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4 years ago

If the system is at dynamic equilibrium, which statement is true?

Chemistry

2 answers:

xxTIMURxx [149]4 years ago

7 0

Answer:

question A

Explanation:

Dafna1 [17]4 years ago

4 0

Where are the following questions

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n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph

olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = 0.111M

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = 0.2501 g of acetic acid

Now, assuming density of solution as 1.00g/mL, 37.54mL weights 37.54g.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)

8 0

3 years ago

The melting point of gallium metal is 29.76⁰C. If a piece of gallium is sliced into three parts, the melting point of each part

tankabanditka [31]

Answer:

29.76ºC

Explanation:

The melting point is a physical property of the matter it doesn't change as you slice the gallium into three different parts. Thus, it's the same value 29.76ºC

8 0

3 years ago

Use the reaction to answer the question.

adelina 88 [10]

Answer:

its either B or C all i got hope that helps

Explanation:

3 0

3 years ago

2H2O2 →2H2O +O2 ∆H=? 2H2 +O2 → 2H2O ∆H=-572 KJ 2H2O2 → 2H2 + 2O2 ∆H=376 KJ

Marat540 [252]

Answer:

WOT in the world is this?

Explanation:

3 0

3 years ago

A hydrated sample weighing 20.0 g is heated in a crucible at 250°C. After cooling the sample now has a weight of 16.5 g.

anzhelika [568]

Given parameters:

Weight of hydrated sample = 20g

Temperature = 250°C

Weight after cooling = 16.5g

Unknown:

Weight of water lost from the sample = ?

Solution:

The weight of water lost from the sample;

Weight of water lost = Weight of hydrated sample - Weight of dry sample

Weight of water lost = 20g - 16.5g = 3.5g

% of water in the sample = $\frac{mass of water}{mass of original sample} x 100$

Input parameters solve;

= $\frac{3.5}{20} x 100$

=17.5%

8 0

3 years ago