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krek1111 [17]
3 years ago
10

If 22.6 grams of an aqueous solution of zinc nitrate, Zn(NO3)2, contains 5.08 grams of zinc nitrate, what is the percentage by m

ass of zinc nitrate in the solution?
____% Zn(NO3)2
Chemistry
1 answer:
sesenic [268]3 years ago
5 0

Answer:

22.5%

Explanation:

To find percentage by mass, you divide the mass of the solute by the mass of the solvent.

5.08 g/22.6 g = 0.2248 = 22.5%

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What is the energy of an electromagnetic wave with a frequency of 8•10^12 Hz?
aniked [119]

Hello!

Find the Energy of the Photon by Planck's Equation, given:

E (photon energy) =? (in Joule)

h (Planck's constant) = 6.626*10^{-34}\:J * s

f (radiation frequency) = 8*10^{12}\:Hz

Therefore, we have:

E = h*f

E = 6.626*10^{-34}*8*10^{12}

E = 53.008*10^{-34+12}

E = 53.008*10^{-22}

\boxed{\boxed{E = 5.3008*10^{-21}\:Joule}}\end{array}}\qquad\checkmark

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8 0
3 years ago
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


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4 0
3 years ago
Which metal does not form cations of differing charges? ( Which metal is NOT a transition metal?)
vitfil [10]

Answer:Na

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3 0
3 years ago
How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 4.46 atm and 45°C in the reaction shown below?
natka813 [3]

Answer:

8.1433 g of XeF₆  are required.

Explanation:

Balanced chemical equation;

XeF₆ (s) + 3H₂ (g)   →  Xe (g) + 6HF (g)

Given data:

Volume of hydrogen = 0.579 L

Pressure = 4.46 atm

Temperature = 45 °C (45+273= 318 k)

Solution:

First of all we will calculate the moles of hydrogen

PV = nRT

n = PV/ RT

n = 4.46 atm × 0.579 L / 0.0821 atm. dm³. mol⁻¹. K⁻¹ × 318 K

n = 2.6 atm . L / 26.12 atm. dm³. mol⁻¹

n = 0.0995 mol

Mass of hydrogen:

Mass = moles × molar mass

Mass =  0.0995 mol × 2.016 g/mol

Mass =  0.2006 g

Now we will compare the moles of hydrogen with XeF₆ from balance chemical equation.

                                         H₂   :  XeF₆

                                          3    :    1

                                 0.0995   : 1/3× 0.0995 = 0.0332 mol

Now we will calculate the mass of XeF₆.

Mass = moles × molar mass

Mass = 0.0332 mol × 245.28 g/mol

Mass = 8.1433 g

4 0
3 years ago
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