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:
Explanation:
When the amount of H2O2 is doubled while KI is kept constant, the rate of reaction doubles.
When the amount of KI is doubled and the amount of H2O2 is halved, the rate stays nearly constant.
2H2O2 (aq) → O2(g) + 2H2O (l) ------------- first order kinetics reaction.
Catalysts are KI, FeCl3 only, KCl is not a catalyst. Order: KI < MnO2 < Pb < FeCl3.
H2O2 + I– -> IO– + H2O (Step 1)
H2O2 + IO– -> I– + H2O + O2 (Step 2)
It can be seen that the iodine ion (provided by the KI solution) is a product as well as a reactant.
02(g)2Fe? (aq) + 2 H(a) 2 H 2 Fe3 (aq) H2O2(aq) + 2 Fe,Taq) H02(aq) 2 Fe (aq) 2 H (aq)
<h2><u>Answer:</u></h2>
A nonpartisan iota of Nitrogen has a mass of 18. There are 7 protons in the core of this iota. What number of neutrons, complete electrons, and valence electrons are available
Nitrogen 15 has a nuclear mass of 15. The mass number is # protons in addition to # of neutrons, so for N-15 mass is 15 and the protons are dependably 7 so there must be 15-7=8 neutrons. N-15 has 7 electrons since it has 7 protons and p = e.
Answer:
Given, 0.29 g of hydrocarbon produces 448ml of CO2 at STP. then, C2H5 is the emperical formula of hydrocarbon . n = 2 , hence, molecular formula will be C4H10
The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
<h3>What is Limiting reagent ?</h3>
The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.
Given chemical equation in balanced form ;
2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).
According to the Chemical equation ;
- The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.
But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
- 6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react
As 80 g NaOH produces 106 g of Na₂CO₃.
Therefore 5 g NaoH will produce ;
106 / 80 x 5 = 6.625 g
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