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3 years ago

Which shows the correct relationship?

1 answer:

8 0

The correct answer is option B. i.e. E°cell > 0 = spontaneous reaction.

The spontaneous reactions always have positive voltage and the energy released in the reaction is used up to do work.

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Based on the results you observed for the iodine test and Benedict’s test, is it better to detect enzyme activity by measuring t

dlinn [17]

Answer:

Is better use the Benedict's test by the increase in the amount of the products if the enzyme is a reductase

Explanation:

The Benedict's test works by the reaction of the reducing sugars with the ion cupric of the reactive. If the enzyme is a reductase (degrades polysaccharides into bi o monosaccharides), it should cut the polysaccharide bond and the products would react with the Benedict's cupric ion

I hope you undestand me

3 0

3 years ago

25g of potassium nitrate are dissolved in 100g of water. What is the percent by mass of potassium nitrate?

trapecia [35]

The the exact mass is 24.740% (according to the internet) but I’d say that it’s 25

8 0

3 years ago

Write the net ionic equation for the reaction that occurs when equal volumes of 0.546 M aqueous acetylsalicylic acid (aspirin) a

Tatiana [17]

Answer:

$C_9H_8O_4+C_2H_3O_2^-\rightarrow C_2H_4O_2+C_9H_7O_4^-$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to figure out the required net ionic equation by firstly writing out the complete molecular equation between aspirin and sodium acetate:

$C_9H_8O_4+NaC_2H_3O_2\rightarrow C_2H_4O_2+NaC_9H_7O_4$

Whereas acetic acid and sodium acetylsalicylate are formed. Now, we write the complete ionic equation whereby sodium acetate and sodium acetylsalicylate are ionized because they are salts yet neither aspirin nor acetic acid are ionized as they are weak acids:

$C_9H_8O_4+Na^++C_2H_3O_2^-\rightarrow C_2H_4O_2+Na^++C_9H_7O_4^-$

Finally, for the net ionic equation we cancel out the sodium spectator ions to obtain:

$C_9H_8O_4+C_2H_3O_2^-\rightarrow C_2H_4O_2+C_9H_7O_4^-$

Regards!

7 0

3 years ago

The density of benzene is 0.879g/mL. The volume (with 1 decimal) of 131.9g sample of benzene is how many mL?

timurjin [86]

Answer:

150.1 mL

Explanation:

Step 1: Given data

Density of benzene (ρ): 0.879 g/mL

Mass of the sample of benzene (m): 131.9 g

Volume of the sample of benzene (V): ?

Step 2: Calculate the volume of the sample of benzene

Density is an intrinsic property. It is equal to the quotient between the mass and the volume of the sample of benzene.

ρ = m/V

V = m/ρ

V = 131.9 g/(0.879 g/mL)

V = 150.1 mL

5 0

3 years ago

g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete

Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899 0.01714 0 0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)

0.00185 | 0 → 0.01714 0.01714 (end)

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0

3 years ago

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