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3 years ago

A compound C6H12 reacts instantly

Chemistry

1 answer:

Scorpion4ik [409]3 years ago

7 0

Answer and Explanation:

Since the compound C6H12 reacts with Br2 to give a dibromide, that means it contains one double bond and also since it gives an aldehyde C2H4O as one of the products, the ozonolysis implies that one end of the double bond is =CHCH3 and because the other product is C4H8O, the original compound has this structure in the attachment below. The original compound is 2-hexene.

(The structure of C6H12Br2 is also in the attachment below)

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Suppose that 98.0g of a non electrolyte is dissolved in 1.00kg of water. The freezing point of this solution is found to be -0.4

Sholpan [36]

Answer:

$\large \boxed{\text{392 u}}$

Explanation:

1. Calculate the molal concentration

The formula for the freezing point depression by a nonelectrolyte is

$\Delta T_{f} = -K_{f}b\\b = -\dfrac{\Delta T_{f}}{ K_{f}} = -\dfrac{-0.465 \, ^{\circ}\text{C}}{\text{1.86 $\, ^{\circ}$C$\cdot$kg$\cdot$mol}^{-1}} = \text{0.250 mol/kg}$

2. Calculate the moles of solute

$\begin{array}{rcl}b & = & \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}\\\\\text{moles of solute} & = & b \times {\text{kilograms of solvent}}\\n & = &\text{0.250 mol/kg} \times \text{1.00 kg}\\ & = & \text{0.250 mol}\\\end{array}$

3. Calculate the molecular mass

$\begin{array}{rcl}\text{Moles} & = &\dfrac{\text{mass}}{\text{molar mass}}\\\\\text{0.250 mol} & = & \dfrac{\text{98.0 g}}{MM}\\\\MM & = & \dfrac{\text{98.0 g }}{\text{0.250 mol}}\\\\& = &\textbf{392 g/mol}\\\end{array}\\\text{The molecular mass of the solute is $\large \boxed{\textbf{392 u}}$}$

3 0

4 years ago

What is solar radiation?

mafiozo [28]

Explanation:

Solar energy is the radiant energy emmitted from the Sun. It is the electromagnetic energy

8 0

3 years ago

What is the ionic equation of Aqueous calcium bromide was mixed with aqueous gold(I) perchlorate, and a crystallized gold(I) bro

Bumek [7]

Answer:- $2AuClO_4(aq) + CaBr_2(aq)\rightarrow 2AuBr(s)+ Ca(ClO_4)_2(aq)$

Explanations:- It's a double replacement reaction where a precipitate of silver(I)bromide is formed. A dpuble replacement reaction in general looks as AB + CD \rightarrow AD + CB

To balance the equation we need to multiply gold compounds on both sides by 2 and the balanced equation is..

$2AuClO_4(aq) + CaBr_2(aq)\rightarrow 2AuBr(s)+ Ca(ClO_4)_2(aq)$

6 0

4 years ago

What would be the change in pressure in a sealed 10.0 l vessel due to the formation of n2 gas when the ammonium nitrite in 2.40

larisa86 [58]

The change in pressure in a sealed 10.0L vessel is 5.28 atm

calculation

The pressure is calculated using the ideal gas equation

That is P=n RT

where;

P (pressure)= ?

v( volume) = 10.0 L

n( number of moles) which is calculated as below

write the equation for decomposition of NH₄NO₂

NH₄NO₂ → N₂ +2H₂O

Find the moles of NH₄NO₂

moles = molarity x volume in liters

= 2.40 l x 0.900 M =2.16 moles

Use the mole ratio to determine the moles of N₂

that is from equation above NH₄NO₂:N₂ is 1:1 therefore the moles of N₂ is also =2.16 moles

R(gas constant) =0.0821 l.atm/mol.K

T(temperature) = 25° c into kelvin = 25 +273 =298 K

make p the subject of the formula by diving both side by V

P = nRT/V

p ={ (2.16 moles x 0.0821 L.atm/mol.K x 298 K) /10.0 L} = 5.28 atm.

3 0

3 years ago

Read 2 more answers

The thiocyanate polyatomic ion, SCN-, is commanly called a pseudohalogen because it acts very much like halide ions. For example

harina [27]

Answer:

A) $\frac{1mol(SCN)_{2}}{2molNaSCN}$

B) 0.025 mol (SCN)₂

C) 2 mol (SCN)₂

D) $\frac{1molMnSO_{4}}{2molH_{2}SO_{4}}$

E) 3.5504 mol H₂SO₄

Explanation:

2NaSCN +2H₂SO₄ + MnO₂ → (SCN)₂ + 2H₂O + MnSO₄ + Na₂SO₄

A) A conversion factor could be $\frac{1mol(SCN)_{2}}{2molNaSCN}$ , as it has the units that we want to convert to in the numerator, and the units that we want to convert from in the denominator.

B) 0.05 mol NaSCN * $\frac{1mol(SCN)_{2}}{2molNaSCN}$ = 0.025 mol (SCN)₂

C) With 4 moles of NaSCN and 3 moles of MnSO₄, the reactant is NaSCN so we use that value to calculate the moles of product formed:

4 mol NaSCN * $\frac{1mol(SCN)_{2}}{2molNaSCN}$ = 2 mol (SCN)₂

D) $\frac{1molMnSO_{4}}{2molH_{2}SO_{4}}$

E) 1.7752 mol MnSO₄ * $\frac{2molH_{2}SO_{4}}{1molMnSO_{4}}$ = 3.5504 mol H₂SO₄

5 0

3 years ago