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Valentin [98]
3 years ago
7

A spaceship moving with an initial velocity of 58.0 meters/second experiences a uniform acceleration and attains a final velocit

y of 153 meters/second. What distance has the spaceship covered after 12.0 seconds?
A,) 6.96 × 102 meters
B.) 1.27 × 103 meters
C.) 5.70 × 102 meters
D.) 1.26 × 102 meters
E.) 6.28 × 102 meters
Chemistry
1 answer:
Leni [432]3 years ago
5 0
Data:

1) <span>Vi = 58.0 m/s

2) Vf = 153 m/s

3) t = 12.0 s

4) D = ?

Formulas:

Uniform acceleration =>

a = (Vf - Vi) / t

d = Vi + a* t^2 / 2

Solution

a = ( 153 m/s - 58.0 m/s) / (12.0 s) = 7.92 m/s^2

d = 58.0 m/s + 7.92 m/s^2 * (12.0s)^2 / 2 = 628 m

628 m = 6.28 * 10^2 m

Answer: option E) 6.28 * 10^2 meters
</span>
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Which model of the atom was used as a result of JJ Thomson's cathode ray
mariarad [96]

Answer:

a. Plum pudding model

Explanation:

The plum pudding model of the atom was proposed by J.J. Thomson. It was the model he derived from his experiment on the gas discharge tube.

J.J Thomson was the first person to discover electrons which he called cathode rays because in the discharge tube, they emanate from the cathode.

  • This led him to suggest the plum pudding model of the atom.
  • The model reflects electrons being surrounded by a volume of negative charges.
6 0
3 years ago
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
Is this correct?<br> Plz answer ASAP!
Angelina_Jolie [31]

YES IT'S @LL CORRECT

3 0
3 years ago
Read 2 more answers
What is the volume occupied by 4.20 miles of oxygen gas (O2) at STP
Drupady [299]

Answer: 94.13 L

Explanation: In STP in an ideal gas there is a standard value for both temperature and pressure. At STP,pressure is equal to 1atm and the temperature at 0°C is equal to 273.15K. This problem is an ideal gas so we use PV=nRT where R is a constant R= 0.08205 L.atm/mol.K.

To find volume, derive the equation, it becomes V=nRT/P. Substitute the values. V= 4.20 mol( 0.08205L.atm/mol.K)(273.15K) / 1 atm = 94.13 L. The mole units, atm and K will be cancelled out and L will be the remaining unit which is for volume.

8 0
3 years ago
Calculate the [oh] for 0.46 Hcl
Bad White [126]

Answer:

2.17 e -14

Explanation:

A strong acid like HCl ionize 100 % in water so [H3O+] = 0.46 M

[OH-] = Kw /  [H3O+]

        =  1.0 e -14 / 0.46

         = 2.17 e -14

3 0
3 years ago
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