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grin007 [14]
3 years ago
12

1. Which is not an example of a physical change to a wooden stick?

Chemistry
1 answer:
Anvisha [2.4K]3 years ago
5 0

The answer is C. burning the stick. This is because burning the stick will change the chemical properties of the wood, and it will not be wood anymore, it will be ash and the rest will be heart or gas.

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When oxygen reacts with hydrogen, it has the capacity to release 29 kilojoules of energy. Inside a fuel cell, oxygen reacts with
Vanyuwa [196]

Answer:

79

Explanation:

7 0
3 years ago
How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume t
ddd [48]

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number  (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

                                  = 2.75 x 10²³ molecules XeF₆.

8 0
3 years ago
Heeeeeellllppp, Iiiimmm sssooo stttuuuccckkk!​
n200080 [17]

Answer:

I think it would be D. None of the above

Explanation:

why I think this is because none of the answers really adds up to it

4 0
3 years ago
A) F-F
WINSTONCH [101]

Answer:

the covalent molecule has OA.A

7 0
2 years ago
What should be the mole fraction of O2 in the gas mixture the diver breathes in order to have the same partial pressure of oxyge
Nutka1998 [239]

The question is incomplete, here is the complete question:

A scuba diver is at a depth of 355 m, where the pressure is 36.5 atm.

What should be the mole fraction of O_2 in the gas mixture the diver breathes in order to have the same partial pressure of oxygen in his lungs as he would at sea level? Note that the mole fraction of oxygen at sea level is 0.209.

<u>Answer:</u> The mole fraction of oxygen in the gas mixture is 0.00573

<u>Explanation:</u>

To calculate the partial pressure, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}      ........(1)

where,

p_A = partial pressure of oxygen at sea level = ?

p_T = total pressure at sea level = 1.00 atm

\chi_A = mole fraction of oxygen at sea level = 0.209

Putting values in equation 1, we get:

p_{O_2}=1.00atm\times 0.209\\\\p_{O_2}=0.209atm

As, partial pressure of the oxygen in the diver's lungs is equal to the partial pressure of oxygen at sea level

We are given:

p_T=36.5atm\\p_{O_2}=0.209atm

Putting values in equation 1, we get:

0.209atm=36.5atm\times \chi_{O_2}\\\\\chi_{O_2}=\frac{0.209}{36.5}=0.00573

Hence, the mole fraction of oxygen in the gas mixture is 0.00573

3 0
3 years ago
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