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IrinaK [193]
3 years ago
8

An acid is a substance that _____.?

Chemistry
1 answer:
evablogger [386]3 years ago
8 0
Thank you for posting your question here. To answer that question, an acid is a substance that contains hydrogen. It usually has a sour taste and also can able to neutralize alkali and reddening blue litmus paper. The hydrogen that it contains can be replaced by a metal.  
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4 years ago
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F a cold object is placed in a warm room, heat will move from the object into the room cold will move from the object into the r
Masja [62]
Cold is the absence of heat! so heat moves into the object from the environment (room) rapidly~ and it slows down by time and water is formed on the surface of the object because of cold and hot air.
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3 years ago
What does 2,3-dibromopentane look like?
Shtirlitz [24]

Answer:

The answer to your question is below

Explanation:

Process

1.- To draw an organic molecule, start from the end of the name (pentane)

Pentane has 5 carbons.

2.- Assign numbers the carbons, you can start from the left or to the right

3.- Attach the branches (bromine) in positions 2 and 3

6 0
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: <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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Explanation:

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3 years ago
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