Something that is bizarre fact about Radon that unique and relatively unknown by the general population??

38 When a sample of a gas is heated in a sealed, rigid container from 200. K to 400. K, the pressure exerted by the gas is

Kryger [21]

Increased by a factor of 200

8 0

3 years ago

Compare and contrast diffusion and osmosis. Include at least one similarity and one difference between the two processes.

Softa [21]

Osmosis is the diffusion of water <span>across a semipermeable membrane (usually cell membrane) from a region of low solute concentration to a more concentrated solution so it can reach equilibrium (balance).

D</span>iffusion is <span>a spontaneous movement of particles from an area of high concentration to an area of low concentration.

Both results in particles moving and help balance out the concentrations.
Also, in osmosis, the water molecules are moving. In diffusion, it is the solutes moving.

I hope this helps and explains well.</span>

4 0

3 years ago

1. calculate the the reaction of gas, F 2 (g) with H 2 O(l) water form and O 2 (e).

Sever21 [200]

hope this helped.:)

Explanation:

Yes, the number of moles of oxygen gas produced by your reaction under those conditions for pressure and temperature will be 0.0025.

Hydrogen peroxide,

H

2

O

2

, decomposes to give water and oxygen gas according to the balanced chemical equation

2

H

2

O

2

(

a

q

)

→

2

H

2

O

(

l

)

+

O

2

(

g

)

You've collected 0.061 L of oxygen gas at 295.15 K and 1 atm, so you've got all the data you need to calculate the number of moles of oxygen gas produced by using the ideal gas law equation

P

V

=

n

R

T

⇒

n

=

P

V

R

T

n

O

2

=

1

atm

⋅

0.061

L

0.082

L

⋅

atm

mol

⋅

K

=

0.0025 moles

So, if this was your first question, then yes, your reaction produced 0.0025 moles of oxygen gas.

I find the second part of your question to be a little confusing. You were given the density of the hydrogen peroxide solution, so are you supposed to use that to determine the theoretical number of moles of oxygen for this reaction?

I'm not sure what

100%

H

2

O

2

=

1.02 g/mL

means, do you have a certain volume of hydrogen peroxide solution?

SIDE NOTE According to the additional information posted by Heather, it turns out that the initial hydrogen peroxide solution had a volume of 5 mL.

Even with the volume of the initial solution, you'd need its percent concentration to try and determine exactly how many moles you had present before the reaction.

Once you know how many moles of hydrogen peroxide you had, assume that all of the react and use the

2

:

1

mole ratio that exists between

H

2

O

2

and

O

2

to get the number of moles of oxygen your reaction could have produced.

7 0

3 years ago

You come home and find that there's been a power outage while food was cooking in the slow cooker. What should you do wit the fo

klasskru [66]

Answer:

Well you can not serve it because it is a slow cooker it cooks slow and the refrigerator uses power.Stove tops use power too, I wouldn't feed it to the dog because it whatever you made could possibly make the dog sick and throwing it away is a waste of money. Out of all of those explanations choose the best one. Which ever one you pick whether right or wrong, think about it.

Explanation:

7 0

3 years ago

A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is

aliina [53]

Answer:

$\large \boxed{\text{67 000 g}}$

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂ = 0

The formula for the heat absorbed or released by an object is

q = mCΔT, where

m = the mass of the sample

C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

q₁ + q₂ = 0

q₁ + m₂C₂ΔT₂ = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹; T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

$q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}$

(b) ΔT

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

$\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\$

$\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}$

7 0

3 years ago