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CaHeK987 [17]
4 years ago
5

Water is unique because no other substance in nature occurs in

Chemistry
1 answer:
Ivahew [28]4 years ago
6 0
Every biome and living substance
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Next to each formula, write the number of atoms of each element found in one unit of the compound
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To answer this question I would have to know the elements in the compound
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A ball has a 17 j of kinetic energy and its mechanical energy is 25 j. Calculate its potential energy
ycow [4]

The potential energy of the ball is 8 J.

<h3>What is mechanical energy?</h3>

Mechanical energy is given as the energy possessed by an object to do work. It is given as the sum of the kinetic energy and potential energy.

Mechanical energy = Kinetic energy + Potential energy

The given ball has:

  • Mechanical energy = 25 J
  • Kinetic energy = 17 J

The potential energy is given as;

25 J = 17 J + Potential energy

Potential energy = 25 J - 17 J

Potential energy =  8 J

The potential energy of the ball is 8 J.

Learn more about potential energy, here:

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2 years ago
How many liters of oxygen are required to completely react with 2.0 liters of CH4 at30 °C and 3.0 atm?CH4(g) + 2O2(g) → CO2(g) +
Dominik [7]

1) Write the chemical equation.

CH_4+2O_2\rightarrow CO_2+2H_2O

2) List the known and unknown quantities.

Sample: CH4.

Volume: 2.0 L.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Moles: <em>unknown</em>.

3) Moles of CH4.

<em>3.1- Set the equation.</em>

PV=nRT

<em>3.2- Plug in the known values and solve for n (moles).</em>

(3.0\text{ }atm)(2.0\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)n=\frac{(3.0\text{ }atm)(2.0\text{ }L)}{(0.082057\text{ }L*atm*K^{-1}*mol^{-1})}=n=0.24\text{ }mol\text{ }CH_4

4) Moles of oxygen that reacted.

The molar ratio between CH4 and O2 is 1 mol CH4: 2 mol O2.

mol\text{ }O_2=0.24\text{ }CH_4*\frac{2\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }CH_4}=0.48\text{ }mol\text{ }O_2

5) Volume of oxygen required.

Sample: O2.

Moles: 0.48 mol.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Volume: <em>unknown</em>.

<em>5.1- Set the equation.</em>

PV=nRT

<em>5.2- Plug in the known values and solve for V (liters).</em>

(3.0\text{ }atm)(V)=0.48\text{ }O_2*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)V=\frac{(0.48\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(303.15\text{ }K)}{3.0\text{ }atm}V=3.98\text{ }L

3.98 L of O2<em> is required to react with 2.0 L CH4.</em>

.

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It does not combine egg and sperm cells in plant.

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