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babunello [35]
3 years ago
10

Each molecule of table sugar, C12H22O11, contains

Chemistry
2 answers:
algol [13]3 years ago
7 0

Answer : The correct option is, (d) 12 atoms of carbon.

Explanation:

The formula of given molecule of table sugar is, C_{12}H_{22}O_{11}

The given molecule is made up of three elements which are, carbon, hydrogen and oxygen.

In the given molecule of table sugar C_{12}H_{22}O_{11}, there are 12 number of carbon atoms, 22 number of hydrogen atoms and 11 number of oxygen atoms.

From the given options we conclude that, the option (d) is the correct option. While the other options are incorrect.

Hence, the correct option is, (d) 12 atoms of carbon.

Delicious77 [7]3 years ago
3 0
D, because C12 means there's 12 atoms of carbon. 
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Answer:

IT IS D because lights are heated and it gives light!

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3 years ago
What information did the scientists miss in 1948 and 1966 that caused them to
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The information from Mariner 6 and Mariner 7

Explanation:

Because they had not been launched yet

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3 years ago
We have air (21% O2 and 79% N2) at 23 bar and 30 C. 4. What is the ideal molar volume (m^3/kmol)? a. b. What is the Z factor? Wh
k0ka [10]

Answer:

The  ideal molar volume is  \frac{V}{n}  =V_z=  0.001095 \ m^3/mol  

The  Z factor is  Z = 0.09997

The  real molar volume is \frac{V_r}{n} = V_k=   0.0001095\ \frac{m^3}{mol}

Explanation:

From the question we are told that

    The pressure is  P  = 23 \ bar =  23 *10^5 Pa

      The temperature is  T  =  30 ^ oC  = 303 \ K

According to the ideal gas equation we have that

          PV  =  nRT

=>      \frac{V}{n}=V_z= \frac{RT}{P}

Where  \frac{V}{n } is the molar volume  and  R is the gas constant with value

            R  =  8.314 \ m^3 \cdot Pa \cdot K^{-1}\cdot mol^{-1}

substituting values

            \frac{V}{n}  =V_z=  \frac{ 8.314 *  303}{23 *10^{5}}

             \frac{V}{n}  =V_z=  0.001095 \ m^3/mol            

The  compressibility factor of the gas is mathematically represented  as

            Z = \frac{P *  V_z}{RT}

substituting values        

          Z = \frac{23 *10^{5} *   0.001095}{8.314 * 303}

          Z = 0.09997

Now the real molar volume is evaluated as

         \frac{V_r}{n} = V_k=  \frac{Z *  RT }{P}

substituting values

             \frac{V_r}{n} = V_k=   \frac{0.09997 *  8.314 *  303}{23 *10^{5}}

             \frac{V_r}{n} = V_k=   0.0001095\ \frac{m^3}{mol}

8 0
3 years ago
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6 0
2 years ago
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Answer:

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Explanation:

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