Give a clear handwritten answer.. tthwं कावछक ली मत्य Tड Pr०वंपट्टवं weउ.न्0 ट जC केCoाटनयेष कि मकटfottowngced equa tion g Theणीणप्योnश कोकल्टतैपवलव४ कवSS o मट्र S उह. पढ 87m०J.H2+ Clz2तले54.781) 72.9gi) 219 g

Given-> Mole of Cl2 = 3.00 mole Molar mass of HCl = 36.46 g/mole

tफ्रिण कावड8 ली मत्च ड Prव्वेपण््dं wh उ.न्0 कd oकी ताक जC के Cागटर्नथेप क मकट Fगीvप्aणश कलललह व मेणल ? ।hिe कवि४ कवडs o+मटर S उ6य6 8 (mo J. 2 सत्य मे०ल Thिe H2+ Cl2 S4.१8 !) 72.9g ) 219g

tफ्रिण कावड8 ली मत्च ड Prव्वेपण््dं wh उ.न्0 कd oकी ताक जC के Cागटर्नथेप क मकट Fगीvप्aणश कलललह व मेणल ? ।hिe कवि४ कवडs o+मटर S उ6य6 8 (mo J. 2 सत्य मे०ल Thिe H2+ Cl2 S4.१8 !) 72.9g ) 219g

Introductory Chemistry: An Active Learning Approach

6th Edition

ISBN:9781305079250

Author:Mark S. Cracolice, Ed Peters

Publisher:Mark S. Cracolice, Ed Peters

Chapter3: Measurement And Chemical Calculations

Section: Chapter Questions

Problem 5E

See similar textbooks

Similar questions

267 °F is equivalent to 389 K 116°C 540 K
With respect to the rules of sig figs explained in the first image, is 8.96 g/cm3 the correct answer or would it need to be two sig figs instead?
How can I round this number? Example: Cp = qmetal = -365.4 cal = 0.0501 cal/g°C m∆T (109.7g)(-66.5°C) help me with this Cp = qmetal = -77.0 cal = 0.9803921cal/g°C m∆T (109.7g)(-66.5°C)
Chemistry (i) For T = 298K ΔG° = -8.314 x 298 x ln (2.83x10-3) = 14.524 KJ/mol (ii) For T = 308.15K ΔG° = -8.314 x 308.15 x ln (9.619x10-3) = 11.891 KJ/mol (iii) For T = 318.15K ΔG° = -8.314 x 318.15 x ln (3.14x10-2) = 9.147 KJ/mol (iv) For T = 328.15K ΔG° = -8.314 x 328.15 x ln (8.82x10-2) = 6.621 KJ/mol (v) For T = 338.15K ΔG° = -8.314 x 338.15 x ln (2.42x10-1) = 3.990 KJ/mol Based on the measured ΔG° values, is this equilibrium spontaneous at room temperature? Which factor, entropy, or enthalpy, has the greater impact on spontaneity in this case? Explain your answers.
2069 J = 4.184 J/g, °C *182.5 g * DeltaT
Calculate ΔH orxnfor the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced]ΔH o/f[CH4(g)] = −74.87 kJ/molΔH o/f[CCl4(g)] = −96.0 kJ/molΔH o/f[CCl4(l)] = −139 kJ/molΔH o/f[HCl(g)] = −92.31 kJ/molΔH o/f[HCl(aq)] = −167.46 kJ/molΔH o/f[Cl(g)] = 121.0 kJ/mol
a) 80°C, 40.0 kPab) 100°C, 120.0 kPac) -10°C, 66.7 kPa
The answer to 6.626*10^-34 Js (7.40*10^6 m/s)(9.11*10^-28 g) is 9.83*10^-17 g right? That's the answer I got but my digital homework says it's wrong, but like, all my math is checking out that it should be what I got?
Also we have a mixture for gas is hydrogen helium neon and nitrogen sealed in a container of -50 Celsius degree so imaginary pick one molecule and follow it around for some time and record speed we determine that on average a small chill travels with speed of approximately 1.9x10^3 km/h . Which molecule or atom did we pick up MM,g/mol Hydrogen 2.02 Helium 4.00 Neon 20.18 Nitrogen 28.01
Given the following data for Mass of test tube and stearic acid = 14.17 gMass of test tube = 11.40 gFreezing point of strearic acid = 69.59o CMass of weighing paper + naphthalene =1.230 gMass of weighing paper = 0.920 gFreezing point solution = 64.00o CKf = 4.5o C/m Determine the following1. mass of stearic acid in g (2 decimal places); 2.77g2. mass of naphthalene in g (2 decimal places); 0.31g3. freezing point depression (2 decimal places); 3.93oC4. molality of solution (3 significant figures); _____5. moles of naphthalene (3 significant figures); _____6. molar mass of naphthalene, experimentally (3 significant figures); _____7. % error if theoretical molar mass of naphthalene is 128.17 g/ mole, USE ABSOLUTE VALUE (3 significant figure); _____
To identify a liquid substance, a student determined itsdensity. Using a graduated cylinder, she measured out a 45-mLsample of the substance. She then measured the mass of thesample, finding that it weighed 38.5 g. She knew that the substancehad to be either isopropyl alcohol 1density 0.785 g>mL2or toluene 1density 0.866 g>mL2. What are the calculated densityand the probable identity of the substance?
H2 (g) + Na (s) → F2 (g) + He (g) →