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Gas mole test. PTg=1mole=6.02x10 23= 22.4L at STP. 1. Wax that can be used as a solid rocket booster is found to contain 85.2 % C and 16.8% H. a) 85.2 g of carbon represents 1pt i) more than 1mole C atoms ii) less than 1 mole C atoms iii) equal to 1 mole C atoms
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PTg=1mole=6.02x1023=22.4L at STP • 1. Wax that can be used as a solid rocket booster is found to contain 85.2 % C and 16.8% H. • a) 85.2 g of carbon represents 1pt • i) more than 1mole C atoms • ii) less than 1 mole C atoms • iii) equal to 1 mole C atoms • b) Explain your answer 1pt
PTg=1mole=6.02x1023=22.4L at STP • 1. Wax that can be used as a solid rocket booster is found to contain 85.2 % C and 16.8% H. • a) 85.2 g of carbon represents • 12.0g C = 1mole C therefore • 85.2g C represents more than • 1mole of C
PTg=1mole=6.02x1023=22.4L at STP • 1. Wax that can be used as a solid rocket booster is found to contain 85.2 % C and 16.8% H. • a) 85.2 g of carbon represents • 12.0g C = 1mole C therefore • 85.2g C represents more than • 1mole of C
PTg=1mole=6.02x1023=22.4L at STP • 1. Wax that can be used as a solid rocket booster is found to contain 85.2 % C and 16.8% H. • a) 85.2 g of carbon represents • 12.0g C = 1mole C therefore • 85.2g C represents more than • 1mole of C
PTg=1mole=6.02x1023=22.4L at STP • 1. Wax that can be used as a solid rocket booster is found to contain 85.2 % C and 16.8% H. • a) 85.2 g of carbon represents • 85.2g C ( 1mole C ) =7.10 mole C • ( 12.0g C )
PTg=1mole=6.02x1023=22.4L at STP • 1. Wax that can be used as a solid rocket booster is found to contain 85.2 % C and 16.8% H. • a) 85.2 g of carbon represents • 85.2g C ( 1mole C ) =7.10 mole C • ( 12.0g C )
PTg=1mole=6.02x1023=22.4L at STP • 1. Wax that can be used as a solid rocket booster is found to contain 85.2 % C and 16.8% H. • a) 85.2 g of carbon represents • 85.2g C ( 1mole C ) =7.10 mole C • ( 12.0g C )
PTg=1mole=6.02x1023=22.4L at STP • 1. Wax that can be used as a solid rocket booster is found to contain 85.2 % C and 16.8% H. • a) 85.2 g of carbon represents • 85.2g C ( 1mole C ) =7.10 mole C • ( 12.0g C )
PTg=1mole=6.02x1023=22.4L at STP • 1. Wax that can be used as a solid rocket booster is found to contain 85.2 % C and 16.8% H. • a) 85.2 g of carbon represents • 85.2g C ( 1mole C ) =7.10 mole C • ( 12.0g C )
PTg=1mole=6.02x1023=22.4L at STP • 1. Wax that can be used as a solid rocket booster is found to contain 85.2 % C and 16.8% H. • a) 85.2 g of carbon represents • 85.2g C ( 1mole C ) =7.10 mole C • ( 12.0g C ) • 7.10 mole C > 1mole C
PTg=1mole=6.02x1023=22.4L at STP • What does the mole concept allow us to do that would impossible to do without it? • The mole concept allows to count particles indirectly which would be impossible to do because particles are extremely small and any measurable sample would contain far too many to count.
PTg=1mole=6.02x1023=22.4L at STP • What does the mole concept allow us to do that would impossible to do without it? • The mole concept allows to count particles indirectly which would be impossible to do because particles are extremely small and any measurable sample would contain far too many to count.
PTg=1mole=6.02x1023=22.4L at STP • What does the mole concept allow us to do that would impossible to do without it? • The mole concept allows to count particles indirectly which would be impossible to do because particles are extremely small and any measurable sample would contain far too many to count.
PTg=1mole=6.02x1023=22.4L at STP • What does the mole concept allow us to do that would impossible to do without it? • The mole concept allows to count particles indirectly which would be impossible to do because particles are extremely small and any measurable sample would contain far too many to count.
