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m_a_m_a [10]
2 years ago
5

The image shows a wheel that's wound up and released. The wheel moves up and down as shown. Identify the position of the wheel w

hen its
potential energy is greatest.
Chemistry
1 answer:
Lesechka [4]2 years ago
7 0

Explanation:

Since the wheel moves up and down, the position that represents the potential energy is that which has the maximum height from the ground.

Potential energy is the energy at rest of a body.

It is given as:

      Potential energy = m x g x h

m is the mass of the body

g is the acceleration due to gravity

h is the height of the body

We can see that mass and height are directly related to the potential energy a body exerts.

The higher the wheel from ground, the higher its potential energy.

learn more

Potential energy brainly.com/question/10770261

#learnwithBrainly

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How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation? 2HBr(aq)+
Mkey [24]

Full Question:

A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?

How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?

2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)

Answer:

13.1 g K2CO3 required to neutralize spill

Explanation:

2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)

Number of moles = Volume * Molar Concentration

moles HBr= 0.42L x .45 M= 0.189 moles HBr

From the stoichiometry of the reaction;

1 mole of K2CO3 reacts  with 2 moles of HBr

1 mole = 2 mole

x mole = 0.189

x = 0.189 / 2 = 0.0945 moles

Mass = Number of moles * Molar mass

Mass = 0.0945 * 138.21  = 13.1 g

3 0
2 years ago
A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27°C and 7.50 * 103 Pa (absolute pressure). The tank has a tight‐f
Nuetrik [128]

Answer: The answer is 68142.4 Pa

Explanation:

Given that the initial properties of the cylindrical tank are :

Volume V1= 0.750m3

Temperature T1= 27C

Pressure P1 =7.5*10^3 Pa= 7500Pa

Final properties of the tank after decrease in volume and increase in temperature :

Volume V2 =0.480m3

Temperature T2 = 157C

Pressure P2 =?

Applying the gas law equation (Charles and Boyle's laws combined)

P1V1/T1 = P2V2/T2

(7500 * 0.750)/27 =( P2 * 0.480)/157

P2 =(7500 * 0.750* 157) / (0.480 *27)

P2 = 883125/12.96

P2 = 68142.4Pa

Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa

8 0
2 years ago
The highest principal enegy level of period 2 elements is 2. Period 3 elements all have six 3p electrons. Period 4 elements have
Dmitrij [34]

Explanation:

The highest principal energy level of period 2 elements is 2. (True)

The highest principle energy level of elements of any period is equal to period number. So the given statement is true.

Period 3 elements all have six 3p electrons – (False)  

For example, sodium and sulfur are period 3 elements but do not contain six 3p electrons. Only Argon of period possesses six p-electrons.

Period 4 elements have an inner electron configuration of [Ar]- (True)

Atomic number of elements belonging to period 4 have atomic number more than 18 (atomic number of Ar). So they must have an inner electron configuration of [Ar]

The valence electrons of group 5A elements are in the 6s subshell – [FALSE]

Elements of group 5A are nitrogen, arsenic, antimony and bismuth. They are p-block elements. Therefore, valence electrons are present in p-subshell.

Group 8A elements have full outer principal s and p subshells –(True)  

Group 8A elements are also known as noble gas element. Octets of noble gases are complete and hence, have full outer principal s and p subshells.

The valence electrons of group 2A elements are in an s subshell – (True)

Elements of group 2A are barium, magnesium, calcium, etc. They belong to s-block elements. Therefore, their valence electrons will be in s-subshell.

The highest principal energy level of period 3 elements is 4 – (False)

The highest principle energy level of elements of any period is equal to period number.

For example, sodium is present in 3rd period, so its principle quantum number will be 3.

Period 5 elements have an inner electron configuration of [Xe] – (False)

Period 5 elements have an inner electron configuration of [Kr]. The first member of this period is rubidium. 37 and that of Kr is 36.  

6 0
2 years ago
A sample of gasoline contains various hydrocarbons, which comprise atoms of carbon, hydrogen, and oxygen. The hydrocarbons mix t
Naddika [18.5K]

Answer:

  • mixture
  • homogenous mixture  (of hydrocarbons)
  • compound

Explanation:

Mixture can be easily separated by physical methods. Homogeneity and heterogeneity of a mixture is determined by whether the components there in are in a single phase and evenly distributed or not.

A solution has a solute evenly dissolved in  solvent to form a liquid substance.

An element is the basic form of  substance which cannot be broke down into any other simpler unit.  

I hope this was helpful.

4 0
2 years ago
What is the reaction energy Q of this reaction? Use c2=931.5MeV/u. Express your answer in millions of electron volts to three si
emmasim [6.3K]

Answer:

Energy= 2.7758 × 10^-11 J ;

71.112×10^-6 kJ.

Mass defect in Kilogram= 3.0885×10^-28 kg.

That is; 3.1×10^-28 kg(to two significant figure).

Explanation:

(Note: Check equation of reaction in the attached file/picture).

STEP ONE: we have to calculate the Mass defect.

Mass defect= Mass of reactants -- Mass of products.

Mass of the products: (140.9144+91.9262+3.060) u.

= 235.8666 u.

Mass of reactants: (1.0087+235.0439) u= 236.0526 u.

Therefore, the Mass defect= (236.0526 -- 235.8666) u

= 0.1860 u.

STEP TWO: Converting the Mass defect to energy;

0.18860 × 1.6605 × 10^27 kg

= 3.0885× 10^-28 kg

STEP THREE: Calculating energy released . Recall(from the question) c^2= 931.5 Mev/u. This is also equals to 9×10^16 m/s.

E=Mc^2.

Where E= energy released, c= speed of light, M= Mass.

Slotting in the values;

E= 3.0885×10^-28 kg × 9×10^16 m/s.

E=2.7758 × 10^-11 J.

Know that;( 1g of uranium × 1 mol of uranium ÷ 235.0439 g of uranium) × (6.002×10^23 atom of uranium/ 1 mol of uranium) × 2.7758× 10^-11.

=7.1112×10^-10 J

= 71.112×10^6 kJ.

3 0
2 years ago
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