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4 years ago

What mass of oxygen reacts with 4.62 g of magnesium to form magnesium oxide according to the following... 2 Mg O2 --> MgO

Chemistry

1 answer:

Goryan [66]4 years ago

7 0

Answer:

3.08 g of Oxygen .

Explanation:

2Mg + O₂ = 2MgO

2 mole 1 mole

molecular weight of Mg = 24

2 mol of Mg = 48 g

48 g of Mg reacts with 32 g of oxygen

4.62 g of Mg will reacts with 3.08 g of Oxygen .

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How will you prepare carbon dioxide gas on industrial scale? explain with chemical equation

r-ruslan [8.4K]

Answer:

Explanation:

https://www.bbc.co.uk/bitesize/guides/zxgkp39/revision/3

Hope it helps

3 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:

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

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

Explain the concept: Law of Multiple Proportions. Short and simple please

cricket20 [7]

In chemistry the law of multiple proportions states that if two elements from more than one compound between them then the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers .

4 0

3 years ago

Read 2 more answers

Suppose that 2.14 grams of H2SO4 is mixed with enough water to make 225 mL of solution. Determine the Molarity (M) of the soluti

Alik [6]

Answer:

0.0970 M

Explanation:

Remember this equation:

mol/M x V

Convert it so that you can get M.

M=mol/V

Convert the 2.14 grams of H2SO4 into mols

=0.0218

Convert mL to L

225/1000

=0.225

Plug it in.

0.0218/0.225

=0.0970 M

6 0

3 years ago

Calculate the molar solubility of ca(io3)2 in each solution below. the ksp of calcium iodate is7.1 × 10−7.

e-lub [12.9K]

In order to find the answer, use an ICE chart:

Ca(IO3)2...Ca2+......IO3-
some.......0..........0 
less.......+x......+2x 
less........x.........2x

Ca(IO₃)₂ ⇄ Ca⁺² + 2 IO⁻³

K sp = [Ca⁺²][IO₃⁻]²
K sp = (x) (2 x)² = 4 x³
7.1 x 10⁻⁷ = 4 x³
x = molar solubility = 5.6 x 10⁻³ M

The answer is 5.6 x 10 ^ 3 M. (molar solubility)

5 0

3 years ago

Read 2 more answers