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Ratling [72]
3 years ago
9

Jules Verne wrote the book Twenty Thousand Leagues Under the Sea. If one league - 5.556 km and one furlong - 660.0 feet, how man

y furlongs did the Nautilus travel? (assume the twenty thousand leagues is good for 3 sig. fig.) ball park-550,000 furlong T
Chemistry
1 answer:
Vlad [161]3 years ago
7 0

Answer:

The Nautilus travel 5.52\times 10^5 furlongs.

Explanation:

Given: 1 league = 5.556 km

1 furlong = 660.0 feet

To find: 20,000 leagues = ? furlongs

Solution:

20,000 leagues = 20,000\times 5.556 km =111,120 km

1 km = 3280.84 feet

111,120 km= 111,120\times 3280.84 feet=364,566,940.8 feet

If ,1 furlong = 660.0 feet.

Then, 1 foot = \frac{1}{660.0} furlong

364,566,940.8 feet=364,566,940.8 \times \frac{1}{660.0}

=552,374.1524 furlongs\approx 5.52\times 10^5 furlongs

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If you travel through a city and find that you travel 5 km in 30 minutes, you could say that your constant speed is 6 km/h. is i
Reika [66]

Answer:

False.

Explanation:

(5 kilometers / 30 minutes), you could then say that in 60 minutes (one hour) you would be traveling at (10 kilometers / 60 minutes)...

Therefore, it is unlikely that you would be traveling at (6 kilometers / hour), so the answer should be... False.

7 0
3 years ago
You are presented with a mystery as part of your practical experiment. You have a solution of Pb(NO3)2
shusha [124]
Ho123 right I’m sorry of I’m wrong
7 0
2 years ago
Following the instructions in your lab manual, you have titrated a 25.00 mL sample of 0.0100 M KIO3 with a solution of Na2S2O3 o
vivado [14]

<u>Answer:</u>

<u>For 1:</u> The amount of potassium iodate that were titrated is 2.5\times 10^{-4} moles

<u>For 2:</u> The amount of sodium thiosulfate required is 1.25\times 10^{-4} moles

<u>Explanation:</u>

  • <u>For 1:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}

Molarity of KIO_3 solution = 0.0100 M

Volume of solution = 25 mL

Putting values in above equation, we get:

0.0100M=\frac{\text{Moles of }KIO_3\times 1000}{25}\\\\\text{Moles of }KIO_3=\frac{0.0100\times 25}{1000}=0.00025mol

Hence, the amount of potassium iodate that were titrated is 2.5\times 10^{-4} moles

  • <u>For 2:</u>

The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:

2KIO_3+Na_2S_2O_3\rightarrow K_2S_2O_3+2NaIO_3

By Stoichiometry of the reaction:

2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate

So, 0.00025 moles of potassium iodate will react with = \frac{1}{2}\times 0.00025=0.000125mol of sodium thiosulfate

Hence, the amount of sodium thiosulfate required is 1.25\times 10^{-4} moles

6 0
3 years ago
When Brian first sorted out his thoughts, he figured this was the first day after the crash and what did he think would happen n
Leya [2.2K]

It should be noted that when Brian first sorted out his thoughts, the first thing that he thought would happen next is that he would probably be rescued that day.

It should be noted that in Chapter 5 of Hatchet, Brian wakes up in the forest and then, he realized that he was desperately thirsty.

When Brian first sorted out his thoughts, he figured this was the first day after the crash and that he would probably be rescued that day. He believed that he was going to wither up and die if he doesn't get anything that he'll drink.

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brainly.com/question/21400963

7 0
2 years ago
At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank
hodyreva [135]

Answer:

\large \boxed{\text{-92 $^{\circ}$C}}

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data  

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}

2. Temperature in degrees Celsius

\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}

8 0
3 years ago
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