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den301095 [7]
4 years ago
12

A 2 kg bowlling ball is 2.5 meters off the ground on a post when it falls just before it reachers the ground it is traveling m/s

assuming that there is no air resistance is true
A;The mechanical energy is conserved

B;The initial potential energy is greater than the final kinetic energy

CThe initial potential energy is less than the final kinetic energy

D;The mechanical energy is not conserved
Chemistry
1 answer:
Alchen [17]4 years ago
6 0
The mechanical energy is conserved 
Choice A
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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
Answer question 13. What is the mechanical advantage of a lever that can lift a 100 N load with a input force of 20 N
vazorg [7]
MA= output force/ input force
MA= 100N/20N
MA= 50
8 0
3 years ago
Gaseous butano (CH,(CH),CH) will react with gaseous oxygen (0) to produce gaseous carbon dioxide (CO2) and goseous woter (11,0).
slega [8]

Answer:

Firstly, We have to convert it in the Miles formula...

No. of moles = Mass given/Molar Mass

So, the final answer be come<em> </em>

<h3><em><u> </u></em><em><u>Ans</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>5</u></em><em><u>0</u></em><em><u>.</u></em><em><u>8</u></em><em><u> </u></em><em><u>gm</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>there</u></em><em><u> </u></em><em><u>same</u></em><em><u> </u></em><em><u>%</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>5</u></em><em><u>0</u></em><em><u>.</u></em><em><u>8</u></em><em><u>%</u></em><em><u> </u></em><em><u>butane</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>react</u></em><em><u>ion</u></em><em><u> </u></em></h3>
5 0
3 years ago
The concentration of potassium is higher in red blood cells than in the surrounding blood plasma. This higher concentration is m
lana [24]

Answer:

active transport

Explanation:

Active transport -

It refers to the movement of the molecules from a region of lower concentration towards a region of higher concentration via a membrane , is referred to as active transport.

The process of active transport need to have some cellular energy .

Hence,

The concentration of potassium is maintained in the red blood cells via the process of active transport.

3 0
3 years ago
What temperature will the water reach when 10.1 g CaO is dropped into a coffee cup containing 157 g H2O at 18.0°C if the followi
Zepler [3.9K]

Answer:

Final temperature attained by water = 34.6°C

Explanation:

The reaction of CaO and H₂O is an <em>exothermic reaction</em>. The equation of reaction is given below:

CaO + H₂O ----> Ca(OH)₂

The quantity of heat given off, ΔH°rxn = 64.8KJ/mol = 64800J/mol

Number of moles of CaO = mass/molar mass, where molar mass of Ca0 = 56g/mol, mass of CaO = 10.1g

Number of moles of CaO = 10.1g/56g/mol =0.179moles

Quantity of heat given off by 0.179 moles = 64800 *0.179 = 11599.2J/mol

Using the formula, <em>Quantity of heat, q = mass * specific heat capacity * temperature rise.</em>

mass of mixture = (10.1 + 157)g = 167.1g, Initial temperature = 18.0°C

Final temperature(T₂) - Initial temperature(T₁) = Temperature rise

11599.2J/mol = 167.1g * 4.18J/g·°C * ( T₂ - 18.0°C)

11599.2 = 698.478T₂ - 12572.604

11599.2 + 12572.604 = 698.478T₂

698.478T₂ = 24171.804

T₂ = 34.6°C

Therefore, final temperature attained by water = 34.6°C

6 0
3 years ago
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