PTg=1mole=6.02x1023=22.4L at STP • 2. How many hydrogen gas molecules (H2) are found in the 16.8g of hydrogen?1pt 16.8g H2 (6.02x1023 H2) = 5.06x1024 H2 ( 2.0g H2 )
PTg=1mole=6.02x1023=22.4L at STP • 2. How many hydrogen gas molecules (H2) are found in the 16.8g of hydrogen?1pt 16.8g H2 (6.02x1023 H2) = 5.06x1024 H2 ( 2.0g H2 )
PTg=1mole=6.02x1023=22.4L at STP • 2. How many hydrogen gas molecules (H2) are found in the 16.8g of hydrogen?1pt 16.8g H2 (6.02x1023 H2) = 5.06x1024 H2 ( 2.0g H2 )
PTg=1mole=6.02x1023=22.4L at STP • 2. How many hydrogen gas molecules (H2) are found in the 16.8g of hydrogen?1pt 16.8g H2 (6.02x1023 H2) = 5.06x1024 H2 ( 2.0g H2 )
PTg=1mole=6.02x1023=22.4L at STP • 2. How many hydrogen gas molecules (H2) are found in the 16.8g of hydrogen?1pt 16.8g H2 (6.02x1023 H2) = 5.06x1024 H2 ( 2.0g H2 )
PTg=1mole=6.02x1023=22.4L at STP • What question does the following factor label solution answer? 2pts __________________________________________________________________________________ • 1.00x109 wax ( 1mole wax ) • (6.02 x 10 23 wax)
PTg=1mole=6.02x1023=22.4L at STP • What question does the following factor label solution answer? 2pts • How many moles of wax are found • in a 1.00x109 wax molecule sample? • 1.00x109 wax ( 1mole wax ) • (6.02 x 10 23 wax)
PTg=1mole=6.02x1023=22.4L at STP • 1.00x109 wax ( 1mole wax ) = • (6.02 x 10 23 wax) • 1.66x10-15 moles of wax
PTg=1mole=6.02x1023=22.4L at STP • 1.00x109 wax ( 1mole wax ) = • (6.02 x 10 23 wax) • 1.66x10-15 moles of wax
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44gH2 4.44L H2 • 4.44mole H2 4.44 x1023 H2
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44gH2 (1moleH2) = 2.22 mole H2 (2.0g H2)
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44gH2 (1moleH2) = 2.22 mole H2 (2.0g H2)
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44gH2 (1moleH2) = 2.22 mole H2 (2.0g H2)
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44gH2 (1moleH2) = 2.22 mole H2 (2.0g H2)
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44gH2 (1moleH2) = 2.22 mole H2 (2.0g H2)
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44 LH2 (1moleH2) = .198 mole H2 (22.4L H2)
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44 LH2 (1moleH2) = .198 mole H2 (22.4L H2)
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44 LH2 (1moleH2) = .198 mole H2 (22.4L H2)
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44 LH2 (1moleH2) = .198 mole H2 (22.4L H2)
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44 LH2 (1moleH2) = .198 mole H2 (22.4L H2)
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44 x1023H2(1moleH2) = .738 mole H2 (6.02x1023H2) • Largest is 4.44 mole H2
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44 x1023H2(1moleH2) = .738 mole H2 (6.02x1023H2) • Largest is 4.44 mole H2
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44 x1023H2(1moleH2) = .738 mole H2 (6.02x1023H2) • Largest is 4.44 mole H2
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44 x1023H2(1moleH2) = .738 mole H2 (6.02x1023H2) • Largest is 4.44 mole H2
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44 x1023H2(1moleH2) = .738 mole H2 (6.02x1023H2) • Largest is 4.44 mole H2
PTg=1mole=6.02x1023=22.4L at STP • 3. Which sample of hydrogen gas is the largest? You must convert to one common quantity to answer • this question. • 4.44 x1023H2(1moleH2) = .738 mole H2 (6.02x1023H2) • Largest is 4.44 mole H2
PTg=1mole=6.02x1023=22.4L at STP • 4a) What is the density of H2 (g) at STP • ( 273 K and 101.3 kPa)? 2pts • D=m = 2.0g H2 (per mole) = .0893g/L • V 22.4 L H2 (per mole)
PTg=1mole=6.02x1023=22.4L at STP • 4a) What is the density of H2 (g) at STP • ( 273 K and 101.3 kPa)? 2pts • D=m = 2.0g H2 (per mole) = .0893g/L • V 22.4 L H2 (per mole)
PTg=1mole=6.02x1023=22.4L at STP • 4a) What is the density of H2 (g) at STP • ( 273 K and 101.3 kPa)? 2pts • D=m = 2.0g H2 (per mole) = .0893g/L • V 22.4 L H2 (per mole)
PTg=1mole=6.02x1023=22.4L at STP • 4a) What is the density of H2 (g) at STP • ( 273 K and 101.3 kPa)? 2pts • D=m = 2.0g H2 (per mole) = .0893g/L • V 22.4 L H2 (per mole)
PTg=1mole=6.02x1023=22.4L at STP • b) A H2 balloon has a volume of 1.5 L at • 20oC. What is its volume at 0oC if the • number of gas molecules in the balloon • and the pressure of the H2 remain • constant? 1pt
PTg=1mole=6.02x1023=22.4L at STP • b) A H2 balloon has a volume of 1.5 L at • 20oC. What is its volume at 0oC if the • number of gas molecules in the balloon • and the pressure of the H2 remain • constant? 1pt • V1=1.5 L v2 = ? • T1=20oC+273 =293KT2=0oC+273 =273K • 1.5 L ( 273 k) = 1.40L • (293 k )
PTg=1mole=6.02x1023=22.4L at STP • b) A H2 balloon has a volume of 1.5 L at • 20oC. What is its volume at 0oC if the • number of gas molecules in the balloon • and the pressure of the H2 remain • constant? 1pt • V1=1.5 L v2 = ? • T1=20oC+273 =293KT2=0oC+273 =273K • 1.5 L ( 273 k) = 1.40L • (293 k )
PTg=1mole=6.02x1023=22.4L at STP • b) A H2 balloon has a volume of 1.5 L at • 20oC. What is its volume at 0oC if the • number of gas molecules in the balloon • and the pressure of the H2 remain • constant? 1pt • V1=1.5 L v2 = ? • T1=20oC+273 =293KT2=0oC+273 =273K • 1.5 L ( 273 k) = 1.40L • (293 k )
PTg=1mole=6.02x1023=22.4L at STP • b) A H2 balloon has a volume of 1.5 L at • 20oC. What is its volume at 0oC if the • number of gas molecules in the balloon • and the pressure of the H2 remain • constant? 1pt • V1=1.5 L v2 = ? • T1=20oC+273 =293KT2=0oC+273 =273K • 1.5 L ( 273 k) = 1.40L • (293 k )