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3 years ago

8

The placement of carbonyl group of a kerose sugar is at second-carbon only (b) fir + carbon only () first and fast carbon (d) li

st carbon only Chirulit,

1 answer:

olasank [31]3 years ago

6 0

Answer:

(d)

Explanation:

Carbonyl group can be the placement of kerosene sugar

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Simplify (4f+13g)(2w)

Anna35 [415]

Answer:

8fw+26gw

i think this is it

7 0

3 years ago

Read 2 more answers

Using the equations 2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/m

Dafna11 [192]

Answer : The enthalpy for the reaction is 49.1 kJ/mol

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $C_6H_6$ will be,

$6C(s)+3H_2(g)\rightarrow C_6H_6(l)$ $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $2C_6H_6(g)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)$

$\Delta H_1=-6271kJ/mole$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$

$\Delta H_2=-393.5kJ/mole$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$

$\Delta H_3=-483.6kJ/mole$

Now we will reverse the reaction 1 and divide by 2, multiply reaction 2 by 6 and reaction 3 by 3/2 then adding all the equations, we get :

(1) $6CO_2(g)+3H_2O(l)\rightarrow C_6H_6(g)+\frac{15}{2}O_2(g)$

$\Delta H_1=-\frac{-6271kJ/mole}{2}=3135.5kJ/mol$

(2) $6C(s)+6O_2(g)\rightarrow 6CO_2(g)$

$\Delta H_2=6\times (-393.5kJ/mole)=-2361kJ/mol$

(3) $3H_2(g)+\frac{3}{2}O_2(g)\rightarrow 3H_2O(l)$

$\Delta H_3=\farc{3}{2}\times (-483.6kJ/mole)=-725.4kJ/mol$

The expression for enthalpy of formation of $C_6H_6$ will be,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(3135.5kJ/mole)+(-2361kJ/mole)+(-725.4kJ/mole)$

$\Delta H=49.1kJ/mole$

Therefore, the enthalpy for the reaction is 49.1 kJ/mol

7 0

3 years ago

An alkaline battery produces electrical energy according to the following equation.

Gnesinka [82]

Answer:

Part a: limiting reactant MnO₂

Part b: 12.43 g of Zn(OH)₂

Explanation:

Part a

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Limiting reactant = ?

Given Reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

To determine the limiting reactant first we will look at the reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

1 mol 2 mol

Convert moles to mass

molar mass of Zn = 65.4 g/mol

Mass of MnO₂ = 55 + 2 (16) = 55 + 32 = 87 g/mol

So,

Zn(s) + 2 MnO₂(s) + H₂O(l) ----------> Zn(OH)₂(s) + Mn₂O₃(s)

1 mol (65.4 g/mol) 2 mol (87 g/mol)

65 g 174 g

So its clear from the reaction that 65 g Zn react with 174 g of MnO₂.

now if we look at the given amounts the amount MnO₂ is less then the amount of Zn but in actual calculation amount of MnO₂ is more then amount of zinc.

So, for MnO₂ if we calculate the needed amount of zinc

So apply unity formula

65 g Zn react ≅ 174 g of MnO₂

X g of Zn ≅ 21.5 g of MnO₂

Do cross multiplication

X g Zn react = 65 g x 21.5 g / 174 g

X g of Zn ≅ 8.032 g

So, 8.032 g of zinc will react out of 37.0 grams. the remaining will be in excess.

So MnO₂ will be consumed completely an it will be limiting reactant.

____________

part b

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Mass of Zn(OH)₂ produced = ?

Given Reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

Solution:

As from the part A we come to know that MnO₂ is limiting reactant, so the amount of Zn(OH)₂ will depend on the amount of MnO₂.

So first we convert mass of MnO₂ to moles

Formula Used

no. of moles = mass in grams / molar mass

Mass of MnO₂ = 55 + 2 (16)

Mass of MnO₂ = 55 + 32 = 87 g/mol

Put values in the above equation

no. of moles = 21.5 g / 87 g/mol

no. of moles = 0.25 moles

Now,

Look at the reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

2 mol 1 mol

So its clear from the reaction that 2 mole of MnO₂ gives 1 mole of Zn(OH)₂

then how many moles of Zn(OH)₂ will be produce by 0.25 moles of MnO₂

So.

apply unity formula

2 mol of MnO₂ ≅ 1 mole of Zn(OH)₂

0.25 moles of MnO₂ ≅ X mole of Zn(OH)₂

Do cross multiplication

X mole of Zn(OH)₂ = 1 mole x 0.25 mol x / 2 mol

X mole of Zn(OH)₂ ≅ 0.125

Now Conver moles of Zn(OH)₂ to mass

Formula used

mass in grams = no. of moles x molar mass

Molar mass of Zn(OH)₂

Molar mass of Zn(OH)₂ = 65.4 + 2 (16 + 1)

Molar mass of Zn(OH)₂ = 65.4 + 2 (17)

Molar mass of Zn(OH)₂ = 65.4 + 34 = 99.4 g/mol

Put values in above equation

mass in grams = 0.125 mol x 99.4 g/mol

mass in grams = 12.43 g

So,

12.43 g of Zn(OH)₂ will be produce.

5 0

3 years ago

denpristay [2]

Answer:

The bowling ball has more kinetic energy than the tennis ball

Explanation:

Using the formula 1/2 mass × acceleration we found that the tennis ball had a kinetic energy of 0.75 while the bowling ball had a kinetic energy of 10.5 hence the bowling ball has the ability to do more work

5 0

3 years ago

Determine the number of moles of carbon dioxide gas, water, and sodium chloride formed by the reaction of 42.0 grams of sodium b

brilliants [131]

Answer: .45 moles of CO2, H20,and NaCl

Explanation:

5 0

3 years ago