HNO3 is an Arrhenius ______________ and increases the concentration of _________

Which would have the lowest input force?

Afina-wow [57]

Answer:

A long lever with the fulcrum as close as possible to the load

Explanation:

If F be the effort , W be the weight , L₁ be the distance of load from fulcrum and L₂ be the distance of effort from the fulcrum ,

Taking moment of force about the fulcrum , we have

W x L₁ = F x L₂

F = W x ( L₁ / L₂ )

F will be minimum when L₁ will be minimum .

Hence fulcrum should be as close as possible to the load.

5 0

2 years ago

A possible mechanism for the gas phase reaction of NO and H2 is as follows: Step 1 2NO N2O2 Step 2 N2O2 + H2 N2O + H2O Step 3 N2

mel-nik [20]

Answer: Option (d) is the correct answer.

Explanation:

Steps involved for the given reaction will be as follows.

Step 1: $2NO \Leftrightarrow N_{2}O_{2}$ (fast)

Rate expression for step 1 is as follows.

Rate = k $[NO]^{2}$

Step 2: $N_{2}O_{2} + H_{2} \rightarrow N_{2}O + H_{2}O$

This step 2 is a slow step. Hence, it is a rate determining step.

Step 3. $N_{2}O + H_{2} \rightarrow N_{2} + H_{2}O$ (fast)

Here, $N_{2}O_{2}$ is intermediate in nature.

All the steps are bimolecular and it is a second order reaction. Also, there is no catalyst present in this reaction.

Thus, we can conclude that the statement step 1 is the rate determining step, concerning this mechanism is not directly supported by the information provided.

4 0

2 years ago

g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the

weqwewe [10]

Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

5 0

2 years ago

Convert 512 kilograms to milligrams. 0.000512 0.512 512,000 512,000,000

olga55 [171]

The answer is D! 512000000

5 0

2 years ago

Read 2 more answers

What is the molarity (molar concentration, unit = M) of K+ found in 200 mL 0.2 M K2HPO4 solution?

enyata [817]

Answer:

0.4 M

Explanation:

The process that takes place in an aqueous K₂HPO₄ solution is:

K₂HPO₄ → 2K⁺ + HPO₄⁻²

First we calculate how many K₂HPO₄ moles are there in 200 mL of a 0.2 M solution:

200 mL * 0.2 M = 40 mmol K₂HPO₄

Then we convert K₂HPO₄ moles into K⁺ moles, using the stoichiometric coefficients of the reaction above:

40 mmol K₂HPO₄ * $\frac{2mmolK^+}{1mmolK_2HPO_4}$ = 80 mmol K⁺

Finally we divide the number of K⁺ moles by the volume, to calculate the molarity:

80 mmol K⁺ / 200 mL = 0.4 M

5 0

2 years ago

HNO3 is an Arrhenius ______________ and increases the concentration of __________ when added to water.

HNO3 is an Arrhenius ____ and increases the concentration of when added to water.