All categories

2 years ago

What should the IUPAC name for a binary covalent compound lack? prefixes Roman numerals an -ide ending the name of a nonmetal.

Mathematics

1 answer:

vitfil [10]2 years ago

5 0

The IUPAC name for a binary covalent compound lack prefixes such as mono, di, tri tetra etc.

A binary compound is a compound that is composed of only two elements. Many binary compounds could be made up of a metal and a nonmetal or even two nonmetals as the case may be. Examples of binary compounds include; SO2, NaH, K2S etc.

We must note that a binary covalent compound uses Roman numerals to indicate the oxidation state of the central atom in the compound. For instance, SO2 is called Sulfur IV oxide.

Hence, the IUPAC name for a binary covalent compound lack prefixes such as mono, di, tri tetra etc.

Lear more about IUPAC: brainly.com/question/11587934

You might be interested in

Factor the following polynomial completely <br><br> 4x^2-100

Murrr4er [49]

Answer:

4 [ ( x - 5) ( x + 5) ]

Step-by-step explanation:

4x²-100

4 (x²-25)

4 (x²-5²)

4 [( x - 5) ( x + 5)]

4 0

3 years ago

Given f(x)= -5x + 1, solve for x when f(x) = 6.

aniked [119]

-29

-5(6) + 1
-30 + 1
-29

hope this helps !

7 0

3 years ago

Read 2 more answers

Reduce -8 + b^2 by 5 + b^2

ozzi

Answer:-13

Step-by-step explanation:

-8 + b^2 - 5 - b^2

=============

-13

8 0

3 years ago

Read 2 more answers

Consider rolling two fair dice one 3-sided the other 5-sided

Ne4ueva [31]

Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose

$1\leq x\leq 3,\quad 1\leq y \leq 5$

We have $P(X=x)=\frac{1}{3}$ and $P(Y=y)=\frac{1}{5}$ , because the dice are fair.

Now we use the assumption of independence to claim that

$P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}$

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:

2 in a unique way (1+1)
3 in two possible ways (1+2, 2+1)
4 in three possible ways
5 in three possible ways
6 in three possible ways
7 in two possible ways
8 in a unique way

This implies that the probabilities of the outcomes of $W=X+Y$ are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5

8 0

3 years ago

Amy jogs 1/3 of a mile in 1/15 of an hour, while john takes 1/30 of an hour to jog 1/5 of a mile. If they continue at this rate,

jolli1 [7]

Step-by-step explanation:

Given that,

Amy jogs 1/3 of a mile in 1/15 of an hour and john takes 1/30 of an hour to jog 1/5 of a mile.

Speed of Amy,

$v_1=\dfrac{\dfrac{1}{3}\ miles }{\dfrac{1}{15}\ h}=5\ mph$

Speed of John,

$v_2=\dfrac{\dfrac{1}{5}\ miles }{\dfrac{1}{30}\ h}=6\ mph$

John will jog farther in one hour as he is moving with faster speed. The speed of John is 1 mph more than Amy.

6 0

3 years ago

What should the IUPAC name for a binary covalent compound lack? prefixes Roman numerals an -ide ending the name of a nonmetal.​

What should the IUPAC name for a binary covalent compound lack? prefixes Roman numerals an -ide ending the name of a nonmetal.