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2 years ago

Activity: A Ferris wheel with a diameter of 60.0 m is moving at a speed of 2.09 m/s. What is the centripetal

Chemistry

1 answer:

GarryVolchara [31]2 years ago

5 0

Answer:

Centripetal acceleration of the wheel is $0.145\ m/s^2$ .

Explanation:

We have,

Diameter of a Ferris wheel is 60 m

Radius of the wheel is 30 m

Speed of the wheel is 2.09 m/s

It is required to find the centripetal acceleration of the wheel. The formula of centripetal acceleration is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(2.09)^2}{30}\\\\a=0.145\ m/s^2$

So, the centripetal acceleration of the wheel is $0.145\ m/s^2$ .

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energy= heat capacity *mass * temperature difference
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Vilka [71]

Answer:

Air can be described as: Mass and Mixture of Gases

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3 years ago

An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g

vodomira [7]

Answer:

A) 2.69 M

B) 0.059

Explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

molarity = mol solute/ 1 L solution

molarity= $\frac{33.8 g solute}{100 g solution}$ x $\frac{1.15 g solution}{1 ml}$ x $\frac{1 mol solute}{145.6 g solute}$ x $\frac{1000 ml}{1 L}$

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

mass of water= 100 g - 33.8 g = 66.2 g

Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x $\frac{1 mol solute}{145.6 g}$ = 0.232 mol

mol H20= 66.2 g H₂O x $\frac{1 mol H2O}{18 g}$

Finally, the mol fraction of solute (Xsolute) is calculated as follows:

Xsolute= $\frac{mol solute}{total mol}= \frac{mol solute}{mol solute + mol H2O}=\frac{0.232 mol}{0.232 mol + 3.677 mol}$

Xsolute= 0.059

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2 years ago

Question 10

bixtya [17]

The reaction that should be followed is
Na2SO4 + Ca(NO3)2 --> CaSO4 + 2NaNO3
first calculate the limiting reactant
mol Na2SO4 = 0.075 L (1.54×10−2 mol / L) = 1.155x10-3 mol

mol Ca(NO3)2 = 0.075 L (1.22×10−2 mol / L) = 9.15x10-4 mol
so the limiting reactant is the Ca(NO3)2

so all of the Ca2+ will be precipitated, percentage unprecipitated = 0.00 %

5 0

3 years ago