2. Which equation shows the relationship between the Kelvin and Celsius temperature scales?

Chemistry

1 answer:

givi [52]3 years ago

7 0

Answer: I think the right answer is c

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A beaker contains 100.0 mL of pure water. A second beaker contains 100.0 mL of seawater. The two beakers are left side by side o

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Answer: The beaker in which water level has decreased more is the one that contains pure water.

Explanation: Even when water is not heated it can evaporate, that is because molecules move at different speed and those that are closer to surface can break the gass pressure and scape as water vapor.

The movement of molecules of water can be affected by the presence of solutes (salts or other soluble compounds) and increase the evaporation temperature. For this particular case, molecules of salt atract the water molecules (remember salt has ions and water is a dipole with partial charges) and do not let the water move freely which avoid them to scape to the surface.

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3 years ago

Chemical reactions _____.

aalyn [17]

1) Chemical reactions produce new substances

Ans (d)

2) The symbol, Δ in a chemical reaction means that heat of supplied to the reaction.

Ans (a)

3) A skeleton equations does not show the reaction stoichiometry, i.e. the relative amounts of reactants and products.

Ans (d)

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3 years ago

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A and B react together in this reversible reaction. A + 3B --> C + 2D. A mixture of 10 mol of A and 10 mol of B were left to

alexira [117]

Not multiple choice?

4 0

4 years ago

Help me solve. Djdjdjfjdkdjdjdkdkckc

Ket [755]

From what I would think, it is D.

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3 years ago

The atomic radius of a gold atom is 144×10^-12. The volume of a gold atom can be calculated using the volume of a sphere. The de

Alekssandra [29.7K]

Given:
Atomic radius of gold = 144 x 10⁻¹² m = 144 pm
Density of gold = 19.3 g/cm³ = 19.3 x 10⁻³ kg/cm³
Mass of gold sample = 1.40 g = 1.4 x 10⁻³ kg

Calculate the volume of an atom.
v = [(4π)/3]*(144 x 10⁻¹² m)³ = 1.2508 x 10⁻²⁹ m³

Calculate the mass of an atom.
$m=(1.2508\tmes10^{-29}\,m^{3})*(19.3\times 10^{-3}\, \frac{kg}{cm^{3}} )*(10^{2}\, \frac{cm}{m})^{3} \\ = 2.4142 \times 10^{-25} \, kg$

Calculate the number of atoms in a sample of mass 1.40 g.
$N = \frac{1.40 \time 10^{-3} kg}{2.4142 \times 10^{-25} \, \frac{kg}{atom} } =5.8 \times 10^{21} \, atoms$

Answer: 5.8 x 10²¹ atoms

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3 years ago