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3 years ago

9

Why might calcium also be called the calcium of citric acid

1 answer:

Stella [2.4K]3 years ago

6 0

Calcium citrate malate contains the calcium salt of citric acid and malic acid

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Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because

olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$

So, $H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

The ICE table can be given as:

$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

= 0.30225 OR - 0.72225

So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).

6 0

3 years ago

How toxic is uranium

Serhud [2]

Answer:

Inhaling large concentrations of uranium can cause lung cancer from the exposure to alpha particles. Uranium is also a toxic chemical, meaning that ingestion of uranium can cause kidney damage from its chemical properties much sooner than its radioactive properties would cause cancers of the bone or liver.

Explanation:

6 0

3 years ago

Read 2 more answers

A chemist needs 1.36 mol of strontium bromide, SrBr2, for a reaction. What mass of strontium chloride should the chemist use?

yaroslaw [1]

Mass SrBr2 = 247.42 g/mol x 1.36 mol
<span>= 336 g</span>

5 0

3 years ago

Read 2 more answers

Water from a lake changes to a gas state in the process of what?

boyakko [2]

Of heating. Or when the lake is exposed to boil because of the temperature.

7 0

4 years ago

Read 2 more answers

At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o

Damm [24]

Answer : The partial pressure of $H_2$ at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of $HBr$ = $1.0\times 10^{-2}atm$

The partial pressure of $H_2$ = $2.0\times 10^{-4}atm$

The partial pressure of $Br_2$ = $2.0\times 10^{-4}atm$

$K_p=4.2\times 10^{-9}$

The balanced equilibrium reaction is,

$2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$

Initial pressure 1.0×10⁻² 2.0×10⁻⁴ 2.0×10⁻⁴

At eqm. (1.0×10⁻²-2p) (2.0×10⁻⁴+p) (2.0×10⁻⁴+p)

The expression of equilibrium constant $K_p$ for the reaction will be:

$K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}$

Now put all the values in this expression, we get :

$4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}$

$p=-1.99\times 10^{-4}$

The partial pressure of $H_2$ at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of $H_2$ at equilibrium is, 1.0 × 10⁻⁶

4 0

4 years ago