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Masteriza [31]
3 years ago
15

A chemist has 500 ml of a 30% acid solution. She adds x milliliters of a 10% acid solution. Which statement is true about the gr

aph of the function that represents the concentration of the final solution?
Chemistry
1 answer:
rosijanka [135]3 years ago
3 0
Let y be concentration of the final solution:
y = (0.1x + 0.3(500))/(500 + x)
y = (0.1x + 150) / (500 + x)
The numerator contains the total amount of acid and the denominator contains the volume of the solution.
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Do x-rays travel slower than infrared
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3 0
3 years ago
The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism
MatroZZZ [7]

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

<em>Where K' = K1 * K2</em>

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just <em>mechanism A and B are consistent with observed rate law</em>

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just <em>mechanism A is consistent with the observation of J. H. Sullivan</em>

5 0
3 years ago
What is the volume of water in 150ml of the 35% of sucrose with a specific gravity of 1.115?
anzhelika [568]

Answer:

The volume of the water is 108.71 mL

Explanation:

Step 1: Data given

Volume of water =150 mL = 0.150 L

concentration of sucrose solution 35 % w/w this means in 100 grams of water we have 35 grams of sucrose

specific gravity =1.115

Step 2: Calculate the density of the solution

Density = specific gravity * density of water

Density of solution = 1.115 * 1g/ mL

Density of solution = 1.115 g/ mL

Step 3: Calculate mass of the solution

Mass of solution = density ¨volume

Mass of solution = 1.115 g/ mL * 150 mL

Mass of solution = 167.25 grams

Step 4: Calculate mass of sucrose

35 % = 0.35 * 167.25 grams

Mass sucrose = 58.54 grams

Step 5: Calculate mass of water

Mass of water = mass of sample - mass of sucrose

Mass of water = 167.25 - 58.54 = 108.71 grams

Step 6: Calculate volume of water

Volume = mass / density

Volume = 108.71 grams / 1g/ mL

Volume = 108.71 mL = 0.10871 L

The volume of the water is 108.71 mL

5 0
3 years ago
Read 2 more answers
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