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What is the molarity of 10.0 g of kcl in 0.500 l of solution

2 months ago

Solution 1

Guest #5688

2 months ago

Molarity is defined as number of moles of the solute in 1 liter of solution.

Molarity = Moles of solute (mol) / Volume of the solution (L)

moles of KCl = mass (g) / molar mass (g/mol)
= 10.0 g / 74.5 g/mol
= 0.134 mol

Hence molarity = 0.134 mol / 0.500 L
= 0.268 mol/L

📚 Related Questions

Question

Given that the ka for hocl is 3.5 × 10–8, calculate the k value for the reaction of hocl with oh–. a. 3.5 × 10–22 b. 3.5 × 10–8 c. 2.9 × 10–7 d. 3.5 × 106 e. none of these

Solution 1

Answer:

3.5 * 10^6

Explanation:

In this question, we are asked to calculate the the k-value for the reaction of HOCl with OH- given Ka value for HOCl

Firstly, we write the dissociation equation for HOCl

This is as follows;

HOCl ——> H+ + OCl-

Now, we write an expression for Ka

Ka = [H+][OCl-]/HOCl = 3.5 * 10^-8

Now, we write an equation for the reaction of HOCl and OH- ;

HOCl + OH- —-> OCl- + H2O(l)

The K value expression for this can be written as;

K = [OCl-]/[HOCl][OH-]

= [H+][OCl-]/[HOCl] * [1/[H+][OH-]

= ka * 1/Kw = 3.5 * 10^-8 /(1 * 10^-14) = 3.5 * 10^6

Solution 2

Following reaction is involved in above system
HOCl(aq) ↔ H+(aq) + OCl-(aq)
OCl-(aq) + H2O(l) ↔ HOCl(aq) + OH-(aq)

Now, if the system is obeys 1st order kinetics we have
K = [OCl-][H+]/[HOCl] ............. (1)
∴ [HOCl-] / [OCl-] = [H+] (1 / 3.0 * 10-8) ............. (2)

and now considering that system is obeying 2nd order kinetics, we have
K = [OH-][HOCl-] / [OCl-] ................. (3)
Subs 2 in 3 we get
K = [OH-][H+] (1 / 3.0 * 10-8)
we know that, [OH-][H+] = 10-14

∴K = 3.3 * 10-7

Thus, correct answer is e i.e none of these

Question

Describe what changes occur during electron capture. describe what changes occur during electron capture. the mass number and atomic number increases. the mass number and atomic number decreases. the mass number is unchanged and the atomic number decreases. the mass number and atomic number do not change. the mass number is unchanged and the atomic number increases.

Solution 1

The changes that occur during electron capture are the mass number is unchanged and the atomic number decreases. The correct option is C.

What is electron capture?

Electron capture is a process when the electrons of the inner orbital of an atom, are captured but the nucleus and the emission of neutrino happens, and the protons are changed into neutrons.

In electron capture, the element changes into another atom with a change in the number of the proton, usually decreasing. An example is a carbon with proton 6 changes into boron with proton 5.

Thus, the correct option is C, the mass number is unchanged and the atomic number decreases.

Learn more about electron capture, here:

brainly.com/question/17215285

#SPJ2

Solution 2

The mass number is unchanged and the atomic number decreases.

Question

If you add 500 kj of heat to 500 g of water at 50.0°c, how much water is left in the container? the latent heat of vaporization of water is 2.26 ×103 j/g and its specific heat is is 4.190 j/(g∙k)

Solution 1

First, we will get the heat required to raise the 500 g of water from 50°C to 100°C = m*C*ΔT

when m is the mass = 500 g

and C is the specific heat capacity of water = 4.19

ΔT change in temperature = 50 °C

by substitution:

q = 0.5 Kg * 4.19 * 50°C

=104.75 KJ

∴ heat left to boil the water= 500KJ - 104.75KJ = 395 KJ

the heat required to boil water from 100°C to steam = mass *latent heat of vaporization

