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jeyben [28]
3 years ago
6

Chemical formula of binary compound a of oxygen acidic​

Chemistry
1 answer:
Reika [66]3 years ago
8 0

A binary compound of oxygen with another element is called oxide. An oxide is a binary compound of oxygen and another element. Oxygen combines with metals and non-metals to form respective oxides.

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The anticodon is the three-base code on the tRNA molecule that binds to the codon on mRNA. What is the associated codon for the
neonofarm [45]
It is AAU
A-U
U-A
C-G
G-C
5 0
3 years ago
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Gold is currently trading at very high price. Suppose that gold is selling for around $1860/ounce. How
Ugo [173]

Answer:

The answer is "3.81041978"

Explanation:

\to 1 \ OZ= 28.349523125 \ grams\\\\

              =28.349523125\times {1000} \ miligrams\\\\= 28349. 5231  \ miligrams\\

In 1860 = 28349.5231 \ miligrams\\

\to In \$ \ 1 = \frac{28349.5231}{1860} \ \ miligrams\\

             = 15.2416791 \ miligrams

\ In \  1 \ quarter =  \$ \ 0.25

\to \$ \ 0.25 =  15.2416791  \times 0.25 \  miligrams\\

              = 3.81041978

3 0
3 years ago
Why sometimes the electrons of an atom transfer to higher energy level​
Elenna [48]

Answer:

According to Bohr, the amount of energy needed to move an electron from one zone to another is a fixed, finite amount. ... The electron with its extra packet of energy becomes excited, and promptly moves out of its lower energy level and takes up a position in a higher energy level. This situation is unstable, however.

8 0
3 years ago
What is the correct name of this compound?
Reika [66]

Answer:

7. 3–ethyl–6 –methyldecane

8. 5–ethyl–2,2–dimethyl–4–propyl–4 –heptene

Explanation:

It is important to note that when naming organic compounds having two or more different substituent groups, we simply name them alphabetically.

The name of the compound given in the question above can be written as follow:

7. Obtaining the name of the compound.

Compound contains:

I. Decane.

II. 3–ethyl.

III. 6 –methyl.

Naming alphabetically, we have

3–ethyl–6 –methyldecane

8. Obtaining the name of the compound.

Compound contains:

I. 2,2–dimethyl.

II. 4–propyl.

III. 4 –heptene.

IV. 5–ethyl.

Naming alphabetically, we have

5–ethyl–2,2–dimethyl–4–propyl–4 –heptene

8 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
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