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KengaRu [80]
3 years ago
7

Liquid nitrogen is cold and can be used to cool objects to -196°C. If you put the bottle of warm air in liquid nitrogen at this

temperature, what would the volume (mL) of air in the bottle be?
Chemistry
1 answer:
vampirchik [111]3 years ago
6 0

Answer:

Explanation:

It depends on how this is done. If you raise the pressure, the nitrogen will disappear (liquify) and all that will be left will be the 21 % oxygen and the 1% argon.

The process is very complicated because the boiling point of nitrogen keeps on changing. The boiling point is unstable.

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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is
Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

7 0
4 years ago
Analyses of enzymes found in the blood are used as indicators that tissue damage (heart, liver, muscle) has occurred and resulte
Reika [66]

Answer:

True.

Explanation:

Yes, analyses of enzymes found in the blood are used as indicators of tissue damage in the heart, liver, muscle etc has occurred. This leakage of enzymes into the bloodstream tells us whether the tissue is damaged or not. Lactate dehydrogenase is a type of enzyme which is used as indicator which is responsible for the interconverts lactate and pyruvate. The concentration of this enzyme in the blood tells us about tissue damage.

7 0
3 years ago
Read 2 more answers
Does pb(no3)2 + Na3(PO4) = Pb3(PO4)2 + Na(NO3) have a precipitate?
ipn [44]

Answer:

Yes, Pb3(PO4)2.

Explanation:

Hello there!

In this case, according to the given balanced chemical reaction, it is possible to use the attached solubility series, it is possible to see that NaNO3 is soluble for the Na^+ and NO3^- ions intercept but insoluble for the Pb^3+ and PO4^2- when intercepting these two. In such a way, we infer that such reaction forms a precipitate of Pb3(PO4)2, lead (II) phosphate.

Regards!

4 0
3 years ago
The weather report says that if the temperature drops 10°, it will reach 25° below zero. The current temperature T can be found
Neporo4naja [7]

Answer:

The current temperature is -15° (15° below zero)

Explanation:

The temperature drops 10°:

T-10°

It will reach 25° below zero:

T - 10° = -25°

We add 10° in both members of the equation:

T - 10° +10° = -25° +10°

The equation is simplified as follows:

T = -25° +10°

T = -15°

The integer -15° can be expressed as 15° below zero.

8 0
3 years ago
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
3 years ago
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