395KJ = M * 2.26 x 10^3

Mass = 0.17Kg = 170 g

∴ water remain= 500 g - 170 = 330 g

Question

Find the age t of a sample, if the total mass of carbon in the sample is mc, the activity of the sample is a, the current ratio of the mass of 14 6c to the total mass of carbon in the atmosphere is r, and the decay constant of 14 6c is λ. assume that, at any time, 14 6c is a negligible fraction of the total mass of carbon and that the measured activity of the sample is purely due to 14 6c. also assume that the ratio of mass of 14 6c to total carbon mass in the atmosphere (the source of the carbon in the sample) is the same at present and on the day when the number of 14 6c atoms in the sample was set. express your answer in terms of the mass ma of a 14 6c atom, mc, a, r, and λ.

Solution 1

N₀ is the number of C-14 atoms per kg of carbon in the original sample at time = Os when its carbon was of the same kind as that present in the atmosphere today. After time ts, due to radioactive decay, the number of C-14 atoms per kg of carbon is the same sample which has decreased to N. λ is the radioactive decay constant.
Therefore N = N₀e-λt which is the radioactive decay equation,
N₀/N = eλt In (N₀.N= λt. This is the equation 1
The mass of carbon which is present in the sample os mc kg. So the sample has a radioactivity of A/mc decay is/kg. r is the mass of C-14 in original sample at t= 0 per total mass of carbon in a sample which is equal to [(total number of C-14 atoms in the sample at t m=m 0) × ma]/ total mass of carbon in the sample.
Now that the total number of C-14 atoms in the sample at t= 0/ total mass of carbon in sample = N₀ then r = N₀×ma
So N₀ = r/ma. this equation 2.
The activity of the radioactive substance is directly proportional to the number of atoms present at the time.
Activity = A number of decays/ sec = dN/dt = λ(number of atoms of C-14 present at time t) =
λ₁(N×mc). By rearranging we get N = A/(λmc) this is equation 3.
By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).

Question

An industrial chemist introduces 8.1 atm h2 and 8.1 atm co2 into a 1.00-l container at 25.0°c and then raises the temperature to 700.0°c, at which keq = 0.534: h2(g) + co2(g) ⇔ h2o(g) + co(g) how many grams of h2 are present after equilibrium is established?

Solution 1

I reccomend searching it up

Question

Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -

Solution 1

The given complex ion is as follow,

[Ru (CN) (CO)₄]⁻

Where;
[ ] = Coordination Sphere

Ru = Central Metal Atom = Ruthenium

CN = Cyanide Ligand

CO = Carbonyl Ligand

The charge on Ru is calculated as follow,

Ru + (CN) + (CO)₄ = -1
Where;
-1 = overall charge on sphere

0 = Charge on neutral CO

-1 = Charge on CN

So, Putting values,

Ru + (-1) + (0)₄ = -1

Ru - 1 + 0 = -1

Ru - 1 = -1

Ru = -1 + 1

Ru = 0
Result:
Oxidation state of the metal species in each complex [Ru(CN)(CO)₄]⁻ is zero.

Solution 2

The oxidation state of the Ru metal in the complex ${\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}$ is $\boxed{{\text{zero}}}$ .

Further explanation:

Oxidation number:

The oxidation number is used to represent the formal charge on an atom. It also shows that gain or loss of electrons by the atom. Oxidation number can be a positive or negative number but cannot be fractional.

The rules to identify the oxidation state:

(1) The oxidation state of an atom in the elemental form is zero.

(2) The total charge on the species is equal to the sum of the oxidation state of individual atoms.

(3) The oxidation state of halides is -1. For example, fluorine, chlorine, bromine, iodine have -1 oxidation state.

(4)Hydrogen has a +1 oxidation state.

(5) Oxygen has an oxidation state -2.

(6) In a coordination compound, neutral ligands have zero oxidation state and negative ligands such as CN have -1 oxidation.

The given compound is ${\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}$

Here, CN is a negative ligand thus oxidation state is -1 and CO is a neutral ligand thus it has 0 oxidation state. Also, the complex has -1 negative charge.

