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2 years ago

How many grams of chromium(II) nitrate, , must be dissolved to prepare 500. mL of a 0.188 M aqueous solution of the salt?

Chemistry

1 answer:

Wewaii [24]2 years ago

8 0

22.37 g of chromium(II) nitrate, must be dissolved to prepare 500. mL of a 0.188 M aqueous solution of the salt.

According to the definition, the molar concentration of a substance in a solution is the ratio of the number of the moles to the volume of the solution:

c=n/V.

The number of the moles is related to the mass with the molar mass:

n=m/M;

m=n·M.

Thus, given the volume of the solution of chromium(II) nitrate, its concentration and molar mass is 238.011 g/mol we can calculate the mass of chromium(II) nitrate needed for the preparation :

∴ Cr(NO₃)₃ = cVM

= 0.188 M × 0.5 L × 238.011 g/mol

= 22.37 g

Therefore, 22.37 g of chromium(II) nitrate, must be dissolved to prepare 500. mL of a 0.188 M aqueous solution of the salt.

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Dissolved hydrofluoric acid reacts with dissolved sodium hydroxide to form water and aqueous sodium fluoride

GalinKa [24]

Answer:

HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)

Explanation:

Complete question

Dissolved hydrofluoric acid reacts with dissolved sodium hydroxide to form water and aqueous sodium fluoride. What is the net ionic equation

Equilibrium equation between the undissociated acid and the dissociated ions

HF(aq)⇌H+(aq)+F−(aq)

Sodium hydroxide will dissociate aqueous solution to produce sodium cations, Na+, and hydroxide anions, OH−

NaOH(aq)→Na+(aq)+OH−(aq)

Hydroxide anions and the hydrogen cations will neutralize each other to produce water.

H+(aq)+OH−(aq)→H2O(l)

On combining both the equation, we get –

HF(aq)+Na+(aq)+OH−(aq)→Na+(aq)+F−(aq)+H2O(l)

The Final equation is

HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)

5 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 orange line is 3.26 x 10⁻¹⁹ J.

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

the diagram of the emission spectrum has been added.

From the given chart;

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;

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

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

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

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ Js

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 orange line is 3.26 x 10⁻¹⁹ J.

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6 0

2 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

Bond [772]

Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$