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sammy [17]
3 years ago
15

How many moles of n are in 0.195 g of n2o?

Chemistry
2 answers:
Blababa [14]3 years ago
8 0

Explanation:

As molar mass of N_{2}O is 44.013 g/mol and it is given that its mass is 0.195 g.  

Therefore, calculate its number of moles as follows.

              No. of moles = \frac{mass}{\text{molar mass}}

                                    = \frac{0.195 g}{44.013 g/mol}

                                    = 0.0044 mol

Here, in 1 mole of N_{2}O there is 2 mole of N. Thus, moles of N present in it is calculated as follows.

                             2 \times 0.0044 mol

                                 = 0.0088 mol

Thus, we can conclude that in 0.195 g of N_{2}O there are 0.0088 mol of N.

ivann1987 [24]3 years ago
4 0
0.195g N2O x 2moles of N / 28g of N = 5.46 moles of N



2X14= 28g.
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Acid if I remeber correctly
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1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction
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Answer:

H_{comb}=-4406kJ/mol

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

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4 0
3 years ago
Yet a third pair of compounds of manganese and oxygen is 50.48% and 36.81% oxygen respectively. In what small whole number ratio
Mariulka [41]

Answer:

The number ratio is 4:7

Explanation:

Step 1: Data given

Compound 1 has 50.48 % oxygen

Compound 2 has 36.81 % oxygen

Molar mass oxygen = 16 g/mol

Molar mass manganese = 54.94 g/mol

Step 2: Calculate % manganes

Compound 1: 100 - 50.48 = 49.52 %

Compound 2: 100 - 36.81 = 63.19 %

Step 3: Calculate mass

Suppose mass of compounds = 100 grams

Compound 1:

 50.48 % O = 50.48 grams

 49.52 % Mn = 49.52 grams

Compound 2:

36.81 % O = 36.81 grams

63.19 % Mn = 63.19 grams

Step 4: Calculate moles

Compound 1

Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles

Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles

Compound 2

Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles

Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles

Step 5: calculate mol ratio

We will divide by the smallest amount of moles

Compound 1

O: 3.155/0.9013 = 3.5

Mn: 0.9013 / 0.9013 = 1

Mn2O7

Compound 2

O: 2.301 / 1.150 = 2

Mn: 1.150 / 1.150 = 1

MnO2

The number ratio is 2:3.5 or 4:7

7 0
3 years ago
What is the temperature of 1.485 moles of N₂ gas at a pressure of 1.072 atm and a volume of 20 L?
morpeh [17]

Answer:

178.67K

Explanation:

PV=nRT

T=PV/nR

= 1.072atm*20L/1.485mol*0.0821LatmK^-1

=178.67K

8 0
3 years ago
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