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Shalnov [3]
3 years ago
12

A chemist adds of a mercury(I) chloride solution to a reaction flask. Calculate the micromoles of mercury(I) chloride the chemis

t has added to the flask.
Chemistry
1 answer:
Georgia [21]3 years ago
7 0

Answer:

3.383x10⁻³ micromoles of HgCl

Explanation:

<em>The chemist adds 170mL of a 1.99x10⁻⁵mmol/L Mercury (I) chloride, HgCl.</em>

<em />

The solution contains 1.99x10⁻⁵milimoles of HgCl in 1L. That means in 170mL = 0.170L there are:

0.170L × (1.99x10⁻⁵milimoles HgCl / L) = 3.383x10⁻⁶ milimoles of HgCl.

Now, in 1milimole you have 1000 micromoles. That means in 3.383x10⁻⁶ milimoles of HgCl you have:

3.383x10⁻⁶ milimoles of HgCl ₓ (1000micromoles / 1milimole) =

<h3>3.383x10⁻³ micromoles of HgCl</h3>
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Answer

If the temperature is increased , the number of collision per second increases.

Explanation

Temperature is proportional to the average kinetic energy of a sample of a gas according to the equation PV=n R T. An increased in temperature , increases the kinetic energy of the gas particles which in turn rises the velocity of the gas particles hitting the walls of the container. The more the number of particles the higher the collision rate and greater the pressure as long as the volume of container and the temperature are constant.


7 0
4 years ago
Read 2 more answers
Iron-59 has a half-life of 44 days. a radioactive sample has an activity of 0.64 mbq. what is the activity of the sample after 8
myrzilka [38]
Radioactive decay is a pseudo-first order reaction. When you know the half-life of the material, you could use this equation.

A= A₀(1/2)^t/h
where 
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h is the half-life

A = (0.64)(1/2)^88/44 = <em>0.16 mbq</em>
6 0
4 years ago
How many excited electron in the atom with an electron configuration of [kr]5s24d55p6 ?
IRISSAK [1]
The ground-state electronic configuration follows the order of orbitals as shown in the first picture. So, we can deduce that the electronic configuration given is in order. However, each orbital must have a maximum number of electrons to be filled before it can move on to the next orbital. Since the d orbital must hold 10 electrons, that means that it should have held 5 electrons more. The missing 5 electrons were excited to the next 5p orbital. <em>Thus, there are 5 excited electrons.</em>

4 0
3 years ago
141.98g of aluminum oxide is equal to how many moles?
NemiM [27]
The relative mass of Al2O3=2*27+16*3=102
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4 0
2 years ago
A solution contains 1.32×10-2 M lead nitrate and 1.16×10-2 M copper(II) acetate. Solid sodium carbonate is added slowly to this
user100 [1]

Answer:

CuCO₃ is the formula of the substance that will precipitate out first

the concentration of carbonate ion when this precipitation first begins = 1.212 × 10⁻¹¹ M

Explanation:

Pb(NO_3)_2 \ + \ Na_2CO_3 -----> PbCO_3 \ + \ 2NaNO\\\\ksp \ of \ PbCO_3 =  1.6*10^{-13} \\\\ \\Cu (CH_3COO)_2 \ + \ Na_2CO_3 ------> CuCO_3 \ + \ Na(C_2H_3O_2)\\\\ksp \ of  CuCO_3 = 1.3*10^{-10}\\\\As \ ksp \ of CuCO_3 > PbCO_3\\\\Then \ \ CuCO_3 \ will \ precipitate \ out \ first

PbCO_3 ---->   Pb^{2+}_{(aq)} + CO^{2-}_{3(aq)}

[Pb^{2+}] will be [Pb(NO_3)_2] = 1.32 *10^{-2} \ M

ksp =  [Pb^{2+}][CO^{2-}_3]

1.6*10^{-13} = (1.32*10^{-2})(CO^{2-}_3)\\\\(CO^{2-}_3) = \frac{1.6*10^{-13}}{(1.32*10^{-2})}\\\\(CO^{2-}_3) = 1.212 *10^{-11} \ \ M

8 0
3 years ago
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