Answer:
<u>2</u> XE + <u>1</u> AB. --> <u>1</u> AE2 + <u>2</u> XB
Above equation is now balanced
Answer:
-68.4 kJ
Explanation:
<u>The standard enthalpy of vaporization = 23.3 kJ/mol</u>
<u>which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).</u>
To calculate heat released when 50.0 g of ammonia is condensed at -33 °C.
This is the opposite of enthalpy of vaporization which means that same magnitude of heat is released.
<u>Thus, Q = -23.3 kJ/mol</u>
<u>Where negative sign signifies release of heat</u>
Given: mass of 50.0 g
Molar mass of ammonia = 17.034 g/mol
Moles of ammonia = 50.0 /17.034 moles = 2.9353 moles
Also,
1 mole of ammonia when condenses at -33 °C releases 23.3 kJ
2.9412 moles of ammonia when condenses at -33 °C releases 23.3×2.9353 kJ
<u>Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.</u>
Answer:
Explanation:
Hello,
In this case, since at 60 °C, 108 grams of ammonium bromide are completely dissolved in 100 grams of water for a saturated solution, once it is cooled to 30 °C, wherein only 83.2 grams are completely dissolved in 100 grams of water, the following mass will precipitate:
Best regards.
> 2,000
mL of a 5.0 × 10–5% (w/v) sucrose solution
5.0 × 10–3
g/mL * 2000 mL * (1 mol / 342.30 g) = 0.0292 mol
<span>
> 2,000 mL of a 5.0 ppm sucrose solution</span>
5 grams /
1000000 mL * 2000 mL* (1 mol / 342.30 g) = 0.0000292 mol
<span>
> 20 mL of a 5.0 M sucrose solution </span>
5.0 M *
0.020 L = 0.1 mol
Answer:
<span>2,000 mL
of a 5.0 ppm sucrose solution</span>
