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2 years ago

1.Calculate the mass of compound required to prepare 100.00 mL of a 1.40x10-2M CuCl2.Show your work.

Chemistry

1 answer:

Verdich [7]2 years ago

5 0

The concentration of the diluted solution is 0.0625 M.

<h3>What is concentration?</h3>

The term concentration refers to the amount of substance present in solution.

1) To find the mass of the compound;

mass/134 g/mol = 1.40x10-2M × 100/1000

mass = 1.40x10-2M × 100/1000 × 134 g/mol

mass = 0.19 g

2) To find the molarity of the solution;

2.50 g/40 = 100/1000 × M

M = 2.50 g/40 × 1000/100

M = 0.625 M

3) Using;

C1V1 = C2V2

0.625 M × 10.00mL = C2 × 100.00 mL

C2 = 0.625 M × 10.00mL/ 100.00 mL

C2 = 0.0625 M

Learn more about concentration: brainly.com/question/3045247

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Question 34 (1 point)

Grace [21]

Answer:

8.33 atm

Explanation:

Xe is 5 out of (4+5) or 5 / 9 ths of the gas present

5/9 * 15 atm = 8.33 atm

4 0

1 year ago

. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r

Tom [10]

Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

Explanation : Given,

Wavelength = $435.8nm=435.8\times 10^{-9}m$

conversion used : $1nm=10^{-9}m$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength = $435.8\times 10^{-9}m$

c = speed of light = $3\times 10^8m/s$

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

$E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}$

$E=4.56\times 10^{-19}J$

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

3 0

3 years ago

if the spin of one electron in an orbital is clockwise , what is the spin of the other electron in that orbital

nata0808 [166]

Here we have to get the spin of the other electron present in a orbital which already have an electron which has clockwise spin.

The electron will have anti-clockwise notation.

We know from the Pauli exclusion principle, no two electrons in an atom can have all the four quantum numbers i.e. principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s) same. The importance of the principle also restrict the possible number of electrons may be present in a particular orbital.

Let assume for an 1s orbital the possible values of four quantum numbers are n = 1, l = 0, m = 0 and s = $\frac{+}{-}$ $\frac{1}{2}$ .