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3 years ago

Which particle has the least mass?(1) alpha (3) neutron (2) beta particle (4) proton

1 answer:

ivann1987 [24]3 years ago

4 0

Alpha particle has a mass of 4 (Two protons and two neutrons)
Neutron has a mass of 1
Beta particle has a mass of about 0 (Electron)
Proton has a mass of 1
So the answer is (2) Beta particle

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For dilute aqueous solutions in which the density of the solution is roughly equal to that of the pure solvent, the molarity of

EleoNora [17]

Explanation:

Given: Molarity = 0.01 M

Let the volume of the solution = 1 L

Hence, moles = 0.01 moles (Molarity*Volume)

Molar mass of urea = 60 g/mol

So, mass of urea = moles x molar mass = 0.01 moles x 60 g/mol = 0.6 g = 0.0006 kg ,

Given: Density of solvent = Density of solution = 1 Kg/liter (Water).

So,

The mass of solution = vol x density = 1 L x 1 kg/L= 1 kg

Also,

Mass of solution = Mass of solute + Mass of solvent

1 kg= 0.0006 kg + mass of solvent

Mass of solvent = 1-0.0006 = 0.9994 kg

Molality is the moles of solute present in 1 kg of the solvent. So,

Molaity = ( 0.01/0.9994) ≈ 0.01 m

Molarity = Molality

Hence proved.

3 0

4 years ago

Which group of compounds is described as insoluble?

Alenkinab [10]

Answer:

phosphates compounds is described as insoluble

8 0

3 years ago

Read 2 more answers

Question 2 (2 points)

poizon [28]

Answer:

$m_w=439.2g$

Explanation:

Hello!

In this case, since the by-mass percent of a solution is a measure of the mass of the solute over the mass of the solution:

$\%m/m=\frac{m_{solute}}{m_{solution}} *100\%$

As we know the mass of the solution and the by-mass percent, we can compute the mass of glucose in the 480 g of solution:

$m_{solute}=\frac{\%m/m*m_{solution}}{100\%}$

Thus, by plugging in the data, we obtain:

$m_{solute}=\frac{8.5\%*480g}{100\%}=40.8g$

Finally, since the solution is made up of glucose and water, we compute the mass of water as follows:

$m_w=m_{sol}-m_{solute}=480g-40.8g\\\\m_w=439.2g$

Best regards!

7 0

3 years ago

1) ____ NaCl + ____ KOH  ____ NaOH + ____ KCl

GalinKa [24]

Answer:

hope it's helpful to you .

your 9th question is wrong but I did it correctly

it's not BaS2 it's BaS

8 0

4 years ago

Silver chloride, AgCl (Ksp = 1.8 x 10‒10), can be dissolved in solutions containing ammonia due to the formation of the soluble

Sergeeva-Olga [200]

Explanation:

The given reaction will be as follows.

$AgCl(s) \rightarrow Ag^{+}(aq) + Cl^{-}(aq)$ ............. (1)

$K_{sp}$ = $[Ag^{+}][Cl^{-}] = 1.8 \times 10^{-10}$

Reaction for the complex formation is as follows.

$Ag^{+}(aq) + 2NH_{3}(aq) \rightleftharpoons [Ag(NH_{3})_{2}]^{+}(aq)$ ........... (2)

$K_{f}$ = $\frac{[Ag(NH_{3})_{2}]}{[Ag^{+}][NH_{3}]^{2}} = 1.0 \times 10^{8}$

When we add both equations (1) and (2) then the resultant equation is as follows.

$AgCl(s) + 2NH_{3}(aq) \rightarrow [Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq)$ ............. (3)

Therefore, equilibrium constant will be as follows.

K = $K_{f} \times K_{sp}$

= $1.0 \times 10^{8} \times 1.8 \times 10^{-10}$

= $1.8 \times 10^{-2}$

Since, we need 0.010 mol of AgCl to be soluble in 1 liter of solution after after addition of $NH_{3}$ for complexation. This means we have to set

$[Ag^{+}]$ = $[Cl^{-}]$

= $\frac{0.010 mol}{1 L}$

= 0.010 M

For the net reaction, $AgCl(s) + 2NH_{3}(aq) \rightarrow [Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq)$

Initial : 0.010 x 0 0

Change : -0.010 -0.020 +0.010 +0.010

Equilibrium : 0 x - 0.020 0.010 0.010

Hence, the equilibrium constant expression for this is as follows.

K = $\frac{[Ag(NH_{3})^{+}_{2}][Cl^{-}]}{[NH_{3}]^{2}}$

$1.8 \times 10^{-2}$ = $\frac{0.010 \times 0.010}{(x - 0.020)^{2}}$

x = 0.0945 mol

or, x = 0.095 mol (approx)

Thus, we can conclude that the number of moles of $NH_{3}$ needed to be added is 0.095 mol.

3 0

4 years ago