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3 years ago

Some chemical bonds are considered molecules while others are considered compounds.

1 answer:

5 0

Molecule is the general term used to describe any atoms that are connected by chemical bonds. Every combination of atoms is a molecule. A compound is a molecule made of atoms from different elements. All compounds are molecules, but not all molecules are compounds.

One example of a molecule is $Cl_2$ (chlorine).

Examples of a compound is NaCl (sodium chloride) or $H_2O\\$ (water)

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When might a theory or scientific claim be modified or changed?

Ludmilka [50]

Whenever new evidence is discovered we must make changes in order to adapt theories to the current base of knowledge. If we were to simply never change any scientific theory, then they would likely no longer stand as factual credible pieces of information in the future.

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3 years ago

What is the ratio of hydrogen atoms (h) to oxygen atoms (o) in 2 l of water? enter the simplest whole number ratio in order of h

KiRa [710]

Answer:- 2,1

Explanations:- Water is $H_2O$ and from it's formula it is clear that the ratio of H to O moles or atoms is 2:1 means two hydrogen atoms for each oxygen atom. No matter what amount of water we have, the ratio of H atoms to O atoms is always same.

Hence, the answer is 2,1.

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3 years ago

Read 2 more answers

If a gas occupies 4600 mL at 0.9 atm and 195°C, what is the new volume in ml

Bumek [7]

Answer:

The new volume is 2415 mL

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases.

Boyle's law says that the volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure and is expressed mathematically as:

P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

$\frac{V}{T} =k$

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. This can be expressed mathematically in the following way:

$\frac{P}{T} =k$

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

$\frac{P*V}{T} =k$

Having two different states, an initial state and an final state, it is true:

$\frac{P1*V1}{T1} =\frac{P2*V2}{T2}$

In this case:

P1= 0.9 atm
V1=4,600 mL= 4.6 L (being 1 L=1,000 mL)
T1= 195 °C= 468 °K (being 0°C=273°K)

The final state 2 is in STP conditions:

P2= 1 atm
V2= ?
T2= 0°C= 273 °K

Replacing:

$\frac{0.9 atm*4.6L}{468K} =\frac{1 atm*V2}{273K}$

Solving:

$V2=\frac{0.9 atm*4.6L}{468K}*\frac{273K}{1 atm}$

V2= 2.415 L =2,415 mL

The new volume is 2415 mL

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3 years ago

What do acids do in solution?

KiRa [710]

B
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3 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: unknown.

3) Moles of CH4.

3.1- Set the equation.

$PV=nRT$

3.2- Plug in the known values and solve for n (moles).

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

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: unknown.

5.1- Set the equation.

$PV=nRT$

5.2- Plug in the known values and solve for V (liters).

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

3.98 L of O2 is required to react with 2.0 L CH4.

6 0

2 years ago