How many bonds are between the two carbon atoms in C2H2?

3. When two atoms bond together to form a molecule, which parts of the atoms

fenix001 [56]

Answer:

Atoms come together to form molecules because of their electrons. Electrons can join (or bond) atoms together in two main ways. When two atoms share electrons between them, they are locked together (bonded) by that sharing. These are called covalent bonds

5 0

3 years ago

2-butanone is converted into 3-methyl-3-hexanol using a grignard reagent prepared from 1-bromopropane and magnesium metal in thf

Andre45 [30]

Answer:

Here's what I get.

Explanation:

At the end of the reaction you will have a solution of the alcohol in THF.

The microdistillation procedure will vary, depending on the specific apparatus you are using, but here is a typical procedure.

Transfer the solution to a conical vial.
Add a boiling stone.
Attach a Hickman head (shown below) and condenser.
Place the assembly in in the appropriate hole of an aluminium block on top of a hotplate stirrer.
Begin stirring and heating at a low level so the THF (bp 63 °C) can distill slowly.
Use a Pasteur pipet to withdraw the THF as needed.
When all the THF has been removed, raise the temperature of the Al block and distill the alcohol (bp 143 °C).

7 0

4 years ago

A sample of a compound is decomposed in the laboratory and produces 330 g carbon, 69.5 g hydrogen, and 220.2 g oxygen. Calculate

Zarrin [17]

Answer: Empirical formula is $C_2H_5O$

Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.

Step 1 : Converting each of the given masses into their moles by dividing them by Molar masses.

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of Carbon = 12.0 g/mol

Molar mass of Hydrogen = 1.0 g/mol

Molar mass of Oxygen = 16.0 g/mol

Moles of Carbon = $\frac{330g}{12g/mol}=27.5moles$

Moles of Hydrogen = $\frac{69.5g}{1g/mol}=69.5moles$

Moles of Oxygen = $\frac{220.2g}{16g/mol}=13.76moles$

Step 2: Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value

Smallest number of moles = 13.76 moles

$\text{Mole ratio of Carbon}=\frac{27.5moles}{13.76moles}=1.99\approx 2$

$\text{Mole ratio of Hydrogen}=\frac{69.5moles}{13.76moles}=5.05\approx 5$

$\text{Mole ratio of Oxygen}=\frac{13.76moles}{13.76moles}=1$

Step 3: Now, the moles ratio of the elements are represented by the subscripts in the empirical formula

Empirical formula becomes = $C_2H_5O$

7 0

3 years ago

If 3.89 × 1024 atoms of a noble gas is collected and it has a mass of of 848 grams, this element is most likely a. He b. Ne c. A

bearhunter [10]

Answer:

Xenon

Explanation:

Avogadro’s number represent the number of the constituent particles which are present in one mole of the substance. It is named after scientist Amedeo Avogadro and is denoted by $N_0$ .

Avogadro constant:-

$N_a=6.023\times 10^{23}$

Let the molar mass of the element is x g/mol

So,

$6.023\times 10^{23}$ atoms have a mass of x g

Also,

$3.89\times 10^{24}$ atoms have a mass of $\frac{x}{6.023\times 10^{23}}\times 3.89\times 10^{24}$ g

This mass is equal to 848 g

So,

$\frac{x}{6.023\times 10^{23}}\times 3.89\times 10^{24}=848$

x= 131.3 g/mol

This mass correspond to xenon.

3 0

3 years ago

The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo

notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0

3 years ago