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just olya [345]
3 years ago
7

Ice bomb

Chemistry
1 answer:
Ahat [919]3 years ago
5 0

Answer:

Formation of Gas

Explanation:

The solid Carbon Dioxide warms it sublimates to gas. With limited room for the gas to expand, the pressure in the bottle increases.

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The molar solubility of pbi2 is 1.5 103 m.
Vsevolod [243]

Answer: -

Concentration of PbI₂ = 1.5 x 10⁻³ M

PbI₂ dissociates in water as

PbI₂ ⇄ Pb²⁺ + 2 I⁻

So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.

Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =

= 1.5 x 10⁻³ x 2 M

= 3 x 10⁻³ M

PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.

[Pb²⁺] = 1.5 x 10⁻³ M

So solubility product for PbI₂

Ksp = [Pb²⁺] x [ I⁻]²

=1.5 x 10⁻³ x (3 x 10⁻³)²

= 4.5 x 10⁻⁹

8 0
3 years ago
The component that dissolves the other component is called the
zheka24 [161]

Answer:

The component that dissolves the other component is called the solvent. Solute – The component that is dissolved in the solvent is called solute

6 0
3 years ago
Read 2 more answers
How does 2 Na (s) + Cl2 (g) = NaCl (s) represent the coefficient in the chemical equation
Elenna [48]

Answer:

The 2 would be placed in front of the reactant Na and in front of the product NaCl

2Na + Cl2 = 2NaCl

Explanation:

This is because the ratio of elements needs to be balanced on both sides.

On the reactants side, there are 2 Na molecules and 2 Cl molecules

On the products side, there are 2 Na molecules and 2 Cl molecules

So, now the equation is balanced

3 0
3 years ago
Two samples of potassium iodide are decomposed into their constituent elements. The first sample produced 13.0g of potassium and
Leona [35]

Answer:

79.43kg

Explanation:

To adequately solve this problem, we should know the law of constant composition. This is a pointer to the fact that no matter the type of sample, the percentage compositions of potassium and iodine still remains the same. We can use these masses to get a formula and we would know the percentage compositions in whatever mass we are dealing with.

First of all, we add the masses in the first sample. 13 + 42.3 = 55.3

Hence, the percentage composition of the potassium is 13/55.3 * 100 = 23.51%

The percentage composition of the iodine is = 100 - 23.51 = 76.49%

Now, we need to get the formula of the compound. We can get this by dividing the percentage compositions with the atomic masses. The atomic mass of potassium and iodine is 39 and 127 respectively.

Potassium = 23.51/39 =0.603

Iodine = 76.49/127 = 0.602

We then divide by the smaller value to get the formula and this shoes our formula is KI

We can see they have a ratio of 1 to 1, meaning one atom of potassium to one atom of iodine. This further confirms the percentage compositions of 23.5 to 76.5

Now to get the mass of iodine yielded, let us say the mass is xkg

This means x/(x + 24.4) * 100 = 76.5

100x = 76.5( x + 24.4)

100x = 76.5x + 1866.6

100x - 76.5x = 1866.6

23.5x = 1866.6

x = 1866.6/23.5 = 79.43kg

4 0
3 years ago
How many sig-figs in the following measurements?
DerKrebs [107]

Answer:

1.605cm = 4 significant figures

16.050cm = 4 significant figures

16.050cm = 4 significant figures

12 + 12.5 + 125 = 149.5 = 4 significant figures

1.62 × 10^3/2.8 × 10^-5 = 1620/0.000028 = 11 significant figures

4 0
3 years ago
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