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3 years ago

12

You place a balloon in a closed chamber at STP. When the balloon doubles in size, what happens to the chamber pressure? A.) The

pressure stays the same B.) The pressure doubles C.) The pressure reduced by one half

2 answers:

Katarina [22]3 years ago

8 0

It decrease's so C is correct.

larisa86 [58]3 years ago

4 0

If you increase the volume the pressure will decrease. so the best answer is C

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Gnom [1K]

Answer:

the answers 5, also did u finish ur ia yet,

if so you know what to do.

Explanation:

5 0

3 years ago

Read 2 more answers

What is the difference between a scientist theory and a scientific law

Andru [333]

Answer:

A scientific theory is a widely accepted belief on why something happens in the natural world while a scientific law is proven to a fact that shows what happens.

Explanation:

3 0

2 years ago

Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH₄), by the following process:

kykrilka [37]

Answer:

6.75 g of HCN can be produced by the reaction

Explanation:

Complete reaction is:

2NH₃ (g) + 3O₂ (g) + 2CH₄ (g) → 2HCN (g) + 6H₂O (g)

Let's determine the moles of each reactant:

11.5 g . 1mol / 17g = 0.676 moles of ammonia

12 g . 1 mol / 32g = 0.375 moles of oxygen

10.5 g . 1mol/ 16 g = 0.656 moles of methane

Now is all about rules of three:

2 moles of ammonia reacts with 3 moles of O₂ and 2 moles of methane

0.676 moles of NH₃ may react with:

(0.676 . 3) /2 = 1.014 moles of O₂

(0.676 . 2) / 2 = 0.676 moles of methane

Both can be the limiting reactant.

3 moles of O₂ react with 2 moles of NH₃ and 2 moles of methane

0.375 moles of O₂ will react with:

(0.375 .2) / 3 = 0.375 moles

The same amount for methane, 0.375 moles

2 moles of CH₄ reacts with 3 moles of O₂ and 2 moles of NH₃

0.656 moles of methane would react with 0.656 moles of NH₃

(0.656 . 3 ) /2 = 0.437 moles of O₂ I do not have enough O₂

Oxygen is the limiting reactant → We can work with the reaction now.

Ratio is 3:2. 3 moles of oxygen produce 2 moles of cyanide

0.375 moles of O₂ may produce (0.375 .2 ) / 3 = 0.250 moles

If we convert the moles to mass → 0.250 mol . 27 g / 1mol = 6.75 g

4 0

3 years ago

Project 1 Hypothesis Construction

alukav5142 [94]

The hypothesis should be; "Increasing the amount of ice increases the rate at which the ice will melt"

<h3>What is a hypothesis?</h3>

The term hypothesis refers to a tentative explanation that could be applied in order to explain an observation. When we give a hypothesis, it must be subjected to rigorous experimentation thus it can be confirmed or repelled by experiment.

Now, we can see that the question here is; "How does the amount of salt added to ice affect the rate at which the ice will melt?" This implies that the correct hypothesis here ought to have to do with the impact of the amount of ice in the affirmative.

Thus, the hypothesis should be; "Increasing the amount of ice increases the rate at which the ice will melt"

Learn more about hypothesis:brainly.com/question/13025783

#SPJ1

5 0

1 year ago

The average lung capacity of a human is 6.0L.

Darya [45]

Answer:

(a) 0.25 mol

(b) 0.11 mol

(c) 8.77 mol

Explanation:

(a)

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 298 K

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

n = number of moles = ?

Putting values in above equation, we get:

$1.00 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1.00\times 6.0}{0.0821\times 298}=0.25mol$

(b)

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 0.296 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 200 K

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

n = number of moles = ?

Putting values in above equation, we get:

$0.296 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 200K\\\\n=\frac{0.296\times 6.0}{0.0821\times 200}=0.11mol$

(c)

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 30 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 250 K

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

n = number of moles = ?

Putting values in above equation, we get:

$30 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 250K\\\\n=\frac{30\times 6.0}{0.0821\times 250}=8.77mol$

8 0

3 years ago