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Y_Kistochka [10]
4 years ago
7

When the following equation is balanced using the smallest possible integers, what is the coefficent of oxygen gas?

Chemistry
1 answer:
elena55 [62]4 years ago
6 0

By balancing a chemical reaction, you must first balance the atom that is in smaller amount in the chemical species that you have in the reaction. In the combustion reaction presented in the question that atom is carbon (C). Therefore, the first thing to do is multiply the CO2 by 7, since those are the carbon atoms that are in the C7H16O. The reaction would be like this,

C7H16O + O2  →  7CO2 + H2O

You can make a table of the amount of atoms of C, H and O that you have in the reagents and products after you put a coefficient 7 in front of the CO2, in the following way,

R P

C 7 7

H 16 2

O 3 15

Then you balance the hydrogens. It is better that you leave the last oxygens since there is an oxygen molecule alone, so when adding a coefficient to balance it, the quantities of the rest of the atoms in the equation would not be altered.

To balance the hydrogens you add an 8 in front of the H2O molecule in the reagents, since there are 16 hydrogens in the molecule C7H16O and

8H x 2H = 16H.

The reaction and the table would be like this,

C7H16O + O2  →  7CO2 + 8H2O

R P

C 7 7

H 16 16

O 3 22

Finally, you balance the atoms of O. To do this, you add a coefficient of 21/2 to the O2 molecule, this is because (21/2) x2 = 21 and 21 O coming from the O2 molecule plus an O coming from the C7H16O molecule gives a total of 22 O, which are equal to the amount of O we have in the products.

C7H16O + (21/2)O2  →  7CO2 + 8H2O

To bring the coefficients of the reactants and the products to whole numbers multiply all the coefficients by 2. Then the reaction is like,

2 C7H16O + 21O2  →  14CO2 + 16H2O

As you can see when the given reaction is balanced using the smallest possible integers, the coefficent of oxygen gas is 21

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lora16 [44]

Answer:

i=1.62 .

Explanation:

Let, i be the Van't Hoff Factor.

Moles of benzamide,=\dfrac{Mass}{molar\ mass}=\dfrac{70.4}{121.14}=0.58\ mol.

Molality of solution, m=\dfrac{moles  }{mass\ of\ solvent}=\dfrac{0.58}{0.85}=0.68\ molal.

Now, we know

Depression in freezing point, \Delta T=i\times K_f\times m  .....1

It is given that,

\Delta T=2.7^o C\\i=1 ( since\ it\ is\ non\ dissociable\ solutes)\\K_f  ( freezing\ constant)\\

Putting all these values we get,

K_f=3.949\ C/m.

Now, moles of ammonium chloride=\dfrac{70.4}{53.49}=1.316\ mol.

molality =\dfrac{1.316}{0.85}=1.54 molal.\\\Delta T=9.9 .

Putting all these values in eqn 1.

We get,

i=1.62 .

Hence, this is the required solution.

6 0
3 years ago
Mole Conversions<br> What is the mass in grams of<br> 0.625 mol Ba(NO3)2
sweet [91]

0.0024 g of  Ba(NO₃)₂

Explanation:

First we need to determine the molecular weight of barium nitrate Ba(NO₃)₂

molecular wight of Ba(NO₃)₂ = molecular weight of Ba × number of Ba atoms + molecular weight of N × number of N atoms + molecular weight of O × number of O atoms

molecular wight of Ba(NO₃)₂ = 137 × 1 + 14 × 2 + 16 × 6 = 261 g/mol

number of moles = mass /  molecular weight

mass = number of moles × molecular weight

mass of Ba(NO₃)₂ = 0.625 / 261 = 0.0024 g

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6 0
3 years ago
Name two other elements that have properties that are similar to those of the element rubidium and why would they.
Andrei [34K]
Answer is potassium and sodium. They are in the same group and that means they have the same properties. They are all alkali metals
4 0
3 years ago
A series of chemicals were added to some AgNO3(aq). NaCl(aq) was added first to the silver nitrate solution to produce a precipi
OLEGan [10]

Answer:

First, precipitate of AgCl is formed. Second, a soluble complex of silver and ammonia is formed. Third, AgCl is reproduced due to disappearance of ammonia complex in presence of HNO_{3}.

Explanation:

In presence of NaCl, AgNO_{3} forms an insoluble precipitate of AgCl.

Reaction: Ag^{+}(aq.)+Cl^{-}(aq.)\rightarrow AgCl(s)

In presence of NH_{3}, AgCl gets dissolved into solution due to formation of soluble [Ag(NH_{3})_{2}]^{+} complex.

Reaction: AgCl(s)+2NH_{3}(aq.)\rightarrow [Ag(NH_{3})_{2}]^{+}(aq.)+ Cl^{-}(aq.)

In presence of HNO_{3}, [Ag(NH_{3})_{2}]^{+} complex gets destroyed and free Cl^{-} again reacts with free Ag^{+} to produce insoluble AgCl

Reaction: [Ag(NH_{3})_{2}]^{+}(aq.)+2H^{+}(aq.)+Cl^{-}(aq.)\rightarrow AgCl(s)+2NH_{4}^{+}(aq.)

4 0
3 years ago
Calculate the moles of 36.030 g of H2O.
lions [1.4K]

Answer: 1.99 mol h20

Explanation:

you have 36.030 g, trying to get it into mols

to get from g to mols, use the molar mass (periodic table)

36.030 g h20 (1 mol/18.02g) = 1.99 mol h20

4 0
3 years ago
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