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Star_Killer117
2 years ago
0

At which labeled points would the skateboarder have the most kinetic energy?

The diagram shows a side view of a skateboard ramp. An illustration of a skateboarding halfpipe from the end. The top of the curve on the left is labeled W, the top of the curve of the right is labeled Z, part of the way down the curve on the left is labeled X and part of the way down the curve on the right is labeled Y. At which labeled points would the skateboarder have the most kinetic energy? W and X X and Y Y and Z W and Z (This is an edgenuity question)
Chemistry
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n-Butane fuel (C4H10) is burned with the stoichiometric amount of air. Determine the mass fraction of each product. Also, calcul
tia_tia [17]

Answer:

  1. 0.1852
  2. 0.0947
  3. 0.7201
  4. 3.0345 kg CO_{2} / Kg C_{4} H_{10}
  5. 15.3848 Kg air / kg C_{4} H_{10}

Explanation:

Molar masses of each product are :

Butane = 58 kg /kmol

Oxygen = 32 kg/kmol

Nitrogen = 28 kg/kmol

water = 18 kg/kmol

<u><em>1) Calculate the mass fraction of carbon dioxide </em></u>

= ( 4 * 44 ) / ( (5 * 18) + (4 *44 )+ (24.44 * 28) )

= 176 / 950.32

= 0.1852

<em><u>2) Calculate the mass fraction of water </u></em>

= ( 5 * 18 ) /  (( 5* 18 ) + ( 4*44) + ( 24.44 * 28 ))

= 90 / 950.32

= 0.0947

<em><u>3) Calculate the mass fraction of Nitrogen </u></em>

= (24.44 * 28 ) / ((4 * 44 ) + ( 24.44 * 28 ) + ( 5 * 18 ))

= 684.32 / 950.32

= 0.7201

<em><u>4) Calculate the mass of Carbon dioxide in the products</u></em>

Mco2 = ( 4 * 44 ) / 58  = 3.0345 kg CO_{2} / Kg C_{4} H_{10}

<u>5) Mass of Air required per unit of fuel mass burned </u>

Mair = ( 6.5 * 32 + 24.44 *28 ) / 58  = 15.3848 Kg air / kg C_{4} H_{10}

5 0
3 years ago
How many kilowatt-hours of electricity are used to produce 4.50 kg of magnesium in the electrolysis of molten mgcl2 with an appl
ratelena [41]
First, we need to determine the half reaction of magnesium. It would be expressed as:

Mg2+ + 2e- = Mg

Given the mass of magnesium metal that is produced, we calculate the total charge of the electrolysis by the relations we can get from the half reaction. We do as follows:

4.50 kg Mg ( 1000 g / 1 kg ) ( 1 mol / 24.305 g ) ( 2 mol e- / 1 mol Mg ) ( 96500 C / 1 mol e- ) = 35733388.2 C

We are given the applied EMF in units of V. This value is equal to J/C. So, 5 V is equal to 5 J/C.

35733388.2 C (5 J/C) = 178666941 J
178666941 J ( 1 kW-h / 3.6x10^6 J ) = 49.63 kW-h
3 0
3 years ago
What occurs in order to break the bond in a cl2 molecule? explanation please.
alexandr1967 [171]

A cl2 molecule is a diatomic molecule composed of two atoms of identical halogen - chlorine. In this case, this molecule is composed of covalent bonds in which the identical atom- molecule tells that this is also non-polar. To break the bond, energy has to be absorbed to break the intermolecular force that bound the molecule together. 
6 0
3 years ago
Read 2 more answers
Thomas had a volume of 8.5 g of sodium chloride. What is the molar mass.?
alina1380 [7]

Answer:

58.44 g/mol

Explanation:

In this problem, make sure to remember that volume is measured in mL, L or any other units of volume. Remember that g represents grams, and grams is a measure of mass.

However, independent of what mass or what volume we take, molar mass is known to be an intensive property. That is, molar mass doesn't depend on any external conditions or any measurements.

Molar mass solely depends on the chemical structure of a compound and is a constant number at any given conditions.

In this problem, we are given sodium chloride, NaCl. In order to find its molar mass, we need to refer to the periodic table, find the atomic masses of Na and Cl and then add them up to have the molar mass of NaCl:

M_{NaCl} = M_{Na} + M_{Cl} = 22.99 g/mol + 35.453 g/mol = 58.44 g/mol

5 0
3 years ago
How do you find half life like i know the formula with 1/2 but how do you know how many times you do it
maksim [4K]
Can you give the specific problem?
3 0
3 years ago
Read 2 more answers
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