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3 years ago

Speed is zero when a line on a graph has this shape

Chemistry

1 answer:

irga5000 [103]3 years ago

6 0

Speed is zero when a line on a graph is horizontal.

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Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found

Igoryamba

Answer:

The concentrations are :

$[HAsc^-]=0.000702 M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

The pH of the solution is 3.15.

Explanation:

$H_2Asc\rightleftharpoons HAs^-+H^+$ $K_{a1}=1.0\times 10^{-5}$

Initial

c 0 0

Equilibrium

c-x x x

$K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}$

$1.0\times 10^{-5}=\frac{x\times x}{(c-x)}$

$1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}$

Solving for x:

x = 0.000702 M

$[HAsc^-]=0.000702 M$

$HAsc^-\rightleftharpoons As^{2-}+H^+$ $K_{a2}=5\times 10^{-12}$

Initially

x 0 0

At equilibrium ;

(x - y) y y

$K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}$

$5\times 10^{-12}=\frac{y\times y}{(x-y)}$

$5\times 10^{-12}=\frac{y^2}{(x-y)}$

Putting value of x = 0.000702 M

$5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}$

$y=5.92\times 10^{-8} M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

Total concentration of $[H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M$

The pH of the solution :

$pH=-\log[H^+]$

$pH=-\log[7.0206\times 10^{-4} M}=3.15$

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3 years ago

PLEASE HELP ME ITS DUE IN 1 HOUR PLZZZ

Ierofanga [76]

I can’t use evidence from the picture but the green slime from the pond that they are talking about is called Algae, which is alive and can grow quickly with sunlight or any other source of energy such as water and carbon dioxide. Hope this helps!!

5 0

3 years ago

Read 2 more answers

A scientific theory _______.

Butoxors [25]

describes what occurs every time in a given situation

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3 years ago

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7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$