1)Why do clouds usually form high in the air instead of near Earth's surface?

How does the law of conservation of mass apply to this reaction: Mg + HCl > H2 + MgCl2 ?

Gelneren [198K]

Answer:

Explanation:

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

Explanation:

This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Chemical equation:

Mg + HCl → H₂ + MgCl₂

24 g + 36.5 g = 2 g+ 95 g

60.5 g = 97 g

The reaction does not hold the law of conservation of mass, because it is not balanced.

Balanced chemical equation:

Mg + 2HCl → H₂ + MgCl₂

24 g + 73 g = 2 g+ 95 g

97 g = 97 g

this equation completely follow the law of conservation of mass.

7 0

4 years ago

Which of the following is an oxidation? Select all that apply

butalik [34]

Answer:

2Na=Ca(OH)000.1 AgBr=2KF 2KBr=LiNO

7 0

2 years ago

As ice melts and changes from solid to liquid, A) entropy increases. B) kinetic energy decreases. C) an exothermic reaction occu

Mazyrski [523]

Answer:

C) Because it gets hotter

3 0

4 years ago

A solution was prepared by mixing 20.00 mL of 0.100 M and 120.00 mL of 0.200 M. Calculate the molarity of the final solution of

Marat540 [252]

Answer:

0.186M

Explanation:

First, we need to obtain the moles of nitric acid that are given for each solution. Then, we need to divide these moles in total volume (120mL + 20mL = 140mL = 0.140L) to obtain molarity:

<em>Moles Nitric acid:</em>

0.0200L * (0.100mol / L) = 0.00200 moles

0.120L * (0.200mol / L)= 0.02400 moles

Total moles: 0.02400moles + 0.00200moles = 0.026 moles of nitric acid

Molarity: 0.026 moles / 0.140L

6 0

3 years ago

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

fredd [130]

Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

B)

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0

4 years ago