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konstantin123 [22]
3 years ago
11

in 1965, the mean price of a new car was $2,650 and the standard deviation was $1,000. In 2011, the mean was $30,500 and the sta

ndard deviation was $9,000. If a Ford Mustang cost $2,300 in 1965 and $28,000 in 2011, in which year was it more expensive relative to other cars.
Chemistry
1 answer:
miss Akunina [59]3 years ago
8 0

Answer:

Car were more expensive in 2011

Explanation:

Data provided in the question:

Mean price of the car in 1965 = $2,650

Standard deviation, s = $1,000

Mean price of the car in 2011 = $30,500

Standard deviation in 2011, s = $9,000

Cost in 1965 = $2,300

Cost in 2011 = $28,000

Now,

z score = [ X - mean ] ÷ s

Therefore,

In 1965

z score = [ $2,300 - $2,650] ÷ 1,000

= -0.35

In 2011

z score = [ $28,000 - $30,500 ] ÷ 9,000

= -0.278

Since,

The standard score in 2011 is higher than in 1965

Hence,

Car were more expensive in 2011

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Compute the molar enthalpy of combustion of glucose (C6 H12O6 ): C6 H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O (g) Given that com
lana66690 [7]

Answer:

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

Combustion of 0.305 grams causes a raise of 6.30 °C

Calorimeter has a heat capacity of 755 J/°C

Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g)

Step 3:

ΔH = (m * C * ΔT + c(calorimeter) * ΔT)

with m = mass of the solutin = 0.305 grams

with C = heat capacity of water = 4.184 J/g°C

with ΔT = the change in temperature = 6.30 °C

with c(calorimeter) = 755 J/°C

ΔH = 0.305 * 4.184 *6.30 + 755 * 6.30  = 4764.5 J ( negative because it's exothermic)

Step 4: Calculate moles of glucose

Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = -4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = -2819.3 kJ/moles

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

5 0
3 years ago
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