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kondor19780726 [428]
2 years ago
14

How many grams are in 3.45 moles CO2

Chemistry
1 answer:
Mamont248 [21]2 years ago
6 0
<h3>Hello There!!</h3>

<u> To Find:-</u>

Grams Present in 3.45 Moles of Carbon.

<u>We </u><u>Know:</u><u>-</u>

Molecular Mass of Carbon Dioxide= 44.01 grams/mol or 44 grams/mol

\text{Moles} = \frac{\text{Mass}}{\text{Molecular mass}}

\implies 3.45 \text{moles} =  \frac{ \text{Mass}}{44 \text{g/mol}}  \\  \\  \implies \text{Mass} = {44 \text{g/mol}} \times 3.45 \text{moles} \\  \\ \implies \text{Mass} = {44 \text{g/}  \cancel\text{ mol}} \times 3.45  \cancel\text{moles} \\  \\ \implies \text{Mass} = 44 \text{g} \times 3.45 \\  \\  \implies  \text{Mass} =  \red{151.8 \text{g}}

\therefore \text{Mass of 3.45 Moles of Carbon Dioxide is =  \red{151.8g}}

<h3>Hope This Helps!!</h3>
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Answer:

A) Mass flow rate of air = 22.892 kmol/hr

B)percentage by mass of oxygen in the product gas = 22.52%

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have 0.09 mole of methane(CH4) and the remaining will be the air which is (100% - 9%) = 91% = 0.91

Molar mass of CH4 = 12 + 1(4) = 16 g/mol

We are given the average molecular weight of air = 29 g/mol

Thus;

Average molar mass of air and methane mixture is;

M_avg = (0.09 × 16) + (0.91 × 29)

M_avg = 27.83 g/mol

We are told that air flowing at a rate of 7 × 10² kg/h = 700 kg/h

Thus;

Mass flow rate of CH4 in air mixture = 700kg/h × (0.09CH4)/1 mix × (1/27.83kg/kmol) = 2.264 kmol/hr

Mass flow rate of air in mixture = 2.264kmol/h × 0.91kmol air/0.09kmolCH4 = 22.892 kmol/hr

We are told that the mixture is capable of being ignited if the mole percent of methane is between 5% and 15%.

Thus, for 5% of methane, the air required will be;

2.264kmol/h × 0.95kmol air/0.05kmol CH4 = 43.016 kmol/hr

Now, the dilution air needed will be =

43.016 - 22.892 = 20.124 kmol/hr

Total mass flow rate of mixture =

700kg/hr + (20.124kmol/hr × 29kg/mol) = 1283.596 kg/hr

We are told that air consist of 21 mole% Oxygen (O2).

Molar mass of oxygen = 32

Thus;

Mass fraction of oxygen in the product gas = 43.016kmol/h × (0.21molO2/1mol air) × (32kg oxygen/1kmol oxygen) × (1/1283.596kg/h) = 0.2252

Thus, written in percentage form, we have; 22.52%

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3 0
3 years ago
179.1 g of water is in a Styrofoam calorimeter of negligible heat capacity. The initial T of the water is 16.1oC. After 306.9 g
taurus [48]

Answer:

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Explanation:

Generally the change in temperature of water is evaluated as

                \Delta T = T_2 -T_1

Substituting 16.1°C for T_1 and 27.4°C for T_2

                \Delta T = 27.4 - 16.1

                       =11.3^oC

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                  \Delta T_u = T_3 -T_2

Substituting 27.4°C for T_2 and 94.3°C for T_3

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Substituting 179.1 g  for m , 4.18 J/g.C for c_w(specific heat of water)

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Since there is a decrease in temperature then heat is lost by unknown compound and this can be evaluated as

                    H_u = m_uc_u \Delta T_u

By conservation of energy law

       Heat lost  = Heat gained  

Substituting 306.9 g  for m_u , 8459.6J for H_u

           8459.6 = 306.9 * c_u * 66.9

  Therefore     c_u = \frac{8459.6}{308.9 *66.9}

                           =0.412J/g \cdot C

                   

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4

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8 0
3 years ago
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