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Aneli [31]
2 years ago
7

Scientists tracking a flock of ducks found that, on average, the ducks flew 740 km south in 12 hours. What were the speed and ve

locity of the flock of ducks in meters per second? Show your work.
Chemistry
1 answer:
Verizon [17]2 years ago
5 0

1) 17.1 m/s

2) 17.1 m/s south

Explanation:

1)

The speed of an object is a scalar quantity indicating "how fast" is the object moving, regardless of its direction of motion.

Being a scalar quantity, it has only a magnitude, while it doesn't have a direction.

The speed of an object is given by the formula:

v=\frac{d}{t}

where:

d is the distance covered by the object, which is the total length of the path covered by the object

t is the time taken to cover the distance d

In this problem we have:

d=740 km =740,000 m is the distance covered by the ducks

t=12 h  \cdot 60 \cdot 60 =43,200 s is the time taken

Substituting, we find:

v=\frac{740,000}{43,200}=17.1 m/s

2)

The velocity of an object is a vector quantity indicating how fast the object is moving in a certain direction.

Being a vector, velocity has both a magnitude and a direction.

The magnitude of the velocity is given by:

v=\frac{d}{t}

where

d is the displacement, which is the shortest distance in a straight line between the initial position and the final position of motion

t is the time taken for the displacement d to occur

And the direction is equal to the direction of the displacement.

In this problem:

d=740 km = 740,000 m south is the displacement

t=12 h  \cdot 60 \cdot 60 =43,200 s is the time taken

Substituting, we find:

v=\frac{740,000}{43,200}=17.1 m/s south

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When heating a flammable or volatile solvent for a recrystallization, which of these statements are correct? More than one answe
otez555 [7]

Explanation:

A volatile substance is defined as the substance which can easily evaporate into the atmosphere due to weak intermolecular forces present within its molecules.

Whereas a flammable substance is defined as a substance which is able to catch fire easily when it comes in contact with flame.

Hence, when we heat a flammable or volatile solvent for a recrystallization then it should be kept in mind that should heat the solvent in a stoppered flask to keep vapor away from any open flames so that it won't catch fire.

And, you should ensure that no one else is using an open flame near your experiment.

Thus, we can conclude that following statements are correct:

  • You should heat the solvent in a stoppered flask to keep vapor away from any open flames.
  • You should ensure that no one else is using an open flame near your experiment.
3 0
2 years ago
Rutherford, geiger, and marsden’s experiment demonstrated that the volume of the nucleus is roughly what fraction of the volume
trasher [3.6K]
Rutherford, Geiger and Marsden's experiment proved that every atom has a nucleus and that this nucleus is of positive charge and contains the most of the mass of the atom. 0.005% of the volume occupied by the electrons is the volume of the nucleus.
4 0
2 years ago
A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro
Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on  combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q

Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J

Heat released on  combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows  :

Heat released on  combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q'

Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J

Heat released on  combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0
2 years ago
Complete combustion of a compound containing hydrogen and carbon produced 2.641 g of carbon dioxide and 1.442 grams of water as
klio [65]

  The empirical   formula  of hydrocarbon is C₃H₈

The  molecular formula of hydrocarbon   is C₆H₁₆


<u><em> Empirical  formula  calculation</em></u>

Hydrocarbon  contain  carbon and hydrogen

  Step 1:  find the  mass  carbon (C) in carbon dioxide (CO₂)  and hydrogen (H )  in water

mass of  of element = molar mass  of element/ molar mass molecule x total mass of    molecule

From periodic table the molar mass  of C =12,    for CO₂ = 12+( 16 x2) =44 g/mol,     for H = 1.00 g/mol,    for H₂O = (2 x1)+16 = 18 g/mol

mass of C = 12/44 x 2.641 =0.7203 g

since there are 2 atom  of H in H₂O the molar mass of H = 1 x2 = 2 g/mol

mass of H  is therefore =  2/18 x 1.442 =0.1602 g


Step 2:  find the moles of C and H

moles = mass÷ molar mass

moles of C = 0.7203 g÷ 12 g/mol = 0.060  moles

moles of H  =  0.1602÷ 1 g/mol = 0.1602 moles


Step 3: find the mole ratio  of C and H by dividing  each  mole by smallest mole ( 0.06)

for C = 0.06/0.06 =1

  For H = 0.1602/0.06 =2.67

multiply   by 3  to remove the decimal

For C = 1 x3 =3

For H = 2.67 x3 =8

therefore the empirical formula = C₃H₈


<u><em>The molecular formula calculation</em></u>

[C₃H₈]n  = 88.1 g/mol

[12 x 3)+( 1 x8)]n =88.1 g/mol

44 n = 88.1

divide both side by 44

n=2

therefore [C₃H₈]₂   = C₆H₁₆



7 0
2 years ago
Read 2 more answers
Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine pr
denis-greek [22]

Answer:

194 g/mol.

Explanation:

Hello,

In this case, one first must compute the mass of each element as shown below:

C=18.13gCO_2*\frac{12gC}{44gCO_2} =4.945gC\\H=4.639gH_2O*\frac{2.016gH}{18.0152gH_2O}=0.519gH\\N=2.885gN_2\\O=10.0g-4.945g-0.519g-2.885g=1.651gO

Next, the corresponding moles:

C=4.945gC*\frac{1molC}{12gC}=0.412mol\\H=0.519gH*\frac{1molH}{1gH}=0.519mol\\N=2.885gN*\frac{1molN}{14gN}=0.206molN\\O=1.648gO*\frac{1molO}{16gO} =0.103molO

Then, each element's subscripts is found to be:

C=\frac{0.412}{0.103}=4\\H=\frac{0.519}{0.103}=5\\N=\frac{0.206}{0.103} =2\\O=\frac{0.103}{0.103}=1

Therefore, the empirical formula is:

C_4H_5N_2O

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

C_8H_10N_4O_2

Which has a molar mass of 194 g/mol being correctly contained in the given interval.

Best regards.

8 0
2 years ago
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