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NeX [460]
2 years ago
15

Does riding a bike with larger tires help you go faster even if you pedal at the same rate A:if the person moves faster, then th

e person is riding a bike with larger tires B: if a person rides a bike with larger tires ,then the person with move faster than on a bike with smaller tires
Chemistry
1 answer:
Alekssandra [29.7K]2 years ago
8 0

Answer:

Yes, bikes with larger tires help you go faster.

B: if a person rides a bike with larger tires ,then the person with move faster than on a bike with smaller tires

Explanation:

Bikes with larger wheels cover more distances in one revolution, when compared to bikes with a much smaller wheel, if the wheels are turning at the same revolution per minutes. Although more effort is put into turning a big wheel, when compared to that for a smaller wheel, but it will go faster if you can keep it spinning at the same rate as a small one.

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Part i. you considered the equilibrium: fe3+(aq) + scn-(aq) fe(scn)2+(aq), and compared the color of the solutions in four test
Alexus [3.1K]

Answer:

The equilibrium will be shifted to lift with the formation of a brown gelatinous precipitate of Fe(OH)₃.

Explanation:

  • Le Chatelier's principle states that <em>"when any system at equilibrium for is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to counteract the effect of the applied change and a new equilibrium is established that is different from the old equilibrium"</em>.
  • The addition of NaOH will result in the formation of Fe(OH)₃ precipitate which has a brown gelatinous precipitate.
  • The formation of this precipitate cause removal and decrease of Fe³⁺ ions.
  • According to Le Chatelier's principle, the system will be shifted to lift to increase Fe³⁺ concentration and reduce the stress of Fe³⁺ removal and readjust the equilibrium again. So, the [Fe(SCN)²⁺] decreases.
  • Increasing [Fe³⁺] will produce a yellow color solution that contains a brown gelatinous precipitate of Fe(OH)₃.
3 0
2 years ago
Complete combustion of 7.40 g of a hydrocarbon produced 22.4 g of CO2 and 11.5 g of H2O. What is the empirical formula for the h
cluponka [151]
<span>C2H5 First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2. Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass. moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule. Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen. moles C = 0.50899 moles H = 0.638361 * 2 = 1.276722 We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon. total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185 7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked. Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen. 0.50899 / 1.276722 = 0.398669 0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5. Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is C2H5</span>
3 0
3 years ago
A technician needs 500.0mL of a 0.500 M MgCl2 solution for some lab procedures. The stock bottle is labeled 4.00 M MgCl2. What v
mihalych1998 [28]

Answer:

Explanation:

To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;

    Number of moles  = molarity x volume

Parameters given;

        volume of solution = 500mL  = 0.5L

          molarity of solution = 0.5M

   Number of moles  =  0.5 x 0.5  = 0.25moles

Now to know the volume stock to take;

           Volume of stock = \frac{number of moles }{molarity}

  molarity of stock  = 4M

                 volume  = \frac{0.25}{4}   = 0.0625L or 62.5mL

4 0
3 years ago
What is the role of meteorites in the formation of life?
CaHeK987 [17]

Answer:

Meteorites are consensually considered to be involved in the origin of life on this Planet for several functions and at different levels: (i) as providers of impact energy during their passage through the atmosphere; (ii) as agents of geodynamics, intended both as starters of the Earth’s tectonics and as activators of local hydrothermal systems upon their fall; (iii) as sources of organic materials, at varying levels of limited complexity; and (iv) as catalysts.

Explanation:

4 0
2 years ago
What is the molarity of a 3.0 solution of magnesium chloride prepared with 9.0 molecules of MgCl2 A. 6.0 M B. 3.0 M C. 27.0 M D.
Gala2k [10]

Answer:

Option B. 3.0 M

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 3.0 L

Mole of MgCl₂ = 9 moles

Molarity =?

Molarity can simply be defined as the mole of solute per unit litre of the solution. Mathematically, it can be expressed as:

Molarity = mole of solute /Volume of solution

With the above formula, we can obtain the molarity of the solution as follow:

Volume of solution = 3.0 L

Mole of MgCl₂ = 9 moles

Molarity =?

Molarity = mole of solute /Volume of solution

Molarity = 9 / 3

Molarity = 3 mol/L = 3.0 M

Thus, the molarity of the solution is 3 M

8 0
2 years ago
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