Answer:
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This question is describing the following chemical reaction at equilibrium:
And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:
Thus, by recalling the Van't Hoff's equation, we can write:
Hence, we solve for the enthalpy change as follows:
Finally, we plug in the numbers to obtain:
![\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B-8.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%20%2Aln%280.25%2F9%29%7D%7B%5B%5Cfrac%7B1%7D%7B%2875%2B273.15%29K%7D%20-%5Cfrac%7B1%7D%7B%2825%2B273.15%29K%7D%20%5D%20%7D%20%5C%5C%5C%5C%5C%5C%5CDelta%20H%3D4%2C785.1%5Cfrac%7BJ%7D%7Bmol%7D)
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Answer:
6.02 × 10²³ atoms
Explanation:
The number 6.02 × 10²³ is called Avogadro number. It is the number of atoms, ions and molecules in one gram atoms of an element, one gram ions of substance and one gram molecule of a compound.
For example:
32 g of oxygen = one mole = 6.02 × 10²³ atoms O.
1.008 g of hydrogen = one mole = 6.02 × 10²³ atoms of H.
or
18 g of H₂O =one mole = 6.02 × 10²³ molecules of H₂O
44 g of CO₂ = one mole = 6.02 × 10²³ molecules of CO₂
or
62 g of NO₃⁻ = one mole of NO₃⁻ = 6.02 × 10²³ ions of NO₃⁻
Answer:
142.82 g
Explanation:
The following data were obtained from the question:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Density of gol= 19.3 g/cm³
Mass of gold =.?
Next, we shall determine the volume of the gold. This can be obtained as follow:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Volume of gold =.?
Volume of gold = (Volume of water + gold) – (Volume of water)
Volume of gold = 19.4 – 12
Volume of gold = 7.4 mL
Finally, we shall determine the mass of the gold as follow:
Note: 1 mL is equivalent to 1 cm³
Volume of gold = 7.4 mL
Density of gol= 19.3 g/cm³ = 19.3 g/mL
Mass of gold =?
Density = mass /volume
19.3 = mass of gold /7.4
Cross multiply
Mass of gold = 19.3 × 7.4
Mass of gold = 142.82 g
Therefore, the mass of the gold pebble is 142.82 g
Answer: There are 
molecules in 6.73 moles of phosphorus trichloride
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
1 mole of 
contains = molecules
Thus 6.73 moles of contains = 
molecules
Thus there are molecules in 6.73 moles of phosphorus trichloride
