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Andrew
23 days ago
6

The enthalpy of Sodium is 235 calories. The enthalpy Chlorine is 435 calories. The enthalpy of Sodium chloride 670 joules, what

is the change in enthalpy for this reaction?
Chemistry
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Find the specific heat of sikver in J/g C if 38.5 cal is required to heat 25.0g of silver from 31.5 C to 58.7 C (1cal=4.184 J)
alisha [2963]

Answer:

The specific heat value for silver is 0.236 J/g-C.

Explanation:

Silver has a mass of 25 grams.

The temperature shifts from 31.5° C to 58.7° C.

The required heat is equivalent to 25 g.

To determine silver's specific heat, the following equation applies:

Q=mc\Delta T

Where c represents the specific heat of silver.

c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{38.5 \times 4.184}{25(58.7-31.5)}\\\\c=0.236\ J/g-C

Thus, the specific heat of silver is 0.236 J/g-C.

7 0
3 months ago
A 0.652-g sample of a pure strontium halide reacts with excess sulfuric acid. the solid strontium sulfate formed is separated, d
lorasvet [2795]

Answer:

The original halide's formula is SrCl₂.

Explanation:

  • The chemistry reaction's balanced equation is:

SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X indicates the halide.

  • Based on the equation's stoichiometry, 1.0 mole of strontium halide yields 1.0 mole of SrSO₄.
  • The moles of SrSO₄ (n = mass/molar mass) = (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
  • The moles of SrX can thus be calculated as 4.11 x 10⁻³ moles based on stoichiometry from the balanced equation.
  • n = mass / molar mass, thus n =  4.11 x 10⁻³ moles and mass = 0.652 g.
  • The molar mass of SrX₂ is calculated using mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
  • The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
  • Calculating the atomic mass of halide X, we find = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2  g/mole = 35.5 g/mole.
  • This identifies the atomic mass of Cl.
  • Consequently, the original halide's formula is SrCl₂.
4 0
3 months ago
The ideal gas law tends to become inaccurate when Group of answer choices the pressure is raised and the temperature is lowered.
VMariaS [2998]

Answer: Option (a) is the correct answer.

Explanation:

Under conditions of low pressure and high temperature, gas molecules exhibit negligible attractions or repulsions among themselves. Hence, gases behave ideally in these scenarios.

Conversely, at low temperatures, there is a reduction in the kinetic energy of gas molecules, while high pressure compels the molecules to be closer together.

Thus, attractive forces emerge between molecules in conditions of low temperature and high pressure, causing gases to be termed real gases.

Therefore, we conclude that the ideal gas law becomes less accurate when pressure increases and temperature decreases.

5 0
3 months ago
A sample of helium gas has a volume of 0.180L, a pressure of 0.800a and a temperature of 29 C. What is the new temp or gas of vo
Anarel [2989]

Solution:

The gas's new temperature is 604K

Justification:

Assuming standard temperature and pressure, we can determine the gas's temperature using the ideal gas law;

Step 1: Formulate the general gas law equation

P1V1/T1 = P2V2/T2

Step 2: Insert the values, converting as needed to standard units.

P1 = 0.800 atm

V1 = 0.180 L

T1 = 29°C = 273 + 29 = 302K

P2 = 3.20 atm

V2 = 90 mL = 90 * 10^-3 L = 0.09 L

Step 3: Solve for T2

The new gas temperature T2 is calculated as:

T2 = P2V2T1/(P1V1)

T2 = 3.20 * 0.09 * 302 / (0.800 * 0.180)

T2 = 86.976 / 0.144

T2 = 604K

The gas's new temperature is 604K.

7 0
3 months ago
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