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Digiron
4 months ago
6

Which procedure cannot be performed on a hot plate, requiring a Bunsen burner instead

Chemistry
1 answer:
Tems11 [2.7K]4 months ago
4 0

Answer: The process of heating a crucible to eliminate moisture from a hydrate.

Explanation:

The available choices are:

a. Heating a solvent to aid in the dissolution of a solute.

b. Heating a solid in isolation to remove moisture.

c. Bringing water to a boil for use in a water bath.

d. Heating a crucible to eliminate moisture from a hydrate.

Possible actions that can be done on a hot plate include:

a. Heating a solvent to assist a solute in dissolving.

b. Heating a solid in isolation to dry it.

c. Heating water to boiling for a water bath.

However, it's important to note that using a hot plate for heating a crucible to remove water from a hydrate is not advisable. Silica or ceramic materials are not meant to be heated on a hot plate.

Consequently, the correct procedure is heating a crucible to remove water from a hydrate.

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Hydrogen gas has a density of 0.090 g/L, and at normal pressure and -1.72 C one mole of it takes up 22.4 L. How would you calcul
Anarel [2989]

Answer:

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Explanation:

Assuming all calculations occur at standard pressure and a temperature of -1.72°C :

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Where

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6 0
4 months ago
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Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L
lorasvet [2795]

Answer:

The molality is 1.15 m.

Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

Calculate moles of H₂SO₄ from molarity:

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Mass of solvent (water) based on density:

m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg

Therefore, molality is:

m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m

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4 months ago
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The value of delta G at 141.0 degrees celsius for the formation of phosphorous trichloride from its constituent elements,
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The appropriate answer is option E. Gibbs free energy can be expressed using the equation: ΔG = ΔH - TΔS, where ΔH denotes the change in enthalpy of the reaction, T is the reaction temperature, and ΔS signifies entropy change. For our calculations, we have ΔH = -720.5 kJ/mol which converts to -720500 J/mol (given that 1 kJ = 1000 J), ΔS = -263.7 J/K, and T = 141.0°C, which equals 414.15 K. Consequently, the Gibbs free energy for the specified reaction at 141.0°C is calculated as -611.3 kJ/mol.
6 0
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