Astatine, At, is below iodine in Group 17 of the Periodic Table. Which statement is most likely to be correct?

Cân bằng phương trình Mg+HCl->MgCl2+H2 bằng phương pháp thăng bằng

nignag [31]

quá trình oxi hóa: Mg0 -> Mg2+ + 2e-

quá trình khử: 2H+ + 2e- -> H2 (cần 2 H+ để tạo 1 H2)

----> chỉ cần thêm 2 vào trước HCl (vì H+ là do HCl cung cấp)

----> Mg + 2HCl -> MgCl2 + H2

yay :)

8 0

1 year ago

Enough of a monoprotic acid is dissolved in water to produce a 1.64 M solution. The pH of the resulting solution is 2.82 . Calcu

Lady bird [3.3K]

Answer:

Ka = 1.39x10⁻⁶

Explanation:

A monoprotic acid, HX, will be in equilibrium in an aqueous medium such as:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

<em>Where Ka is:</em>

Ka = [H⁺] [X⁻] / [HX]

<em>Where [] is the molar concentration in equilibrium of each specie. </em>

The equilibrium is reached when some HX reacts producing H+ and X-, that is:

[HX] = 1.64M - X

[H⁺] = X

[X⁻] = X

As pH is 2.82 = -log [H⁺]:

[H⁺] = 1.51x10⁻³M:

[HX] = 1.64M - 1.51x10⁻³M = 1.638M

[H⁺] = 1.51x10⁻³M

[X⁻] = 1.51x10⁻³M

And Ka is:

Ka = [1.51x10⁻³M] [1.51x10⁻³M] / [1.638M]

6 0

3 years ago

Given:

bulgar [2K]

Answer:

The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

CH₄ = 12 + (1x4) = 16 g/mol

59.7 g of CH₄ $= \frac{59.7}{16} = 3.73125 \ moles$

1 mole of methane (CH₄) released 890 kilojoules of energy

3.73125 moles of methane (CH₄) will release ?

= 3.73125 moles x -890 kJ/mol

= -3320.81 kJ

Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

5 0

2 years ago

I have 2 samples of solid chalk (aka calcium carbonate). Sample A has a total mass of 4.12 g and Sample B has a total mass of 19

IRINA_888 [86]

Answer:

A) Sample B has more calcium carbonate molecules

Explanation:

M = Molar mass of calcium carbonate = 100.0869 g/mol

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

For the 4.12 g sample

Moles of a substance is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}$

Number of molecules is given by

$nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}$

For the 19.37 g sample

$n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}$

Number of molecules is given by

$nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}$

$1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}$

So, sample B has more calcium carbonate molecules.

The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.

7 0

2 years ago

43 moles feci3 are dissolved in 0.64l of solution what is the molarity of the solution

nasty-shy [4]

Molarity=moles/liter
molarity=43/0.64
molarity=67.19moles/litre

4 0

3 years ago