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lisabon 2012 [21]
3 years ago
11

The chemical formula for table sugar is c12h22o11. what can you tell from this formula

Chemistry
1 answer:
guajiro [1.7K]3 years ago
6 0
The chemical formula for table sugar is C12H22O11.  From the formula, you can say that it is an organic molecule where in every molecule contains 12 atoms of carbon, 22 atoms of hydrogen and 11 atoms of oxygen.  There are 45 atoms in total in one molecule of sugar. 
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According to the VSEPR model, the arrangement of electron pairs around NH3 and CH4 is A. the same, because in each case there ar
aleksklad [387]

Answer:

Option "B" is correct.

Explanation:

According to VSEPR theory, There are repulsion forces exists among the bond pair - bond pair or bond pair - lone pair of electrons. In NH_{3} and  CH_{4}, the number of electron pairs are same but methane has all the four bond pairs where in ammonia, three bond pairs and one lone pair exists. And thus there are repulsion forces possible in between the lone pair and bond pair of electrons thus the arrangement of electron pairs around both the molecules is same or different depending up on the conditions leading to maximum repulsion.  

3 0
3 years ago
In which location would you expect to find the greatest species diversity?!
Crank
I believe the answer is Canada!
5 0
3 years ago
Br2(g) cl2(g)⇌2brcl(g) δh∘f for brcl(g) is 14. 6 kj/mol. Δs∘f for brcl(g) is 240. 0 j/mol
max2010maxim [7]

The Change in Gibb's free energy, ΔG for the reaction at 298K is; -56.92KJ.

<h3>Gibb's free energy of reactions</h3>

It follows from the Gibb's free energy formula as expressed in terms of Enthalpy and Entropy that;

  • ΔG = ΔH - TΔS

On this note, it follows that;

  • ΔG = 14.6 - (298× 0.24)

Hence, the Gibb's free energy for the reaction is;

  • ΔG = 14.6 - 71.52
  • ΔG = -56.92KJ

Remarks: The question requires that we determine the Gibb's free energy for the reaction at 298K.

Read more on Gibb's free energy;

brainly.com/question/13765848

5 0
2 years ago
Read 2 more answers
The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperatur
vazorg [7]

Answer:

the activation energy Ea = 179.176 kJ/mol

it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

T_1 = 100^0 C = 100+273 = 373 \ K \\ \\  T_2 = 113^0 C = 113 + 273 = 386 \ K

R_1 = \dfrac{1}{7}

R_2 = \dfrac{1}{49}

Thus; \dfrac{R_2}{R_1} = 7

Because at 113.0°C; the rate is 7 time higher than at 100°C

Hence:

In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})

1.9459 = \dfrac{Ea}{8.314}* 9.0292  *10^{-5}

1.9459*8.314 = Ea * 9.0292*10^{-5}

16.1782126= Ea * 9.0292*10^{-5}

Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}

Ea = 179.176 kJ/mol

Thus; the activation energy Ea = 179.176 kJ/mol

b)

here;

T_2 = 386 \  K  \\ \\T_1 = (89.8 + 273)K = 362.8 \ K

In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})

In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})

In (\dfrac{R_2}{R_1}) = 0.00357

\dfrac{R_2}{R_1}= e^{0.00357}

\dfrac{R_2}{R_1}= 1.0035

where ;

R_2 = \dfrac{1}7{}

R_1 = \dfrac{1}{t}

Now;

\dfrac{t}{7}= 1.0035

t = 7.0245 mins

Therefore; it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

4 0
3 years ago
What is the observation of heating copper carbonate???.?
damaskus [11]

Answer:

1)This type of reaction is called thermal decomposition . Copper carbonate is green and copper oxide is black. You can see a colour change from green to black during the reaction. The carbon dioxide produced can be detected using limewater, which turns milky. 2) The extremely low pressure of the partial vacuum inside the can made it possible for the pressure of the air outside the can to crush it. A can is crushed when the pressure outside is greater than the pressure inside, and the pressure difference is greater than the can is able to withstand.

Explanation:

hope this helps you :)

7 0
3 years ago
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