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3 years ago

9

Pat paid $85.50 for renting a scooter for 6 hours. What was the rate per hour for renting the scooter? (Input only numeric value

s and decimal point, and report prices to two decimal places, such as 12.30.) $______ per hour HELP ASAP PLEASE

1 answer:

wlad13 [49]3 years ago

3 0

I believe the answer would be 14.25 an hour because $85.50/ 6 would be 14.25 but i could be wrong.

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To demonstrate the formation of iron (iii) chloride from iron fillings

Pavel [41]

Iron (iii) chloride is obtained by vapor condensation from the reaction between chlorine gas and iron fillings.

<h3>How can iron (iii) chloride be formed from iron fillings?</h3>

Iron (ii) chloride can be formed from iron fillings in the laboratory as follows:

Iron fillings + Cl₂ → FeCl₃

Chlorine gas is introduced into a reaction vessel containing iron fillings and the iron (iii) chloride vapor formed is obtained by condensation.

In conclusion, iron (iii) chloride is formed by the the direct combination of iron fillings and chlorine gas.

Learn more about iron (iii) chloride at: brainly.com/question/14653649

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2 years ago

"What volume of" a 0.300 M BaF2 "solution is needed to prepare" 240.0 mL of a 0.100 M F- solution

zhannawk [14.2K]

Use M x V = M' x V'

0.300 x V = 0.100 x 250

V = .......... ml

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4 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

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3 years ago

You are given a solution that is 518 mM lactose. You need to make up 4.5 L of 16.7 mM solution. What volume do you need to trans

OLga [1]

Answer:

The volume you need to transfer from the stock solution is 0.145 l

Explanation:

Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:

number of moles in volume to transfer = number of moles in the final solution

Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:

Ci * Vi = Cf * Vf

where:

Ci = concentration of the stock solution.

Vi = volume of the stock solution to be transferred.

Cf = concentration of the final solution

Vf = volume of the final solution

Then, replacing with the data:

518 mM * Vi = 16.7 mM * 4.5 l

Vi = 16.7 mM * 4.5 l / 518 mM

<u>Vi = 0.145 l or 145 ml</u>

Notice that any concentration unit can be used, as long as the units of the concentration of the stock and final solution are the same.

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3 years ago

How much energy moves onto the next level, in an energy pyramid

dybincka [34]

10% is normally how it works .
For example like 1,000 is the top the next level would be 100 then the next would be 10

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3 years ago