<span>I’ve answered this
question before so if these are the choices to the question presented:
An oxygen atom double-bonded to a carbon atom, with a hydrogen atom
single-bonded to the same carbon atom. </span><span>
<span>A hydrogen atom covalently bonded to an oxygen atom, which is
covalently bonded to a carbon in the carbon chain. </span>
<span>A carbon atom single-bonded between two other carbon atoms,
with an oxygen atom double-bonded to the central carbon atom as well. </span>
<span>An oxygen atom single-bonded between two carbon atoms within
a carbon chain.
Then, the answer would be “a hydrogen atom covalently bonded to an oxygen atom,
which is covalently bonded to a carbon in the carbon chain.<span>”</span></span></span>
Answer:
Low level (altitude) clouds that are wispy and made of ice crystals.
precipitation and surface runoff
Hail
Nitrogen
Stratosphere
Decreases
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Answer: The volume of 0.640 grams of
gas at Standard Temperature and Pressure (STP) is 0.449 L.
Explanation:
Given: Mass of
gas = 0.640 g
Pressure = 1.0 atm
Temperature = 273 K
As number of moles is the mass of substance divided by its molar mass.
So, moles of
(molar mass = 32.0 g/mol) is as follows.

Now, ideal gas equation is used to calculate the volume as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.

Thus, we can conclude that the volume of 0.640 grams of
gas at Standard Temperature and Pressure (STP) is 0.449 L.
Answer : The mass of
needed are, 1.515 grams.
Explanation :
First we have to calculate the mole of
.

Now we have to calculate the moles of
.
The balanced chemical reaction will be,
produced from 1 mole of 
So, 0.005 mole of
produced from 0.005 mole of 
Now we have to calculate the mass of 


Therefore, the mass of
needed are, 1.515 grams.