Read the sentence from the introduction [paragraphs 1-3].

What is the constant parameters of pressure and volume? The relationship between the two? Are they directly proportional or inve

OverLord2011 [107]

Now I'm just going to assume you mean Charles law. So when working with gases, there are 4 properties: pressure, volume, temp, and quantity. The simple gas laws deal with 2, while leaving the other 2 constant. If Charles' Law changes temp and volume, what 2 stay constant? Pressure and quantity

6 0

3 years ago

Alveoli are tiny sacs of air in the lungs. Their average diameter is 4.50 × 10−5 m. Calculate the uncertainty in the velocity of

Arisa [49]

<u>Answer:</u> The uncertainty in the velocity of oxygen molecule is $4.424\times 10^{-5}m/s$

<u>Explanation:</u>

The diameter of the molecule will be equal to the uncertainty in position.

The equation representing Heisenberg's uncertainty principle follows:

$\Delta x.\Delta p=\frac{h}{2\pi}$

where,

$\Delta x$ = uncertainty in position = d = $4.50\times 10^{-5}m$

$\Delta p$ = uncertainty in momentum = $m\Delta v$

m = mass of oxygen molecule = $5.30\times 10^{-26}kg$

h = Planck's constant = $6.627\times 10^{-34}kgm^2/s^2$

Putting values in above equation, we get:

$4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg}=4.424\times 10^{-5}m/s$

Hence, the uncertainty in the velocity of oxygen molecule is $4.424\times 10^{-5}m/s$

8 0

3 years ago

im making a poster for chemistry. the topic is acid and base. i have to make a creative title to go with the poster. any ideas?

REY [17]

The correct answer for the question that is being presented above is this one: Right now, I haven't seen the poster but there are things that i have in mind. It can be "The Game of Acids and Bases," or you can another one like, "Unity Amidst Diversity."

3 0

3 years ago

what salt is formed in the reaction of aluminum with hydrochloric acid? write the formula of the salt. formula:

Alenkasestr [34]

Aluminum chloride is formed in the reaction of aluminum with hydrochloric acid. The formula of aluminum chloride is AlCl₃.

Aluminium chloride, additionally referred to as Aluminium trichloride, is an inorganic compound with the formula AlCl₃. It bureaucracy hexahydrate with the system [Al(H₂O)₆]Cl₃, containing six water molecules of hydration. Each are colourless crystals, however samples are regularly contaminated with iron(III) chloride, giving a yellow shade.

Aluminum Chloride is used to control immoderate sweating. This medicinal drug can be used for other purposes; ask your fitness care provider or pharmacist when you have questions.

Aluminum chloride is corrosive and nerve-racking to the eyes, skin, and mucous membranes. Can be dangerous if swallowed. Ingestion of big quantities may also motive phosphate deficiency. Aluminum chloride is a compound of aluminum and chloride it truly is widely utilized in petroleum refining and the producing of many products. similarly, antiperspirants and cosmetic astringents use this compound.

Learn more about aluminum chloride here:- brainly.com/question/6061451

#SPJ4

3 0

1 year ago

A 3.24-gram sample of NaHCO3 was completely decomposed in an experiment. 2NaHCO3 → Na2CO3 + H2CO3 In this experiment, carbon dio

Stolb23 [73]

Answer:

a) mass of the H2CO3 produced:

given:

Mass of sample = 3.24 g

Mass of Na2CO3 obtained after decomposition = 2.19 g

Solution :

Molar mass of NaHCO3 = 84

reaction:

2NaHCO3 → Na2CO3 + H2CO3

so it is clear that 2 mole of NaHCO3 gives 1 mole of Na2CO3 and H2CO3

Now, ICE table for the reaction is :

NaHCO3 Na2CO3 H2CO3

I 3.24/84 0 0

C -2x +x +x

E 3.24/84 -2x x x

As NaHCO3 is completely decomposed so final Concentration of NaHCO3 is zero.

=> 3.24/84 -2x = 0

=> 2x = 3.24/84

=> x = 1.62/84

The new ICE table is :

NaHCO3 Na2CO3 H2CO3

I 3.24/84 0 0

C -2x = -2(1.62/84) +x = 1.62/84 +x = 1.62/84

E 0 1.62/84 1.62/84

From the above ICE table,

it is found that (1.62/84 ) moles of H2CO3 is obtained.

Since,

The molar mass of H2CO3 is 62

=> Mass of H2CO3 obtained = moles × molar mass

=> Mass of H2CO3 obtained = (1.62 /84 ) × 62

= 1.19 grams

Mass of H2CO3 experimentally :

Mass of reactants = mass of products

=> Mass of sample = mass of Na2CO3 + mass of H2CO3

=> Mass of H2CO3 = mass of sample - mass of Na2CO3

= 3.24 - 2.19 = 1.05 g

b) Experimental mass = 1.05 g

Theoretical mass = 1.19 g

Percentage yield of H2CO3 = Experimental mass × 100 / Theoretical mass

= 1.05 × 100 /1.19

= 88.23 %

6 0

3 years ago