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3 years ago

What is the total Mass of Reactants in Alpha Decay

Chemistry

2 answers:

sweet [91]3 years ago

8 0

Answer:

Calcite Calcium carbonate

Calgon Calcium hexametaphosphate

Calomel Mercurous chloride

Carbolic acid Phenol

Explanation:

⭐ sorry ⭐

Naddik [55]3 years ago

6 0

Ey do.pe bro if u keep posting free pts ur gonna get ur account deleted or banned.

Btw what did u mean by trolling ppl ?

Have a nice day do.pe bro lol

Ur my new bestie lm.ao

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A closed container is filled with oxygen. The pressure in the container is 245 kPa . What is the pressure in millimeters of merc

Zanzabum

Answer:

1837.65 mmHg is the pressure in millimeters of mercury.

Explanation:

The expression for the conversion of Pressure (kPa) to pressure (mmHg) is shown below as:-

Pressure (kPa) = 7.501 x Pressure (mmHg)

The pressure value given = 245 kPa

It can be expressed in millimeters of mercury as:-

Pressure = 7.501 x 245 mmHg = 1837.65 mmHg

<u>1837.65 mmHg is the pressure in millimeters of mercury.</u>

7 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

What were the names of the minerals and types of rocks?

padilas [110]

Types of rocks: Igneous, sedimentary, and metamorphic
There are over 200 names of minerals, I'm not sure what you want for that

8 0

3 years ago

Read 2 more answers

Prepare 15 mg/dl working standard solution from a stock solution of 20 mg/dl. State the volume of diluent and dilution.

Anastasy [175]

Preparing 15 mg/gl working standard solution from a 20 mg/dl stock solution will require the application of the dilution principle.

Recalling the principle:

initial volume x initial molarity = final volume x final molarity

Since we were not given any volume to work with, we can as well just take an arbitrary volume to be prepared. Let's assume that the stock solution is 10 mL and we want to prepare 15 mg/gl from it:

Applying the dilution principle:

10 x 20 = final volume x 15

final volume = 200/15

= 13.33 mL

This means that in order to prepare 13.33 mL, 15 mg/l working standard solution from 10 ml, 20 mg/dl stock solution, 3.33 mL of the diluent must be added to the stock solution.

More on dilution principle can be found here: brainly.com/question/11493179

4 0

3 years ago

how much current would be measured in a circuit if the light bulb has a resistance of 6 ohms and a voltage of 36 volts

Stels [109]

Answer:

The right response is "6 A". A further explanation is given below.

Explanation:

The given values are:

Resistance,

R = 6 ohms

Voltage,