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Nina [5.8K]
3 years ago
15

Create a story and describe the motion (in meters) of an object that has traveled a distance but has no displacement. Need Help

ASAP
Chemistry
1 answer:
vekshin13 years ago
3 0

Answer:

One complete revolution around a circular path.

Explanation:

Let us take the case of a car moving in a circular track of radius r metres.

In one revolution, the car covers the length(distance) equal to the perimeter of the circle.

In this case, distance traveled = 2\pir metres

But after one complete revolution, the car reaches the same position as it was at the beginning of the motion.

Hence, the initial and final points coincide or the car hasn't changed it's position w.r.t the initial point.

So in this case, the displacement is zero.

Hence, revolution of a car around a circular path is an example of an object traveling a distance but having no displacement.

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Plz help. I just need someone to check answer. :(
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These substances can be separated by distillation, so your answer is A.
5 0
3 years ago
Formula of a substance in water is followed by what symbol
snow_lady [41]
The substance is followed by H2O
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3 years ago
A sample of gas in a 14.6 L flexible container is at 25.0oC and 1.00atm. What is the volume of the sample when heated to 220.0oC
inn [45]

Answer: 24.1 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=14.6L\\T_1=25.0^oC=(25+273)K=298K\\V_2=?\\T_2=220.0^0C=(220+273)K=493K

Putting values in above equation, we get:

\frac{14.6}{298K}=\frac{V_2}{493}\\\\V_2=24.1L

Thus the volume of the sample when heated to 220.0oC and the pressure is constant is 24.1 L

7 0
3 years ago
A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to
katen-ka-za [31]
I wrote the answer on this paper and here is the calculations step by step

very sry the "V" must be replaced with 285 and the 285 must be replaced with "V" and the answer is 231.09 cm3. sorry for the inconvenience.

6 0
3 years ago
Read 2 more answers
Using the following equation 5KNO2+2KMnO4+3H2SO4=5KNO3+2MnSO4+K2SO4+3H20. Starting with 2.47 grams of KNO2 and excess KMnO4 how
Aleonysh [2.5K]

Given equation : 5KNO_{2} + 2KMnO_{4} + 3H_{2}SO_{4}\rightarrow 5KNO_{3} + 2MnSO_{4} + K_{2}SO_{4} + 3H_{2}O

Given information = 2.47 grams KNO2 and excess KMnO4 and we need to find grams of water (H2O).

Since KMnO4 is in excess, so grams of water(H2O) can be calculated using grams of KNO2 with the help of stoichiometry.

To find grams of water(H2O) from grams of KNO2 , we need to follow three steps.

Step 1. Convert 2.47 grams of KNO2 to moles of KNO2.

Moles = \frac{grams}{molar mass}

Molar mass of KNO2 = 85.10 g/mol

Moles = 2.47 gram KNO2\times \frac{1 mol KNO2}{85.10 gram KNO2}

Moles = 0.0290 mol KNO2

Step 2. Convert moles of KNO2 to moles of H2O using mole ratio.

Mole ratio are the coefficient present in front of the compound in the balanced equation.

Mole ratio of KNO2 : H2O is 5 : 3 (5 coefficient of KNO2 and 3 coefficient of H2O)

0.0290 mol KNO2\times \frac{3 mol H2O}{5 mol KNO2}

Mole = 0.0174 mol H2O

Step 3. Convert mole of H2O to grams of H2O

Grams = Moles X molar mass

Molar mass of H2O = 18.00 g/mol

Grams = 0.0174 mol H2O\times \frac{18 g H2O}{1 mol H2O}

Grams of water = 0.313 grams H2O

Summary : The above three steps can also be done in a singe setup as shown below.

2.47 gram KNO2\times \frac{(1 mol KNO2)}{(85.10 gram KNO2)}\times \frac{(3 mol H2O)}{(5 mol KNO2)}\times \frac{(18.00 gram H2O)}{(1mol H2O)}

In the above setup similar units get cancelled out and we will get grams of H2O as 0.313 grams water (H2O)

8 0
3 years ago
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