ . Assumption: the behavior of this gas is ideal.

Explanation:

Initial state of the gas:

$T_\text{initial} = -12.5\; \rm ^\circ C = (-12.5 + 273.15)\; \rm K = 260.65\; \rm K$ .
$P_\text{initial} = \; \rm 10^9\; \rm kPa = 10^{12}\; \rm Pa$ .

STP state:

$T_\text{STP} = 0\; \rm ^\circ C = 273.15\; \rm K$ .
$P_\text{STP} = 10^5\; \rm Pa$ .

The initial state of this gas can be changed to STP state in two steps:

First, reduce the pressure from $10^{12}\; \rm Pa$ to $10^5\; \rm Pa$ .
Second, increase the temperature from $260.65\; \rm K$ to $273.15\; \rm K$ .

Assume that the gas acts like an ideal gas at all time. Also, assume that the number of gas particles did not change.

Assume that temperature stays the same when the pressure changes from $10^{12}\; \rm Pa$ to $10^5\; \rm Pa$ . By Boyle's Law, volume is inversely proportional to pressure when all other factors stay the same. In other words,

$\begin{aligned}V_\text{intermediate} &= V_\text{initial} \cdot \displaystyle \frac{P_{\text{initial}}}{P_{\text{STP}}} \\ &= 80.0\; \rm mL \times \frac{10^{12}\; \rm Pa}{10^5\; \rm Pa} = 8.00 \times 10^8\; \rm mL\end{aligned}$ .

After that, assume that pressure stays the same when the temperature changes from $\rm 260.65\; \rm K$ to $\rm 273.15\; \rm K$ . By Charles's Law, volume is proportional to pressure when all other factors stay the same. In other words,

$\begin{aligned}V_\text{STP} &= V_\text{intermediate} \cdot \displaystyle \frac{T_{\text{STP}}}{T_{\text{initial}}} \\ &= 8.00 \times 10^8\; \rm mL \times \frac{273.15\; \rm K}{260.65\; \rm K} \\ &\approx 8.38\times 10^8\; \rm mL = 8.38 \times 10^5\; \rm L\end{aligned}$ .

7 0

3 years ago