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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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2 years ago

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In order to balance the equation, we can do as follows.There are 2 B - atoms on both sides of the equation, but only 2 H - atoms, and one O - atom on LHS, so we have to balance it by putting 6 in front of water and 2 in front of Boric acid and 3 in front of hydrogen sulphide gas, so that we have 2 B - atoms, 3 - S atoms, 12 H - atoms on both sides of the equation, and it is balanced. Balanced equation is given as,

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3 years ago

Convert 22.4 kg/L to kg/mL

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3 years ago

Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 °C? (HINT: the act

algol13

Answer:

D) 2.3 x 10⁻¹ s⁻¹

Explanation:

The rate constant is related to the activation energy through the formula:

k= Ae^(-Eₐ /RT)

where A is the collision factor, Eₐ the activation energy, R is the gas constant ( 8.314 J/Kmol ) , and T is the temperature (K)

So a plot of lnk versus 1/T ( Arrehenius plot ) gives us a straight line with slope equal -Eₐ/R and intercept lnA

lnk = -(Eₐ/T)(1/T) + lnA

which has the form y= mx + b

In this problem, we can use the data provided to:

a) Using a calculator determine the slope and intercept and then calculate the value of rate constant at 320 ºC, or

b) Plot the data and determine the equation of the best line , and answer the question for k @ 320 ºC by reading the value from the plot.

Once you do the plot, the resulting equation is:

y = - 19 x 10³ x + 30,582 ( R² = 0.999 )

So for T = 320 + 273 K = 593 K

Y = 19 x 10³ X + 30.58

So for T = (320 + 273)K = 593 K

Y = -19 x 10³ ( 1/593) + 30.58 = -32.04 +30.58 = - 1.46

and then since

y = lnk ⇒ e^y = k

k= e^-1.46 = 2.3 x 10⁻¹ s⁻¹

Note: there is an error of transcription in the value for T = 472.1 ( 1/T = 2.118 x 10⁻³ and not 2.228 x 10⁻³). You can recognize this mistake if you plot the data and notice it produces an outlier.

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