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Zielflug [23.3K]
3 years ago
9

How much nano3 is needed to prepare 225 ml of a 1.55 m solution of nano3?

Chemistry
1 answer:
andrezito [222]3 years ago
5 0

Answer:

= 29.64 g  NaNO3

Explanation:

Molarity is given by the formula;

Molarity = Moles/Volume in liters

Therefore;

Number of moles = Molarity × Volume in liters

                             = 1.55 M × 0.225 L

                             = 0.34875 moles NaNO3

Thus; 0.34875 moles of NaNO3 is needed equivalent to;

   = 0.34875 moles × 84.99 g/mol

   = 29.64 g

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At T = 250 °C the reaction PCl5(g) PCl3(g) + Cl2(g) has an equilibrium constant in terms of pressures Kp = 2.15. (a) Suppose the
Ganezh [65]

Answer:

To the right

Explanation:

Step 1: Given data

  • Partial pressure of PCl₅ (pPCl₅) = 0.548 atm
  • Partial pressure of PCl₃ (pCl₃) = 0.780 atm
  • Partial pressure of Cl₂ (pCl₂) = 0.780 atm

Step 2: Write the balanced equation

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

Step 3: Calculate the pressure reaction quotient

Q_p = \frac{pPCl_3 \times pCl_2 }{pPCl_5} = \frac{0.780 \times 0.780 }{0.548} =1.11

Step 4: Determine whether the reaction proceeds to the right or to the left as equilibrium is approached

Since <em>Qp < Kp</em>, the reaction will proceed to the right to attain the equilibrium.

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3 years ago
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Which element has chemical properties that are most similar to the chemical properties of flourine
aleksandrvk [35]
The elements in group 7, which fluorine is in also, would have similar properties due to the valence electrons and the configuration. <span />
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In a mixture of carbon dioxide in water ( a soft drink ) the Carbon Dioxide is the solute
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Answer:

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Explanation:

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4 0
3 years ago
Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

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2 years ago
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