Ask question

Ask question

All categories

3 years ago

12

Adam is trying to identify a solid. He places a small amount of the substance in an acid solution and observes whether or not a

reaction takes place. This test is based on a ___Radial____ property.

1 answer:

sp2606 [1]3 years ago

6 0

i dont tbh what you looking for

You might be interested in

What is the 2nd most abundant element in the earths crust

Elza [17]

Silicon is the second most abundant element in the Earth´s crust and makes up around 27% of it.

3 0

3 years ago

Example of a negative impact of acids in our everyday world

Marizza181 [45]

The acid clouds fog the important nutrients like leave and needles . this loss of nutrient make it easier for infections , insects , and cold weather to damage trees and forests

4 0

3 years ago

What parts are present in every chemical equation?

WITCHER [35]

The reactants, products, coefficients, subscripts. ( I forgot the rest lol)

5 0

2 years ago

Rank the following from highest to lowest priority according to the Cahn-Ingold-Prelog system: -CH2CH3, -CHCH2, -CCH, -CH3.

mario62 [17]

Answer:

The order will be:

CCH > CHCH₂ > CH₂CH₃> CH₃

Explanation:

According to Cahn-Ingold-Prelog system we rank the groups based on the atomic number of directly attached atom with the chiral carbon.

For example: between C and H, we rank Carbon first.

If the same atoms are attached for different groups then we prioritized based on the second element with highest atomic number.

For example:

Among CH₃ and C₂H₅, the priority will be given to C₂H₅.

If an atom is double or triple bonded to the directly attached atom then each pi bond is considered to be a new atom.

Hence CH=CH₂ means, that there are two carbons attached to CH carbon.

So the order based on above selection rules will be:

CCH > CHCH₂ > CH₂CH₃> CH₃

7 0

3 years ago

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:

LiRa [457]

Explanation:

(a) As the given chemical reaction equation is as follows.

$OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}$

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

Rate = $K \times [OCl^{-}] \times [l^{-}]$

(b) Since, the rate equation is as follows.

Rate = $K [OCl^{-}][l^{-}]$

Let us assume that ( $[OCl^{-}] = [l^{-}]$ )

Putting the given values into the above equation as follows.

$1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2$

$1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})$

K = $\frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}$

= $60.4 M^{-1}sec^{-1}$

Hence, the value of rate constant for the given reaction is $60.4 M^{-1}sec^{-1}$ .

(c) Now, we will calculate the rate as follows.

Rate = $K [OCl^{-}][l^{-}]$

= $60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})$

= $6.52 \times 10^{5}$

Therefore, rate when $[OCl^{-}] = 1.8 \times 10^{3}$ M and $[I^{-}]= 6.0 \times 10^{4}$ M is $6.52 \times 10^{5}$ .

8 0

2 years ago