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2 years ago

How many grams of C6H12O6 are needed to be dissolved in water to make 100. grams of a 250. ppm solution?

Chemistry

1 answer:

Vesna [10]2 years ago

4 0

Answer:

0.025 g C6H12O6

Explanation:

ppm = (g solute/ g solution)* 10^6

g solute= (ppm * g solution)/ 10^6

g solute = (250 ppm * 100 g)/10^6

g solute=0.025 g C6H12O6

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What are the various forms of energy?

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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).

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2 years ago

Question 25———-A, YA are two isotopes of element A.

lakkis [162]

Answer:

Option C. 1

Explanation:

Step 1:

Determination of the Neutron of both isotopes. This is illustrated below.

For isotope y xA:

Mass number = y

Atomic number = x

Neutron =..?

Atomic number = proton number = x

Mass number = Proton + Neutron

y = x + Neutron

Rearrange

Neutron = y – x

For isotope (y + 1) xA:

Mass number = y + 1

Atomic number = x

Neutron =.?

Atomic number = proton number = x

Mass number = Proton + Neutron

y + 1 = x + Neutron

Rearrange

Neutron = y + 1 – x

Step 2:

Determination of the difference between the neutron number of both isotopes. This is illustrated below:

For isotope y xA:

Neutron number = y – x

For isotope (y + 1) xA:

Neutron number = y + 1 – x

Difference in neutron number

=> (y + 1 – x) – (y – x)

=> y + 1 – x – y + x

Rearrange

=> y – y + 1 – x + x

=> 1

Therefore, the difference in the neutron number of both isotopes is 1

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2 years ago

Which statement describes an example of destruction interference?

Marianna [84]

Answer: B

Step by Step Explanation:

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