All categories

2 years ago

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

You might be interested in

What is the expensive property of a substance

olya-2409 [2.1K]

is there any choices?

6 0

3 years ago

PLease help me with this question

lukranit [14]

The answer is 6 because the number of chlorine is 2 so if you have 6 moles of chlorine the answer is 6

7 0

3 years ago

Read 2 more answers

what is it called when a scientists personal opinion affects the way experimental results are reported?

Monica [59]

I believe the word that you are looking for is "subjective"

-to be influenced by personal opinion or feeling-

PLEASE RATE AS THE BRAINLIEST ANSWER! THANK YOU! :)

4 0

3 years ago

Read 2 more answers

A chemist must prepare of 800.0 ml potassium hydroxide solution with a pH of 13.00 at 25Â°.

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