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quester [9]
2 years ago
10

A chemist prepares a solution of copper(II) sulfate CuSO4 by measuring out 31.μmol of copper(II) sulfate into a 150.mL volumetri

c flask and filling the flask to the mark with water.
Required:
Calculate the concentration in mol/L of the chemist's copper(II) sulfate solution.
Chemistry
1 answer:
baherus [9]2 years ago
3 0

Answer:

The concentration of the copper (II) sulfate solution is 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM

Explanation:

The concentration of a solution is the amount of solute dissolved in a given volume of solution. In this case, the concentration of the copper(II) sulfate solution in micromoles per liter (symbol ) is the number of micromoles of copper(II) sulfate dissolved in each liter of solution. To calculate the micromoles of copper(II) sulfate dissolved in each liter of solution you must divide the total micromoles of solute by the number of liters of solution.

Here's that idea written as a formula: c= n/V

where c stands for concentration, n stands for the total micromoles of copper (II) sulfate and V stands for the total volume of the solution.

You're not given the volume of the solution in liters, but rather in milliliters. You can convert milliliters to liters with a unit ratio: V= 150. mL * 10^-3 L/ 1 mL = 0.150 L

Next, plug in μmol and liters into the formula to divide the total micromoles of solute by the number of liters of solution: c= 31 μmol/0.150 L = 206.66 μmol/L

Convert this number into scientific notation: 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM

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How many atoms are in 0.54 moles of chromium?
Levart [38]

Answer:

3.25*10^23  atoms

5 0
3 years ago
A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

Now put all the given values in above formula, we get:

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0
3 years ago
Read 2 more answers
POINTS!!!!
Setler79 [48]
C) 3 Moles
This is because of the molecular structure shown in the image

5 0
3 years ago
The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10
Ugo [173]

Answer:

The correct option is a.

Explanation:

2N_2O_5\rightarrow 4NO_2 + O_2

125 kPa

125kpa - 2x                            4x    x

Total pressure after reaction = 176 kPa

125 kPa - 2x + 4x + x = 176 kPa

x = 17

125 kpa - 2x = 125 kPa - 2(17) = 91 kPa

Initial pressure of the dinitrogen pentoxide ,(at t=0) =P_o= 125 kPa

Final pressure of the dinitrogen pentoxide, (at t = t) = P = 91 kPa

The rate constant is = k = 2.8\times 10^{-3} min^{-1}

t=\frac{2.303}{k}\log\frac{P_o}{P}

t=\frac{2.303}{2.8\times 10^{-3} min^{-1}}\log\frac{125 kPa}{91 kPa}

t=113.3969 minutes\approx 113 minutes

It will take 113 minutes for the total pressure to reach 176 kPa.

8 0
3 years ago
Complete the electron-dot structure of s-allylcysteine, showing all lone-pair electrons.
erastova [34]
The structural formula of <span>s-allylcysteine is shown in the picture (top figure). To create its Lewis structure, draw all its bonds between elements. Each single bond contains two electrons. There is an octet rule that must be obeyed by most elements. Each element should be surrounded with 8 electrons. The hydrogen is exempted of this rule. So, there are 4 lone pairs for the S atom, 1 lone pair for the N atom, and 2 lone pairs each for the 2 O atoms.</span>

7 0
3 years ago
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