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My name is Ann [436]
3 years ago
7

Danielle is designing a system to help race cars slow safely and rapidly at her local racetrack. The racetrack is a straight, 1,

000-meter stretch of asphalt. Half of the track is devoted to races and the other half is designed to let cars decelerate after crossing the finish line.
Danielle has designed a pair of small parachutes that are automatically released out the back of a car once it crosses the finish line. When her system works properly, the race car driver does not need to apply the brakes to bring the car to a stop before running out of track.

When Danielle tests a prototype of her design, it brings a race car to a stop 275 meters behind the finish line. After taking some measurements, she decides that her design causes the driver to experience too much force. To correct this, she could
A.
make the parachutes larger in order to increase the drag forces on the car and bring it to a stop 200 meters after crossing the finish line.
B.
add a third parachute in order to increase the drag forces on the car and bring it to a stop 175 meters after crossing the finish line.
C.
make the parachutes smaller in order to decrease the drag forces on the car and bring it to a stop 400 meters after crossing the finish line.
D.
remove one of the parachutes in order to decrease the drag forces on the car and bring it to a stop 550 meters after crossing the finish line.
Chemistry
1 answer:
xeze [42]3 years ago
3 0

Answer:

The answer is C

Explanation:

She should make the parachutes smaller in order to decrease the drag forces on the car and bring it to a stop 400 meters after crossing the finish line.

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C. A reaction that absorbs heat

Explanation:

I do not know Latin, but roughly speaking 'Endo' means inside, and 'thermic' means heat.

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P is the pressure in atmospheres (atm), V is the volume in liters (L), n is the number of moles, R is the gas constant (0.0821 L
yKpoI14uk [10]

Answer:

The mass of the neon gas  m = 1.214 kg

Explanation:

Pressure = 3 atm = 304 k pa

Volume = 0.57 L = 0.00057 m^{3}

Temperature = 75 °c = 348 K

Universal gas constant = 0.0821 \frac{L . atm}{mol K}

We have to change the unit of this constant. it may be written as

Universal gas constant = 8.314 \frac{KJ}{mol K}

Gas constant for neon = \frac{8.314}{20} = 0.41 \frac{KJ}{kg K}

From ideal gas equation,

P V = m R T ------- (1)

We have all the variables except m. so we have to solve this equation for mass (m).

⇒ 304 × 10^{3} × 0.00057 = m × 0.41 × 348

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⇒ m = 1.214 kg

This is the mass of the neon gas.

4 0
3 years ago
Which of the following describes the relationship correctly?
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1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.
solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2) The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = S=3.1\times 10^-2M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

                            S     2S

The solubility product of the lead(II) chloride = K_{sp}

K_{sp}=[Pb^{2+}][Cl^-]^2

K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}

The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion =1\timed 0.000010 M=0.000010 M

Solubility of aluminium hydroxide in aluminum nitrate solution = S

Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)

                            S     3S

The solubility product of the aluminium nitrate = K_{sp}=1.0\times 10^{-33}

K_{sp}=[Al^{3+}][OH^-]^3

1.0\times 10^{-33}=(0.000010+S)\times (3S)^3

S=1.6\times 10^{-10} M

The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3.

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

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = \frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol

Volume of the solution = 0.250 L

[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

[Cl^-]=[NaCl]=0.00024 M

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

n=0.12 M]\times 0.250 L=0.030 mol

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

Solubility of lead(II) chloride = K_{sp}=1.2\times 10^{-4}

Ionic product of the lead chloride in solution :

Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}

Q_i ( no precipitation)

The given statement is false.

3 0
4 years ago
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