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3 years ago

When energy is added to a substance at constant volume and pressure, its temperature increases except ________.

Chemistry

1 answer:

Travka [436]3 years ago

7 0

A. When the substance is in its gaseous state.

Explanation:

When a substance is expanding against its constant volume and pressure, its temperature increases except when the substance is in gaseous state and not in liquid or solid state. So the internal energy increase in the system not only increases and maintaining the volume and pressure of the system remains constant in its gaseous phase. In the first law of Thermodynamics, it is used specifically that to especially in the case of gaseous system.

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You check the air pressure in your car tires on Monday morning and the pressure

olga55 [171]

In warmer weather gases tend to expand and take up more room, thus increasing pressure. but in colder weather they will condense or contract and take up less space, therefore lowering the pressure of the tire in this situation.

5 0

3 years ago

Ppredict the identity of the precipitate in the below reaction: BaCl2(aq) + K2SO4(aq) →

Annette [7]

Answer: The precipitate formed is $BaSO_4$

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid or precipitated form are represented by (s) after their chemical formulas.

A double displacement reaction in which one of the product is formed as a solid is called as precipitation reaction.

The balanced chemical equation is:

$K_2SO_4(aq)+BaCl_2(aq)\rightarrow BaSO_4(s)+2KCl(aq)$

3 0

3 years ago

A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m

Lana71 [14]

Answer:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol$

Moles of hydrogen gas produced = 0.01225 mol

$2M+2xHCl\rightarrow 2MCl_x+xH_2$

Moles of metal = $\frac{0.225 g}{27.0 g/mol}=8.3333 mol$

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

$\frac{8.3333}{0.01225 mol}=\frac{2}{x}$

x = 2.9 ≈ 3

$2M+6HCl\rightarrow 2MCl_3+3H_2$

$MCl_3\rightarrow M^{3+}+Cl^-$

Formulas for the oxide and sulfate of M will be:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$ .

3 0

4 years ago

WILL GIVE BRANLIEST TO FIRST ANSWER!!! I know the correct answer is either a or d. Can anyone explain which one it is and why? C

oee [108]

Answer: The IUPAC name of $CH_3C\equiv CCH_2CH_2Cl$ is 5-chloro-2-pentyne

Explanation:

1. First select the longest possible carbon chain. For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.

2. The longest possible carbon chain should contain all the bonds and functional groups.

3. The numbering is done in such a way that the carbon containing the functional group or substituent gets the lowest number. Triple bond is given priority over substituent halogen.

4. The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne.

Thus the IUPAC name of $CH_3C\equiv CCH_2CH_2Cl$ is 5-chloro-2-pentyne

4 0

3 years ago

Predict the boiling point of water at a pressure of 1.5 atm.

Lina20 [59]

Answer:

100.8 °C

Explanation:

The Clausius-clapeyron equation is:

$ln\frac{P_{1} }{P_{2}} =$ -Δ $\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}} )$

Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'

Isolating for T2 gives:

$T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}$

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C

7 0

3 years ago