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4 years ago

When a counterfeit detection pen is used on an authentic bill, what color does it turn?

Chemistry

1 answer:

NeX [460]4 years ago

3 0

Answer:

Explanation:

the answer is C, sometimes it can be brown

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The solubility of copper(i) chloride is 3.91 mg per 100.0 ml of solution. calculate ksp for cucl (cucl=99.00 g mol-1).

garri49 [273]

Convert Mg to grams
1g =1000mg what about 3.91 Mg
= 3.91mg x 1g/1000mg= 3.91 x10^-3 g
moles= mass/molar mass
that is 3.91 x10^-3g /99 g/mol=3.95 x10^-5moles
concentration= moles / vol in liters

that is 3.95 x10^-5/100 x1000= 3.94 x10^-4M

equation for dissociation of CUCl= CUCl----> CU^+ +Cl^-

Ksp=(CU+)(CI-)
that is (3.95 x10^-4)(3.95 x10^-4)
Ksp= 1.56 x10^-7

3 0

3 years ago

Read 2 more answers

At 100 degrees Celsius, the ion product for pure water is Kw= 51.3*10^-14. What is the pOH of water at this temperature? A) 7.00

Levart [38]

7.86 is the pOH of water at this temperature of 100 degrees celsius.

Option E is the right answer.

Explanation:

Data given:

Kw = 51.3 x $10^{-14}$

pOH = ?

we know that pure water is neutral and will have pH pf 7.

The equation for relation between Kw and H+ and OH- ion is given by:

Kw = [H+] [OH-}

here the concentration of H+ ion and OH- ion is equal

so, [H+]= [OH-]

Putting the values in the equation of Kw

pKw = -log[Kw]

pKw = -log [51.3 x $10^{-14}$ ]

pKw = 12.28

since H+ ion OH ion concentration is equal the pH of water is half i.e. 6.14

Now, pOH is calculated by using the equation:

14 = pOH + pH

14- 6.14 = pOH

pOH = 7.86

8 0

3 years ago

Two elements in the same period have the same number of _____ _____ in their electron clouds.

shutvik [7]

Answer:

outer electrons

Explanation:

3 0

3 years ago

Read 2 more answers

What is 11/6 + 11/6 in fractions

Andreyy89

3 2/3 is your answer

6 0

3 years ago

Read 2 more answers

Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol

prohojiy [21]

Hey there!:

From the given data ;

Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar = g/L * mol/g

[E]t = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹

Hope that helps!

4 0

3 years ago