HELP ASAP 10 POINTS

Will the calculated Molarity of NaOH be too high or too low or unaffected if the following happen: when you answer the question,

vodomira [7]

Answer:

Explanation:

The result will be affected.

The mass of KHP weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).

Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.

During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.

Thus, NaOH will appear at a higher concentration than it actually is.

7 0

3 years ago

Choose the letter of the choice that best completes the statement or answers the question. Elements in the same group have the s

pav-90 [236]

Elements in the same group have D. Same number of valence electrons.

6 0

3 years ago

Read 2 more answers

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

3 years ago

A shiny, smooth piece of black plastic and an identical piece of white plastic are exposed to visible light and sound waves. Whi

Nana76 [90]

Answer:

The awnser is D

Explanation:

Since the black pieces absorb light better than white pieces, the white pieces will reflect light better. But since color doesn't affect sound, the sound waves are reflected the same.

7 0

3 years ago

A chemist has 3.55x1022 molecules of nitrogen monoxide. How

Alina [70]

Answer:

<h2>0.059 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.55 \times {10}^{22} }{6.02 \times {10}^{23} } \\ = 0.05897...$

We have the final answer as

<h3>0.059 moles</h3>

Hope this helps you

6 0

3 years ago