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myrzilka [38]
3 years ago
14

. If one mole each of CH4, NH3, H2S, and CO2 is added to 1 liter of water in a flask (1 liter of water = 55.5 moles of H2O), how

many moles of hydrogen, carbon, oxygen, nitrogen, and sulfur are in the flask? Round your answers to the nearest whole number.
How many moles of Hydrogen, Carbon, Oxygen, Nitrogen, and sulfur will there be?
Chemistry
1 answer:
denpristay [2]3 years ago
6 0
1 litre of water is = 55.5 moles of water.
water is H2O
so, in water:
moles of oxygen = 55.5
moles of hydrogen = 2 x 55.5 = 111

Now, 1 mole each of <span>CH4, NH3, H2S, and CO2 are added:
For CH4: 
moles of C = 1
moles of H = 4 x 1 = 4

For NH3:
moles of N = 1
moles of H = 3 x 1 = 3

For H2S:
moles of H = 2 x 1 = 2
</span>moles of S = 1
<span>
For CO2:
</span>moles of C = 1
moles of  = 2 x 1 = 2
<span>
Now, add the total moles of each atom:
Hydrogen = 111 + 4 + 3 +1 = 119 moles
Oxygen = 55.5 + 2 = 57.5
Carbon = 1+1 = 2
Sulfur = 1
nitrogen = 1

</span>
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natulia [17]

Answer:

The answer should be 1000 kg / m3

8 0
3 years ago
Which parts of atoms can interact (react) to form chemical bonds?
EastWind [94]
The valence electrons are the parts of an atom that make interactions and make chemical bonds.
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3 0
3 years ago
If 4.12 l of a 0.850 m-h3po4 solution is be diluted to a volume of 10.00 l, what is the concentration of the resulting solution?
statuscvo [17]

The process in which the concentration of the solution is lessened  by the addition of water is said to be dilution and equation of dilution relates the initial concentration and volume of stock solution with the final concentration and volume of the solution.

Formula is given by:

C_{1}V_{1}=C_{2}V_{2}   (1)

where,

C_{1} is the initial concentration

V_{1} is the initial volume

C_{2} is the final concentration

V_{2} is the final volume

Now,

C_{1} = 0.850 M

V_{1} = 4.12 L

C_{2}  =?

V_{2} = 10.00 L

Substitute the give values in formula (1),

0.850 M\times 4.12L=C_{2}\times 10.00 L

C_{2} =\frac{0.850 M\times 4.12L}{10.00 L}

= 0.3502 M

Thus, the final concentration of theH_{3}PO_{4} solution = 0.3502 M












5 0
3 years ago
Calculate the rcf (recipe conversion factor) for a new yield of 80 oz. (5 lbs) blueberry muffin batter. the original yield of th
SCORPION-xisa [38]

The rcf (recipe conversion factor)=  0.6

<h3>What is the rcf (recipe conversion factor)?</h3>
  • The conversion factor approach is the most typical technique for modifying recipes.
  • Finding a conversion factor and multiplying the ingredients in the original recipe by that factor are the only two steps needed to do this.
  • Remember that the conversion factor will be larger than 1 if you are raising your amounts to be sure you are finding it correctly.
  • The factor will be less than 1 if your amounts are being decreased. Use the conversion factor approach if you come across a recipe that is written in a standard format.
  • The production of phenolic monomers from lignin is effective and selective when done using reductive catalytic fractionation (RCF).

How this is calculated?

Conversion Factor = New Yield ÷ Old Yield

 =80/134

 =0.5970

 =0.6(rounded to nearest tenth)

To know more about the recipe conversion factor (RCF),refer:

brainly.com/question/23841906

#SPJ4

8 0
2 years ago
The normal boiling point of a liquid is 282 °C. At what temperature (in
ElenaW [278]

Answer:

The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C

Explanation:

Here we make use of the Clausius-Clapeyron equation;

ln\left (\frac{p_{2}}{p_{1}}  \right )=-\frac{\Delta H_{vap}}{R}\cdot \left (\frac{1}{T_{2}}-\frac{1}{T_{1}}  \right )

Where:

P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K

P₂ = 0.2 atm = The substance vapor pressure at temperature T₂

\Delta H_{vap} = The heat of vaporization = 28.5 kJ/mol

R = The universal gas constant = 8.314 J/K·mol

Plugging in the above values in the Clausius-Clapeyron equation, we have;

ln\left (\frac{0.2}{1}  \right )=-\frac{28.5 \times 10^3}{8.3145}\cdot \left (\frac{1}{T_{2}}-\frac{1}{555.15}  \right )

\therefore T_2 = \frac{-3427.95}{ln(0.2)-6.175}

T₂ = 440.37 K

To convert to Celsius degree temperature, we subtract 273.15 as follows

T₂ in °C = 440.37 - 273.15 = 167.22 °C

Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.

5 0
3 years ago
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