All categories

2 years ago

The volume of 1.0 g of nitrogen in a balloon is increased, it causes

1 answer:

7 0

The volume of 1.0 g of nitrogen in a balloon is increased. It causes: the mass of the gas to decrease the density of the gas to increase and the pressure of the gas to decrease the temperature of the gas decrease.

<h3>What is volume?</h3>

How much space an object or substance takes up.

If you increase the temperature of the gas in the balloon, the pressure will initially increase. The balloon will expand until the pressure inside drops back down and is equal to the pressure outside.

If you increase the atmospheric pressure, the balloon will compress until the pressure inside equals the new pressure outside.

If you remove gas from the balloon, the pressure will drop, causing the balloon to compress until the pressure is again equal to the pressure outside.

if you place the balloon underwater, it will be under greater pressure, causing the balloon to compress until the inside pressure equals the new outside pressure.

Learn more about the volume here:

brainly.com/question/1578538

#SPJ1

You might be interested in

An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.

Tju [1.3M]

Answer:

a) $\Delta S$

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

$\delta S=\frac{\delta Q}{T}$

$\Delta S=\frac{Q}{T}$

The energy balance:

$\delta U=[tex]\delta Q- \delta W$

If the process is isothermical the U doesn't change:

$0=[tex]\delta Q- \delta W$

$\delta Q= \delta W$

$Q= W$

The work:

$W=\int_{V1}^{V2}P*dV$

If it is an ideal gas:

$P=\frac{n*R*T}{V}$

$W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV$

Solving:

$W=n*R*T*ln(V2/V1)$

Replacing:

$\Delta S=\frac{n*R*T*ln(V2/V1)}{T}$

$\Delta S=n*R*ln(V2/V1)}$

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

$\Delta S$

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0

4 years ago

How many liters of CO are in 1 mol of CO

Art [367]

Answer:

44.01amu?

Explanation:

8 0

3 years ago

PLZ HELP!!! ILL MARK BRAINLIEST!!! You read an article written by an amateur ghost hunter. It includes the following paragraph.

djverab [1.8K]

No, it is not a scientific claim. If a ghost hunter catches a voice on video like ghost hunters do you can either believe it or not. Ghosts are more of a i believe it you dont have to type of thing becasue they have not been scientifically proven to exist.

4 0

3 years ago

Read 2 more answers

25 points!!! please Which types of elements areconsidered neutraland Where are these types of elements found on the periodic tab

posledela

Elements which composed of an equal amount of three components protons, neutrons and electrons are considered neutral elements. example of neutral elements are hydrogen, helium, lithium and beryllium...

7 0

3 years ago

Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . If of water i

Kaylis [27]

Answer:

87.9%

Explanation:

Balanced Chemical Equation:

HCl + NaOH = NaCl + H2O

We are Given:

Mass of H2O = 9.17 g

Mass of HCl = 21.1 g

Mass of NaOH = 43.6 g

First, calculate the moles of both HCl and NaOH:

Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles

Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles

Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:

Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles

Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles

From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:

Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g

% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%

3 0

4 years ago