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trasher [3.6K]
2 years ago
5

Calculate the average speed between 6 hours and 9 hours

Chemistry
2 answers:
Dominik [7]2 years ago
8 0

Answer:

60 Miles per Hour

Explanation:

You included no scenario so I will develop a hypothetical that you can reference.

Assuming a car traveled a distance of 450 miles from one place to another in 9 hours with the car travelling 180 of those miles between 6 hours and 9 hours, what was the average speed between those 6 and 9 hours.

The car travelled 180 miles in 6hours - 9 hours.  First find out how many hours this is:

= 9 - 6

= 3 hours

Speed = Distance / Time

= 180 / 3 hours

= 60 miles per hour

lyudmila [28]2 years ago
5 0

Answer:

7.5 hours

Explanation:

avarage speed = min speed + max speed ÷ 2

= 6 + 9 ÷ 2

= 15 ÷ 2

= 7.5

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The answer is 6 because the number of chlorine is 2 so if you have 6 moles of chlorine the answer is 6
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A chemist must prepare of 800.0 ml potassium hydroxide solution with a pH of 13.00 at 25°.
ArbitrLikvidat [17]

Answer:

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

Explanation:

The pH of the solution = 13.00

pH + pOH = 14

pOH = 14 - pH = 14 - 13.00 = 1.00

pOH=-\log[OH^-]

1.00=-\log[OH^-]

[OH^-]=10^{-1.00} M=0.100 M

KOH(aq)\rightarrow K^+(aq)+OH^-(aq)

[KOH]=[OH^-]=[K^+]=0.100 M

Molariy of the KOH = 0.100 M

Volume of the KOH solution = 800 mL= 0.800 L

1 mL = 0.001 L

Moles of KOH = n

Molarity=\frac{Moles}{Volume(L)}

0.100 M=\frac{n}{0.800 L}

n = 0.0800 mol

Mass of 0.0800 moles of KOH :

0.0800 mol × 56 g/mol = 4.48 g

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

4 0
3 years ago
onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0
3 years ago
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