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aniked [119]
3 years ago
9

The tip of a match is ignited as it is struck against the matchbox. Why is this a chemical change? a.The color of the match tip

has changed. b.The match changes size as it burns. c.The tip of the match is flammable and is set on fire. d.The match is melting.
Chemistry
2 answers:
Rama09 [41]3 years ago
4 0

Answer:

C

Explanation:

irga5000 [103]3 years ago
3 0

It's (C.: The tip of the match is flammable and is set on fire.

Hope this helps! I got the answer right on my test!

:)

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Which of the following statements explains how the fossil record provides evidence that evolution has occurred?
olasank [31]
I think it will be letter A tbh
Hope this helps you out
7 0
3 years ago
Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide w
jek_recluse [69]

The question is incomplete, here is the complete question:

Phosgene, COCl_2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:  

CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

The value of K_c for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which p_{CO}=p_{Cl_2}=0.265atm and p_{COCl_2}=0.000atm ?

<u>Answer:</u> The equilibrium partial pressure of CO, Cl_2\text{ and }COCl_2 is 0.257 atm, 0.257 atm and 0.008 atm respectively.

<u>Explanation:</u>

The relation of K_c\text{ and }K_p is given by:

K_p=K_c(RT)^{\Delta n_g}

K_p = Equilibrium constant in terms of partial pressure

K_c = Equilibrium constant in terms of concentration = 5.79

\Delta n_g = Difference between gaseous moles on product side and reactant side = n_{g,p}-n_{g,r}=1-2=-1

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

T = Temperature = 570 K

Putting values in above equation, we get:

K_p=5.79\times (0.0821\times 570)^{-1}\\\\K_p=0.124

We are given:

Initial partial pressure of CO = 0.265 atm

Initial partial pressure of chlorine gas = 0.265 atm

Initial partial pressure of phosgene = 0.00 atm

The given chemical equation follows:

                      CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

<u>Initial:</u>            0.265      0.265

<u>At eqllm:</u>        0.265-x    0.265-x        x

The expression of K_p for above equation follows:

K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}

Putting values in above equation, we get:

0.124=\frac{x}{(0.265-x)\times (0.265-x)}\\\\x=0.0082,8.59

Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure

So, the equilibrium partial pressure of CO = (0.265-x)=(0.265-0.008)=0.257atm

The equilibrium partial pressure of Cl_2=(0.265-x)=(0.265-0.008)=0.257atm

The equilibrium partial pressure of COCl_2=x=0.008atm

Hence, the equilibrium partial pressure of CO, Cl_2\text{ and }COCl_2 is 0.257 atm, 0.257 atm and 0.008 atm respectively.

6 0
3 years ago
Find the mass in grams of 2.00x10^23 molecules of F2
earnstyle [38]
Hello!

To find the mass in grams of 2.00 * 10²³ molecules of F₂ you'll need to apply the Avogadro's number, which tells the number of molecules in a mol of a substance, and the Molar Mass, which tells the weight of a mole of a substance. These two quantities are used in the following conversion factor:

2,00*10^{23} moleculesF_2* \frac{1 mol F_2}{6,02*10^{23} moleculesF_2 }* \frac{37,997 g F_2}{1 mol F_2} \\  \\ =12,51 g F_2

So, the mass of 2.00 * 10²³ molecules of F₂ is 12,51 g

Have a nice day!
4 0
3 years ago
What mass of carbon dioxide can be produced from 4 moles of sodium bicarbonate according to the following unbalanced reaction,
KonstantinChe [14]
The answer is on the paper

6 0
3 years ago
The temperature in your town is 31°F. The radio announcer says that the temperature will drop 15 degrees. What will the temperat
hoa [83]

Answer:

The temperature of the town will be 15°F.

The equation showing the final temperature after drop say x degrees can be written as:

T'=(T-x)^oF

Explanation:

The current temperature in our town = T = 31°F

Temperature drop suggested by radio announcer = 15°

Temperature of the town after temperature drop = T'

T'=T-15^oF

T'=31^oF-15^oF=16^oF

The equation showing the final temperature after drop say x degrees can be written as:

T'=(T-x)^oF

3 0
3 years ago
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