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tensa zangetsu [6.8K]
2 years ago
6

Britta traveled 1250 km on her road trip and used 209 L of gasoline, filling her tank atan average of 1.14 euros per liter. Pier

ce traveled 1405 km o. His trip and used 175 L of gasoline, filling his tank at an average cost of 1.23 euros per liter. Who paid more per kilometer driven cor their trip?
Chemistry
1 answer:
icang [17]2 years ago
4 0

Answer:

  • Britta paid more per kilometer driven

Explanation:

<u>1) Britta:</u>

  • Distance traveled: 1,250 Km
  • Gasoline used: 209 liter
  • Gasoline price: 1.23 euros / liter

Cost per km = total cost of gasoline / distance traveled

Cost per km = gasoline used × gasoline price / distance traveled

Cost per km = 209 liter × 1.14 (euro / liter) / 1,250 km =  0.19 euro / km

<u>2) Pierce:</u>

  • Distance traveled: 1,405 km
  • Gasoline used: 175 liter
  • Gasoline price = 1.23 euros / liter

Cost per km = 175 liter × 1.23 (euro / liter) / 1,405 km =  0.15 euro / km

<u>3) Comparison:</u>

  • 0.19 > 0.15 ⇒ Britta paid more per kilometer driven
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VladimirAG [237]
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<span>H2C=O---------H-OH </span>

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<span>Formaldehyde in aqueous solution is in equilibrium with its hydrate. </span>

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5 0
1 year ago
A student has two samples of NaCl, each one from a different source. Assume that the only potential contaminant in each sample i
bija089 [108]

Answer:

The correct option is;

A. Which sample has the higher purity

Explanation:

The information given relate to the presence of two samples of NaCl, from different sources

The only potential contaminant in each of the sources = KCl

The content of the sample = NaCl

The molar mass of NaCl = 58.44 g/mol

The molar mass of KCl = 74.5513 g/mol

Let the number of moles of KCl in the sample = X

For a given mass of NaCl, KCl mixture, we have;

The molar mass of potassium = 39.0983 g/mol

The molar mass of chlorine = 35.453 g/mol

The molar mass of sodium ≈ 23 g/mol

Therefore;

Each mole of KCl, will yield 35.453 g/mol per 74.5513 g/mol of KCl

While each mole of NaCl will yield 35.453 g/mol per 58.44 g/mol of NaCl

Therefore, the pure sodium chloride sample will yield more chlorine per unit mass of sample.

As such if the two samples have the same mass, the sample with the contaminant of KCl will yield less mass of chlorine per unit mass of the sample, from which the student will be able to tell the purity of the solution.

The sample with the higher purity will yield  a higher mass chlorine per unit mass of the sample.

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2 years ago
2Na2O2 + 2CO2 → 2Na2CO3 + O2
ahrayia [7]

the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment

theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations

we need to first find the theoretical yield

2Na₂O₂ + 2CO₂ ---> 2Na₂CO₃ + O₂

molar ratio of Na₂O₂ to Na₂CO₃ is 2:2

number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed

number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol

therefore number of Na₂CO₃ moles formed is - 0.10 mol

mass of Na₂CO₃ expected to be formed is - 0.10 mol x 106 g/mol = 10.6 g

therefore theoretical yield is 10.6 g

percent yield = actual yield / theoretical yield  x 100%

81.0  % = actual yield / 10.6 g x 100 %

actual yield = 10.6 x 0.81

actual yield = 8.59 g

therefore actual yield is 8.59 g

7 0
2 years ago
Read 2 more answers
Q1) A vapor-compression refrigeration system operates on the cycle of Fig. 9.1. The refrigerant is 1,1,1,2-Tetrafluoroethane. Gi
hoa [83]

Answer:

i)   0.5071 (kg/s)

ii)  -1407.1 kj/kg

iii)  204.05 Kw

iv)  5.881

v)    9.238

Explanation:

Given Data:

evaporation temperature ( T ) = 4°c = 277.15 K

Condensation Temperature ( T ) = 34°c = 307.15 K

<em>n</em> ( compressor efficiency ) = 0.76

refrigeration rate = 1200 kJ.s^-1

i) determine the circulation rate of the refrigerant

m = \frac{Q}{H2 - H1}  = \frac{Q}{H2 - H4\\}  ------- 1

Q = 1200 Kj.s^-1

H2 = entropy at step 2 = 2508.9 (kJ / kg ) ( gotten from Table F )

H4 = entropy at step 4 = 142.4 ( kJ/ kg )

back to equation 1

m ( circulation rate of refrigerant ) = 0.5071 (kg/s)

ii) heat transfer rate in the condenser

Q = m ( H4 - H3 )

    = 0.5071 ( 142.4 - 2911.27 )

    = -1407.1 kj/kg

where H3 = H2 + ΔH23 = 2911.27 (kj/kg) ( as calculated )

iii) power requirement

w = m * ΔH23

   = 0.5071 (kg/s) * 402.37 (kj/kg) =  204.05 Kw

where: ΔH23 = \frac{H'3 - H2 }{0.76} = \frac{2814.7-2508.9}{0.76} = 402.37 (kj/kg)

iv) coefficient of performance of a cycle

W = Qc / w

  = 1200 Kj.s^-1/ 204.05 kw

  = 5.881

v) coefficient of performance of a Carnot refrigeration cycle

w_{carnot} = \frac{T2}{T4 - T2}

            =  277.15 / ( 307.15 - 277.15 )

            = 9.238

4 0
2 years ago
Air fills a room with a volume of 5600 L. Atmospheric pressure is 740 torr. What will be the pressure if all of the gas is pumpe
Lapatulllka [165]

Answer:

P_2 = 6.9\times 10^3 kpa

Explanation:

Assuming that temperature is constant

According to Boyle's Law, at constant temperature pressure is inversly proportional to the volume  and mathematically it can be expressed as:

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P_1=740torr

P_2=?

V_1=5600L

V_2=80L

from the first equation after putting all the value

we get,

P_2=51800torr =6906098.7Pa=6.91 \times 10^3 kpa

P_2 = 6.9\times 10^3 kpa

8 0
2 years ago
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