A gaseous mixture contains O2 and N2 in the ratio 1:4 by weight. What will be the ratio of the number of molecules?
Ratio of O2 : N2 by weight=1:4.Let the weight of O2= 1 gm.The weight of N2=4 gm.The ratio of the No. Of molecules will be the ratio of the Number of their moles ie.,= 1/32 :4/28= 28: (4 x32) = 7 : 32= 7:32 .Always use simple means to solve problems like this.Don't give direct formulas to calculate, though it may give answers directly. The matter is that even an average ‘STUDENT' should understand.O.K.
A mixture of gases consisting of 7.5 g of CO2, and 10 g of SO2, is in a 2.5 L container at 25 C. What is the total pressure inside the container?
Assuming the mixture to behave as an ideal gas.molecular weight of CO2 = 12+16+16 = 44molecular weight of SO2 = 32+16+16 = 64moles of CO2 = 7.5/44 = 0.17045moles of SO2 = 10/64 = 0.15625n= moles of CO2 + moles of SO2 = 0.15625 + 0.17045 = 0.3267V= 2.5L = 2.5*10^-3 m^3T= 25C = 298KPV = nRTP = nRT/V = 0.3267*8.314*298/(2.5*10^-3) = 323769.10896 Pa = 3.1953atm
What is the effect of adding inert gas in a reaction?
When an inert gas is added to an equilibrium at constant volume, the pressure increases but the no.of moles per unit volume remains the same. Hence the equilibrium does not change. However at constant pressure addition of an inert gas increases the overall volume due to which no. of moles per unit volume of each component present at equilibrium decreases.Hence to maintain the equilibrium constant, the reaction goes in a direction where no. of moles are higher.
If air is heated in a sealed container what will be change in density?
None. You see density = [math]mass/volume[/math] You must know that mass cannot change as the container is sealed, so no molecules can leak out. You must also that volume of the container is constant. So, when mass and volume are same as they were before heating, then does density changes after heating? No!!BUT, I think what made you confuse here was the thought that air becomes less dense on heating, but that is valid of open spaces, because the air can expand to the extent it wants to.The things changing here is the kinetic energy of the molecules of air and their temperature, the pressure inside the container. :)
If 10 g of ice at - 10° C is added to 50 g of water at 15° C, what is the temperature of the mixture?
You need to understand 3 concepts to solve this answer.Firstly, You have to use the principle of calorimetry (measuring heat content of a system) to get the answer for this question.The principle goes like this:heat gained by a system(ice) = heat given by a system (water).That means the heat absorbed by ice will be equal to the heat given by the water.Secondly,now since you know the principle, next step is to know the formula of heat given or taken, that isQ= m*c*(t2-t1)where,m=mass of the substancec=heat capacity of substance (ice=2.1kJ/kgK, water=4.2kJ/kgK)t2-t1=change in temperature of the substanceNow the last but not the least concept, you must know what is latent heat, you see while there is a change of state that is when solid converts to liquid or liquid to water there is no change in temperature with the absorption of heat and that value for ice to water is 336kJ/kg.So this is the process is which is gonna happen, as the ice comes in contact with water, its temperature will rise upto 0 deg celsius. At that temperature which is the melting point of ice, it converts into water without any change in temperature. After complete conversion it again absorbs heat to reach a state where the converted ice and the water are at same temperatures (thermal equillibrium).So your final formula would be:Heat given by water = heat absorbed by ice + latent heat + heat absorbed by water (melted ice)mw*cw*(tw-T) = mice*cice*(tice-0) +336 +mmw*cmw*(T-0)The only unknown here is T =mixture temperature.Hope you got the science. :)
What is the number of moles of oxygen in one litre of air containing 21% of oxygen by volume under standard conditions?
21%of 1litre = 210ml. 22.4 l of oxygen consists of 6.023 *10^23 molecules therfore 210 mole consists of 210/22400*6.023 * 10^23 which is equal to 5.64 * 10^21 molecules. One mole equals 6.023*10^23 molecules . Therefore 5.64*10^21 molecule equals 0.00923 moles
How much volume at STP does 1g of hydrogen occupy?
In many textbooks you must have read the statement,“One mole of any ideal gas occupies 22.4 L at STP.”.Well, not since 1982. The above statement was accurate till 1982 as it was derived from the Ideal Gas Law using the definition of STP till then. In 1982, the definition of STP was slightly modified: Instead of using the pressure as 1 atm or [math]1.01325\times10^5[/math] Pa, we now use exactly [math]10^5[/math] Pa.Now the volume is approximately 22.8 L.Let’s see for ourselves.1g of [math]H[/math] = 1 mol of [math]H[/math] = 0.5 mol of [math]H_2[/math]By the current definition, STP or Standard Temperature and Pressure is,Pressure [math]P = 10^5 Pa = 10^5 \dfrac{N}{m^2}[/math][math] [/math]Temperature [math]T = 273.15 K[/math]Let’s start with the basics;Ideal Gas Law or the Equation of State states that:[math]PV = nRT[/math],where [math]V[/math] is the volume we need to find,[math]n = [/math]the molar count [math]= .5[/math] moland [math]R = [/math]Universal Gas Constant [math]= 8.314 J.K^{-1}.mol^{-1} [/math]Thus we can rewrite the equation of state to[math]V = \dfrac{nRT}{P}[/math]Now substituting the factors by their values we get,[math]V = \dfrac{.5\times(8.314)\times(273.15)}{10^5} m^3[/math][math]\implies V = 0.0114 m^3[/math]As we know that [math]1 m^3 = 10^3 litres[/math] or [math]10^3[/math] L[math]\implies V = 11.4[/math] L (ANS)Thus we can tell that 1 mole of the gas will occupy 22.8 L.P.S. Try using [math]P = 1.01325\times10^5[/math] Pa and see that the volume for 1 mole of gas will be 22.4 L, which is an obsolete value.Your teachers may not have noticed it because they are too busy teaching you science just as religion is taught or as an animal is trained for circus.