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WITCHER [35]
2 years ago
14

1. What is the potential energy of a 8 kg ball that is at a height of 20 m?

Chemistry
1 answer:
weqwewe [10]2 years ago
4 0

Answer:

1. 1568 J

2. 0 J

3. 1176 J

Explanation:

PE = mgh

(PE = Potential Energy) = (m = mass)(g = gravitational force which is 9.8)(h = height)

1. (3)(9.8)(20) = 1568 J

2. PE = (3)(9.8)(0) = 0 J

3. (5)(9.8)(24) =  1176 J

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For an ideal gas condition, what is the mass (g) of N2 if the pressure is 2.0 atm, the volume is 25 mL and the temperature is 29
Alchen [17]

Answer:

THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g

Explanation:

In an ideal condition

PV = nRT or PV = MRT/ MM where:

M = mass = unknown

MM =molar mass = 28 g/mol

P = pressure = 2 atm

V = volume = 25 mL = 0.025 L

R = gas constant = 0.082 L atm/mol K

T = temperature = 290 K

n = number of moles

The gas in the question is nitrogen gas

Molar mass of nitrogen gas = 14 * 2 = 28 g/mol

Then equating the variables and solving for M, we have

M = PV MM/ RT

M = 2 * 0.025 * 28 / 0.082 * 290

M = 1.4 / 23.78

M = 0.0589 g

The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g

3 0
3 years ago
Is wine a homogenous mixure
Fiesta28 [93]

Tap water and rain water are both homogeneous, even though they may have different levels of dissolved minerals and gases. A bottle of alcohol is a man-made homogeneous mixture, from a fine Italian wine to a glass of Scotch whisky. In the human body, blood plasma is an example of a homogeneous mixture.

4 0
2 years ago
How much heat (in kj) is required to warm 13.0 g of ice, initially at -12.0 ∘c, to steam at 113.0 ∘c? the heat capacity of ice i
olga nikolaevna [1]
The total amount of heat required is the sum of all the sensible heat and latent heats involved in bringing the ice to a desired temperature and state. The latent heat of fusion and vaporization of water 333.55 J/g and 2260 J/g, respectively. Solving for the total amount of heat,
                   total amount of heat = 13.0 g (2.09 J/gC)(12) + 13(333.55 J/g) + 13.0 g (4.18 J/gC)(100 - 0) + (13.0 g)(2260 J/g) + (13 g)(2.01 J/g)(113-100) 
                                        = 39815.88 J 
                                        = 39.82 kJ
5 0
3 years ago
Read 2 more answers
If metal ions in a solution were reduced, what would you expect to see?
marysya [2.9K]

Answer:

Solid metal

Explanation:

The reduced form of metal ions is the metal in elemental state (simple substance). So, if you have a solution with metal ions and they are reduced, you probably will see the deposition of the metal. For example: if you have a solution with sodium ions (Na⁺), and the ions are then reduced, you will see the aparition of a solid phase of metallic sodium (Na(s)), according to the following half-reaction:

Na⁺ + e- → Na(s)

8 0
2 years ago
Write the formula for potassium oxide. why do you not need prefixes in the name ​
Molodets [167]

Potassium oxide: K₂O.

There's no need for prefixes since K₂O is an ionic compound.

<h3>Explanation</h3>

Find the two elements on a periodic table:

  • Potassium- K- on the left end of period four.
  • Oxygen- O- near the right end of periodic two.

Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.

  • Potassium is a metal,
  • Oxygen is a nonmetal.

A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:

  • Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.
  • Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion \text{O}^{2-} when it combines with metals.

The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each \text{O}^{2-} ion is twice that on a \text{K}^{+} ion. Each \text{K}^{+} would pair up with two \text{O}^{2-}. Hence the subscript in the formula: \text{K}_{\bf 2}\text{O}.

There are two classes of compounds:

  • Covalent compounds, which need prefixes, and
  • Ionic compounds, which need no prefix.

Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide \text{CO}_2, the prefix di- indicates that there are two oxygen atoms in the formula \text{CO}_2. However, there's no need for prefix in ionic compounds such as \text{K}_2\text{O}.

7 0
3 years ago
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