Ask question

Ask question

All categories

2 years ago

11

The atom in a chemical bond that attracts electrons more strongly acquires a(n)____________ charge, and the other atom acquires

a(n)__________ charge. If the electron transfer is significant but not enough to form ions, the atoms acquire _____________ and charges. The bond in this situation is called a polar covalent bond.

1 answer:

pshichka [43]2 years ago

4 0

Answer:

The answer to your question is below

Explanation:

The atom in a chemical bond that attracts electrons more strongly acquires a(n)__<u>negative</u>______ charge, and the other atom acquires a(n)_<u>positive</u>_______ charge. If the electron transfer is significant but not enough to form ions, the atoms acquire _<u>dipoles</u>_________ and charges. The bond in this situation is called a polar covalent bond.

You might be interested in

If each NADHNADH generates 3 ATPATP molecules and each FADH2FADH2 generates 2 ATPATP molecules, calculate the number of ATPATP m

Nadusha1986 [10]

Answer:

$128~ATP$

Explanation:

The metabolic pathway by which energy can be obtained from a fatty acid is called <u>"beta-oxidation"</u>. In this route, acetyl-Coa is produced by removing <u>2 carbons</u> from the fatty acid for each acetyl-Coa produced. In other words, for each round, 1 acetyl Coa is produced and for each round 2 carbons are removed from the initial fatty acid. Therefore, the first step is to calculate the <u>number of rounds</u> that will take place for an <u>18-carbon fatty</u> acid using the following equation:

$Number~of~Rounds=\frac{n}{2}-1$

Where "n" is the <u>number of carbons</u>, in this case "18", so:

$Number~of~Rounds=\frac{18}{2}-1~=~8$

We also have to calculate the amount of Acetyl-Coa produced:

$Number~of~Acetyl-Coa=\frac{18}{2}~=~9$

Now, we have to keep in mind that in each round in the beta-oxidation we will have the <u>production of 1 $FADH_2$ and 1 $NADH$ </u>. So, if we have 8 rounds we will have 8 $FADH_2$ and 8 $NADH$ .

Finally, for the total calculation of ATP. We have to remember the <u>yield for each compound</u>:

-) $1~FADH_2~=~2~ATP$

-) $1~NADH~=~3~ATP$

-) $Acetyl~CoA~=~10~ATP$

Now we can do the total calculation:

$(8*2)~+~(8*3)~+~(9*10)=130~ATP$

We have to <u>subtract</u> "2 ATP" molecules that correspond to the <u>activation</u> of the fatty acid, so:

$130-2=128~ATP$

In total, we will have 128 ATP.

I hope it helps!

6 0

2 years ago

An element X forms a compound XH 3 with hydrogen. Another element Y forms a compound YX 2 with X. Given that the valency of hydr

Marina CMI [18]

X $H_{3}$
H has a positive 1 charge. This means that having 3H = +3<span>. This is a neutral compound so x= -3 because X+3H= 0

Y</span> $X_{2}$ is also neutral so 2X+Y= 0
we know X=-3 So, 2(-3)+Y=0
-6+y=0
Y=+6 charge

Answer: The valency of X is -3. The valency of Y is 6

7 0

3 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

2 years ago

What happens to the number of valence electrons as you move from left to right in the periodic table?

fomenos

Valence electrons increases

6 0

2 years ago

How much a material takes time to burn?

Nat2105 [25]

(a)- Time

(b)- Heat produced(i guess)

(c)- Material

this is what I think, hope it helps

5 0

3 years ago