The volume of 0. 250 mole sample of 
gas occupy if it had a pressure of 1. 70 atm and a temperature of 35 °C is 3.71 L.
Calculation,
According to ideal gas equation which is known as ideal gas law,
PV =n RT
- P is the pressure of the hydrogen gas = 1.7 atm
- Vis the volume of the hydrogen gas = ?
- n is the number of the hydrogen gas = 0.25 mole
- R is the universal gas constant = 0.082 atm L/mole K
- T is the temperature of the sample = 35°C = 35 + 273 = 308 K
By putting all the values of the given data like pressure temperature universal gas constant and number of moles in equation (i) we get ,
1.7 atm×V = 0.25 mole ×0.082 × 208 K
V = 0.25 mole ×0.082atm L /mole K × 308 K /1.7 atm
V = 3.71 L
So, volume of the sample of the hydrogen gas occupy is 3.71 L.
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Answer:
lithium
Explanation:
this is because lithium has a valency of 1 and oxygen has a valency of 2 thereby exchanging valency to create Li²0
The correct option is D.
The hydrogen atoms that are attached to the nitrogen atom in the ammonia molecule are capable of forming hydrogen bond. The hydrogen bond that exist in the ammonia molecule is the reason why it shows higher boiling point compare to the other hydrides. Hydrogen bond occur in ammonia because ammonia is one of the most electronegative elements.
ΔG° at 450. K is -198.86kJ/mol
The following is the relationship between ΔG°, ΔH, and ΔS°:
ΔH-T ΔS = ΔG
where ΔG represents the common Gibbs free energy.
the enthalpy change, ΔH
The temperature in kelvin is T.
Entropy change is ΔS.
ΔG° = -206 kJ/mol
ΔH° equals -220 kJ/mol
T = 298 K
Using the formula, we obtain:
-220kJ/mol -T ΔS° = -206kJ/mol
220 kJ/mol +206 kJ/mol =T ΔS°.
-T ΔS = 14 kJ/mol
for ΔS-14/298
ΔS=0.047 kJ/mol.K
450K for the temperature Completing a formula with values
ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol
ΔG° = -220 kJ/mol + 21.14 kJ/mol.
ΔG°=198.86 kJ/mol
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CxHy + O2 --> x CO2 + y/2 H2O
Find the moles of CO2 : 18.9g / 44 g/mol = .430 mol CO2 = .430 mol of C in compound
Find the moles of H2O: 5.79g / 18 g/mol = .322 mol H2O = .166 mol of H in compound
Find the mass of C and H in the compound:
.430mol x 12 = 5.16 g C
.166mol x 1g = .166g H
When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.
Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).
In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).
