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Ray Of Light [21]
1 year ago
9

To burn 1 molecule of C5H12 to form CO2 and H2O (complete combustion), how many molecules of O2 are required?

Chemistry
2 answers:
viva [34]1 year ago
5 0
C5H12 + 8 O2 → 5 CO2 + 6 H2O 

8 molecules of O2 are required.
matrenka [14]1 year ago
5 0

Answer:

8 molecules of oxygen gas are required.

Explanation:

Combustion reaction is the type of chemical reaction in which a hydrocarbon reacts with oxygen gas to give to carbon dioxide and water.

C_5H_{12}+8O_2\rightarrow 5CO_2+6H_2O

According to reaction ,1 mole of pentane reacts with 8 molecules of oxygen gas to give 5 moles of carbon dioxide and 6 moles of water.

So, to burn 1 mole of pentane we will need 8 molecules of oxygen gas.

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How many atoms of oxygen are in 75.6g of silver nitrate (AgNO3)?​
vodomira [7]

Answer:

1.50x10^21 molecules O2

Explanation:

8 0
1 year ago
Acetic acid is a weak acid with a pKa of 4.76. What is the concentration of acetate in a buffer solution of 0.2M at pH 4.9. Give
Ghella [55]

Answer:

[base]=0.28M

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:

pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\

[base]=1.38[acid]=1.38*0.20M=0.28M

Regards.

6 0
1 year ago
The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =
Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>

<em />

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>

Mass of disodium succinate:

0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>

<em></em>

5 0
2 years ago
Alexandra decides to climb Mt. Krumpett, which is 5000 m high. She determines that this will require a total of 1350 kcal of ene
Kisachek [45]
Alexandra requires a total energy of 1350 kcal for the climb
by eating proteins, fats and carbohydrates the amount of calories per gram contributed varies.
Proteins and carbohydrates - 4 calories per gram 
fats - 9 calories and gram
This means that by eating the same mass of fats and proteins/ carbohydrats the calories gained from fats is higher.
each bar contains;
<span>50 g of carbohydrates - 4 calories/g x 50 g = 200 calories
10 g of fat - 9 calories/g x 10 g = 90 calories 
40 g of protein - 4 calories/g x 40 g = 160 calories 
total amount of calories from 1 bar = 200 + 90 + 160 = 450 calories 
energy required = 1 350 000 calories 
bars required = 1 350 000/450 = 3000
alexandra should consume 3000 bars </span>
8 0
2 years ago
A chemist has 2.0 mol of methanol (CH3OH). The molar mass of methanol is 32.0 g/mol. What is the mass, in grams, of the sample?
rodikova [14]

Answer:

\boxed {\boxed {\sf D. \ 64 \ grams }}

Explanation:

Given the moles, we are asked to find the mass of a sample.

We know that the molar mass of methanol is 32.0 grams per mole. We can use this number as a fraction or ratio.

\frac{32 \ g \ CH_3OH}{1 \ mol \ CH_3OH}

Multiply by the given number of moles, which is 2.0

2.0 \ mol \ CH_3OH *\frac{32 \ g \ CH_3OH}{1 \ mol \ CH_3OH}

The moles of methanol will cancel each other out.

2.0 \ *\frac{32 \ g \ CH_3OH}{1 }

The denominator of 1 can be ignored.

2.0 * 32 \ g\ CH_3OH

Multiply.

64 \ g \ CH_3OH

There are 64 grams of methanol in the sample.

3 0
1 year ago
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