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lutik1710 [3]
3 years ago
10

Warning signs are sometimes placed on aerosol cans to prevent people from throwing them into a fire. What would be true about th

e contents of the gas in an aerosol can just after it is placed in a fire?
Volume stays constant.

Pressure stays constant.

Temperature stays constant.
Chemistry
2 answers:
sweet [91]3 years ago
8 0

Answer:

The Volume stays constant.

Explanation:

GarryVolchara [31]3 years ago
3 0

Answer:

The correct option is volume stays constant

Explanation:

When a gas container (in this case an aerosol can) is subjected to heat (from fire), the temperature of the can and subsequently <u><em>the temperature of the gas itself increases</em></u>, an increase in the temperature of the gas cause <u><em>the pressure to also increase;</em></u> as the gas molecules will collide more and faster with each other and against the wall of the can. However, the volume of the gas will remain the same as before it was subjected to the heat - the gas particles do not get destroyed or increased as a result of the heat (law of conservation of matter explains this).

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Which statement is true about the electrons that can be located together in an orbital?
Salsk061 [2.6K]

The two electrons that share an orbital repel each other.

All electrons bear a negative charge. They are held in their orbits by the attractive force of charged protons. The farther away an orbital is to the atomic nucleus the easier it is to expunge an electron from this distant orbital shell.

Explanation:

Because electrons have the same negative charge, they repel each other especially when they occupy the same orbital shell in an atom. To reduce this repulsion, each of the electrons in the orbital shell (remember electrons occupy orbital shells of atoms in 2s) assumes an opposite quantum (M<em>s</em>) spin; one with  – ½ while the other + ½ .

Learn More:

For more about electrons check out;

brainly.com/question/13251728

brainly.com/question/13174681

#LearnWithBrainly

3 0
3 years ago
2.0 L of Carbon dioxide is heated from -25.0 °C to Standard Temperature.
RideAnS [48]

The final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L.

<h3>How to calculate volume?</h3>

The volume of a given gas can be calculated using the Charles law equation as follows:

V1/T1 = V2/T2

Where;

  • V1 = initial volume
  • V2 = final volume
  • T1 = initial temperature
  • T2 = final temperature

  • V1 = 2L
  • V2 = ?
  • T1 = -25°C + 273 = 248K
  • T2 = 273K

2/248 = V2/273

273 × 2 = 248V2

546 = 248V2

V2 = 546/248

V2 = 2.2L

Therefore, the final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L

Learn more about volume at: brainly.com/question/11464844

5 0
2 years ago
Suppose a snack bar is burned in a calorimeter and heats 2,000 g water by 20 °C. How much heat energy was released? (Hint: Use t
Salsk061 [2.6K]

The answer is: 167360

6 0
2 years ago
Read 2 more answers
Guys- my science teacher doesn't teach, he just says go on schoology ​
Free_Kalibri [48]
The relative mass of each element can be found from the periodic table (the larger number). For instance, P2O5, P = 31.0 O = 16.0, thus the formula mass (Mr) is 2(31) + 5(16) = 142 amu (atomic mass unit). I used a not really specific periodic table. Maybe your teacher is referring to open Schoology (a website where teacher can post files or announcements to students in his or her class). Sorry if i got something mistaken.
5 0
2 years ago
Calcium hydroxide is a strong base but is not very soluble ( Ksp = 5.02 X 10-6 ). What is the pH of a saturated solution of Ca(O
Misha Larkins [42]

Answer : The pH of a saturated solution is, 12.33

Explanation : Given,

K_{sp} = 5.02\times 10^{-6}

First we have to calculate the solubility of OH^- ion.

The balanced equilibrium reaction will be:

Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-

Let the solubility will be, 's'.

The concentration of Ca^{2+} ion = s

The concentration of OH^- ion = 2s

The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}][OH^-]^2

Let the solubility will be, 's'

K_{sp}=(s)\times (2s)^2

K_{sp}=(4s)^3

Now put the value of K_[sp} in this expression, we get the solubility.

5.02\times 10^{-6}=(4s)^3

s=1.079\times 10^{-2}M

The concentration of Ca^{2+} ion = s = 1.079\times 10^{-2}M

The concentration of OH^- ion = 2s = 2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M

First we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (2.158\times 10^{-2})

pOH=1.67

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33

Therefore, the pH of a saturated solution is, 12.33

7 0
3 years ago
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