Which of these is given away or accepted by an atom when it reacts with other atoms?

Assuming a car (with a 70-L) gas tank can hold approximately 50,000 (5.00 * 10^4) g of octane(C8H18) or 50,000 (5.00 * 10^4) g o

konstantin123 [22]

Answer:

- From octane: $m_{CO_2}=1.54x10^5gCO_2$

- From ethanol: $m_{CO_2}=9.57x10^4gCO_2$

Explanation:

Hello,

At first, for the combustion of octane, the following chemical reaction is carried out:

$C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O$

Thus, the produced mass of carbon dioxide is:

$m_{CO_2}=5.00x10^4gC_8H_{18}*\frac{1molC_8H_{18}}{114gC_8H_{18}}*\frac{8molCO_2}{1molC_8H_{18}}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=1.54x10^5gCO_2$

Now, for ethanol:

$C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O$

$m_{CO_2}=5.00x10^4gC_2H_6O*\frac{1molC_2H_6O}{46gC_2H_6O}*\frac{2molCO_2}{1molC_2H_6O}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=9.57x10^4gCO_2$

Best regards.

3 0

3 years ago

On the basis of periodic trends, choose the larger atom from each pair (if possible): match the elements in the left column to t

Ne4ueva [31]

The given question is incomplete. The complete question is as follows.

On the basis of periodic trends, choose the larger atom from each pair (if possible): match the elements in the left column to the appropriate blanks in the sentences on the right. make certain each sentence is complete before submitting your answer.

Sn, F, Ge, not predictable, Cr

Of Ge or Po, the larger atom is .........

Of F or Se, the larger atom is ..........

Of Sn or I, the larger atom is .........

Of Cr or W, the larger atom is ........

Explanation:

When we move down a group then there occurs an increase in atomic size of the atoms due to increase in the number of electrons.

Ge is a group 14 element which lies in period 4 and Po is a group 16 element that lies in period 6. As polonium is larger in size as compared to germanium.

Fluorine is a group 17 element and lies in period 2. Selenium is a group 16 element lies in 4. Therefore, selenium is larger in size as compared to fluorine.

Sn is a group 14 element that lies in period 5 and I is a group 17 element that lies in period 5. Hence, I is a larger atom.

Cr is a d-block element which lies in period 4 and W is also a d-block element which lies in period 6. Hence, W is larger in size than Cr.

Thus, we can conclude that given blanks are matched as follows.

Of Ge or Po, the larger atom is Po.

Of F or Se, the larger atom is Se.

Of Sn or I, the larger atom is I.

Of Cr or W, the larger atom is W.

7 0

2 years ago

7.no . <br> with steps please anyone <br> no spam

gulaghasi [49]

Given mass=1.3g
Atomic mass=65u

Molar mass=65g/mol

Now

$\boxed{\sf No\:of\:moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \rm\longmapsto No\:of\;Moles=\dfrac{1.3}{65}=0.02mol$

6 0

2 years ago

Convert 3 moles of h2o into grams

lilavasa [31]

To convert 3 mol H2O to grams, just multiply by the molar mass of H2O.

<span>3 mol H2O * 18 g H2O / 1 mol H2O = 54 g H2O

Hope this helped.</span>

4 0

3 years ago

Need help asap with this chemistry if someone could help me

Burka [1]

Answer:

Structure One:

N: -2
C: 0
O: +1

Structure Two:

N: 0
C: 0
O: -1

Structure Three:

N: -1
C: 0
O: 0.

Structure Number Two would likely be the most stable structure.

All five C atoms: 0
All six H atoms to C: 0
N atom: +1.

The N atom is the one that is "likely" to be attracted to an anion. See explanation.

Explanation:

When calculating the formal charge for an atom, the assumption is that electrons in a chemical bond are shared equally between the two bonding atoms. The formula for the formal charge of an atom can be written as:

$\text{Formal Charge} \\ = \text{Number of Valence Electrons in Element} \\ \phantom{=}-\text{Number of Chemical Bonds} \\\phantom{=} - \text{Number of nonbonding Lone Pair Electrons}$ .

For example, for the N atom in structure one of the first question,

N is in IUPAC group 15. There are 15 - 10 = 5 valence electrons on N.
This N atom is connected to only 1 chemical bond.
There are three pairs, or 6 electrons that aren't in a chemical bond.

The formal charge of this N atom will be $5 - 1 - 6 = -2$ .

Apply this rule to the other atoms. Note that a double bond counts as two bonds while a triple bond counts as three.

Structure One:

N: -2
C: 0
O: +1

Structure Two:

N: 0
C: 0
O: -1

Structure Three:

N: -1
C: 0
O: 0.

In general, the formal charge on all atoms in a molecule or an ion shall be as close to zero as possible. That rules out Structure number one.

Additionally, if there is a negative charge on one of the atoms, that atom shall preferably be the most electronegative one in the entire molecule. O is more electronegative than N. Structure two will likely be favored over structure three.

Similarly,

All five C atoms: 0
All six H atoms to C: 0
N atom: +1.

Assuming that electrons in a chemical bond are shared equally (which is likely not the case,) the nitrogen atom in this molecule will carry a positive charge. By that assumption, it would attract an anion.

Note that in reality this assumption seldom holds. In this ion, the N-H bond is highly polarized such that the partial positive charge is mostly located on the H atom bonded to the N atom. This example shows how the formal charge assumption might give misleading information. However, for the sake of this particular problem, the N atom is the one that is "likely" to be attracted to an anion.

5 0

3 years ago