The expression to calculate the oxidation state in ${\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}$ is,

$\left[{\left({{\text{oxidation state of Ru}}}\right)+\left({{\text{oxidation state of CN}}}\right)+4\left({{\text{oxidation state of CO}}}\right)}\right]=-1$

…… (1)

Rearrange equation (1) for the oxidation state of Ru.

${\text{Oxidation state of Ru}}=\left[{-\left({{\text{oxidation state of CN}}}\right)-4\left({{\text{oxidation state of CO}}}\right)-1}\right]$

…… (2)

Substitute -1for the oxidation ${\text{state}}$ of CN and 0 for the oxidation state of CO in equation (2).

$\begin{aligned}{\text{Oxidation state of Ru}}&=\left[{-\left({-{\text{1}}}\right)-4\left({\text{0}}\right)-1}\right]\\&=0\\\end{aligned}$

The oxidation state of Ru is zero.

Learn more:

1. General statement that is not applied on metals: brainly.com/question/2474874

2. The neutral element represented by the excited state electronic configuration: brainly.com/question/9616334

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Coordination complex

Keywords: Oxidation state, metal complex, ru(cn)(co)4-, formal charge, cynide, carbonyl, zero, hydrogen, oxygen, 0 and -1.

Question

What volume does 43.5 g of n2 occupy at stp? (r = 0.08206 l⋅atm/mol⋅k)?

Solution 1

We are going to use this formula:

PV= n RT

when at STP

p is the pressure = 1 atm

and n is the moles = mass /molar mass = 43.5 g /28g/mol

= 1.56 moles

R is the ideal gas constant = 0.0821

and T is a temperature in Kelvin = 273 K

by substitution, we will get V (the volume)

1 atm * V = 1.56 moles* 0.0821 *273K

∴ V = 34.96 L

Solution 2

Answer is: volume of nitrogen is 34.72 liters.
m(N₂) = 43.5 g.
n(N₂) = m(N₂) ÷ M(N₂).
n(N₂) = 43.5 g ÷ 28 g/mol.
n(N₂) = 1.55 mol, amount of substance.
T = 273 K, standard temperature.
p = 1 atm, standard pressure.
R = 0.08206 L·atm/mol·K, universal gas constant.
Ideal gas law: p·V = n·R·T.

V = n·R·T / p.
V(N₂) = 1.55 mol · 0.08206 L·atm/mol·K · 273 K / 1.00 atm.
V(O₂) = 34.72 L.

Question

What is the ground-state electron configuration for the element cobalt (z = 27)?

Solution 1

Cobalt is a Transition Element and it has a partially filled d-orbital. As given it has atomic number 27. In neutral state Cobalt has 27 electrons. These electrons are filled in following order,

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d⁷

Or it can also be written as,

[Ar] 4s², 3d⁷

Where; [Ar] is Argon with electronic configuration 1s², 2s², 2p⁶, 3s², 3p⁶.

Question

What is the daughter nucleus (nuclide) produced when 90sr undergoes beta decay by emitting an electron? replace each question mark with the appropriate integer or symbol?

Solution 1

Beta decay by emitting an electron is called as β⁻ decay. when this happens, a neutron of the element converts into a proton by emitting an electron. Hence, the mass of daughter nucleus is same as parent atom but atomic number/number of protons is higher by 1 than atomic number of parent atom.

Sr has atomic number as 38.

After β⁻ decay, the daughter nucleus will have atomic number as 38 + 1 = 39.
Hence, the daughter nucleus should be Y (Yttrium). Formula is,

₃₈⁹⁰Sr → ₃₉⁹⁰Y + ₋₁⁰β + energy

Question

Identify the nuclide produced when uranium-238 decays by alpha emission: 238 92u→42he + ?

Solution 1

Alpha decay is an emission of a ₂⁴He nucleus. After an alpha decay, the formed daughter nuclide has the reduced mass number by 4 and reduced atomic number by 2 with compared to the parent atom.

After an alpha decay of ₉₂²³⁸U, the formed daughter atom should be Th (Thorium) which has atomic number as 92 - 2 = 90. The equation is,

₉₂²³⁸U → ₉₀²²⁹Th + ₂⁴He

The sum of mass number and sum of atomic number should be equal in both side.

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