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cluponka [151]
3 years ago
15

How many grams of naoh are there in 500.0 ml of .125?

Chemistry
1 answer:
Neporo4naja [7]3 years ago
5 0
Multiply .5( I converted to Liters) by .125 Molars, which gets .0625, then multiply by the molar mass of NaOH, which is 22.99+16.00+1.01, all that equals 40.00, then multiply 40.00•.0625 which I got to be 2.5g
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Part III.The two reactions involved in quantitatively determining the amount of iodate in solution are: IO3-(aq) 5 I-(aq) 6 H (a
slega [8]

Answer:

\large \boxed{\math{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}

Explanation:

The I₂ is the common substance in the two equations.

(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O

{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻

From Equation (1), the molar ratio of iodate to iodine is

\dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{3}{1}

From Equation (2), the molar ratio of iodine to thiosulfate is

\dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} = \dfrac{2}{1}

Combining the two ratios, we get

\text{Stoichiometric factor} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{IO}_{3}^{-}} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} \times \dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{2}{1} \times \dfrac{3}{1} = \mathbf{\dfrac{6}{1}}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}

7 0
3 years ago
If 16.0 mL of acetone is dissolved in water to make 155 mL of solution what is the concentration expressed in volume/volume % of
Lady_Fox [76]
Percentage by volume of solution is the percentage volume of solute in total volume of solution.
Volume percentage (v/v%) = volume of solute / total volume of solution x 100%
volume of solute - 16.0 mL
total volume of solution - 155 mL 
v/v% = 16.0 / 155 x 100% = 10.32%
this means that in a volume of 100 mL solution, 10.32 mL is acetone.
7 0
3 years ago
Why was the line of best fit method used to determine the experimental value of absolute zero?
BartSMP [9]

Answer:

<u><em></em></u>

  • <u><em>Because the x-intercet of the graph represents volume zero, which indicates the minimum possible temperature or absolute zero.</em></u>

Explanation:

Charle's Law for ideal gases states that, at constant pressure, the <em>temperature</em> and the <em>volume</em> of a sample of gas are protortional.

                       \dfrac{V}{T}=k\\\\V=kT

That means that the graph of the relationship between Temperature, in Kelivn, and Volume is a line, which passes through the origin.

When you work with Temperature in Celsius, and the temperature is placed on the x-axis, the line is shifted to the left  273.15ºC.

Meaning that the Volume at 273.15ºC is zero.

You cannot reach such low temperatures in an experiment, and also, volume zero is not real.

Nevertheless, you can draw the line of best fit and extend it until the x-axis (corresponding to a theoretical volume equal to zero), and read the corresponding temperature.

Subject to the experimental errors, and the fact that the real gases are not ideal, the temperature that you read on the x-axis is the minimum possible temperature (<em>absolute zero</em>) as the minimum possible volume is zero.

3 0
3 years ago
1.
dolphi86 [110]

Answer: D. transverse

Explanation:

Light is a transverse wave, while sound is a longitudinal wave.

8 0
2 years ago
A solution of ammonia NH3(aq) is at equilibrium. How would the equilibrium
Dmitriy789 [7]

Answer:

If NH4+ is removed,  the reaction will shift toward products to replace the product removed.  This means the reaction will shift to the right.

Explanation:

A solution of ammonia NH3(aq) can be written through the following equation:

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

If the concentration of one of the reactants changes, the balance will shift in such a way that the change in concentration is counteracted.

If NH4+ is removed, now there is less product, so the reaction will shift toward products to replace the product removed.  This means the reaction will shift to the right.

This principle is due to the fact that the equilibrium constant for this reaction is constant at a certain temperature. The ratio of reactants and products is fixed. Any change in concentration disrupts the balance and will result in the balance being restored.

7 0
3 years ago
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