Name the element in the second period of the periodic table with 6 valence electrons.

You are given sodium acetate, 1m hcl, nahco3 and na2co3. Determine which of these four you would need and then show calculations

valentina_108 [34]

solution:

the given compounds are sodium acetate, 1M HCL,NaHCO₃ and Na2CO₃

pH of the buffer solution is 4.7

the value of pKa of sodium bicarbonate is 6.37

the value of pKa of acetic acid is 4.7

calculate concentration of acetic acid by using the following forumula

pH=pKa+lag[salt]/[acid]

substitute the pH and Pka values in the formula.

4.7=4.7+log[salt]/[acid]

log[salt]/[acid]=0

thus, the concentration ratio of the salt and acid should be equal to each other.

Thus, concentration of sodium acetate is 0.05M

Concentration of sodium acetate= concentration of acid

= 0.05M

Volume of the buffer solution is 100mL

The buffer solution can be prepared as 0.05M of 50mL sodium acetate will react with 0.05M of 50mL of 0.05M of HCL.

The chemical equation for neutralization of the weak base with strong can be represented as show as

CH₃COONa+HCL-->CH₃COOH+NaCL

5 0

3 years ago

How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with

konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = $a=31 cm^2$

Thickness of the gold plating = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

$V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3$

Density of the gold = $d=19.3 g/cm^3$

$m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g$

Moles of gold = $\frac{29.915 g}{197 g/mol}=0.152 mol$

$Au^{3+}+3e^-\rightarrow Au$

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

$\frac{3}{1}\times 0.152 mol=0.456 mol$ of electrons

Number of electrons = N = $0.456\times \times 6.022\times 10^{23}$

Charge on single electron = $q=1.6\times 10^{-19} C$

Total charge required = Q

$Q=N\times q$

Amount of current passes = I = 8 Ampere

Duration of time = T

$I=\frac{Q}{T}$

$T=\frac{N\times q}{I}$

$=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s$

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

7 0

3 years ago

You make a solution by putting 45.6g of iron lll carbonate into 167ml of water. What is it's molarity?

ludmilkaskok [199]

1. The molar mass of Fe2(CO3)3 is 291.72 g/mol. This means that 45.6 g is equivalent to 0.156 mol. Dividing by the 0.167 L of water gives a solution of 0.936 M.
2. Multiplying (0.672 M)(0.025 L) = 0.0168 mol. The molar mass of Ni(OH)2 is 92.71 g/mol, so multiplying by 0.0168 mol = 1.56 grams. Therefore you would need to dissolved 1.56 g of Ni(OH)2 into 25 mL of water.
3. Fe2(CO3)3 + Ni(OH)2 --> Fe(OH)3 + NiCO3Balancing: Fe2(CO3)3 + 3Ni(OH)2 --> 2Fe(OH)3 + 3NiCO3The reaction quotient is:[Fe(OH)3]^2 * [NiCO3]^3 / [Fe2(CO3)3][Ni(OH)2]^3= (0.05)^2 * (1.45)^3 / (0.936)(0.672)^3= 0.0268Since this is < 1, it implies that the reactants are favored at equilibrium.

4 0

3 years ago

Why did the Mars Climate Orbiter crash?

IRISSAK [1]

Answer:

<h3> The Mars Climate Orbiter crashed, because of a navigation error, and a futile to convert english units to metric units.</h3>

4 0

3 years ago

A 26.08 g mixture of zinc and sodium is reacted with a stoichiometric amount of sulfuric acid. The reaction mixture is then reac

Over [174]

Answer:

Molar percent of sodium in original mixture is 88,50%

Explanation:

The last reaction is:

BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl

The moles of BaCl₂ are:

0,132L × 3,80M = 0,502 moles of BaCl₂

As the amount of BaCl₂ is the maximum possible to produce BaSO₄, the moles of BaCl₂ must be the same than moles of Na₂SO₄.

0,502 moles of BaCl₂ ≡ 0,502 moles of Na₂SO₄

These moles of Na₂SO₄ comes from:

2 Na + H₂SO₄ → Na₂SO₄ + H₂

As the reaction is in stoichiometric amounts, moles of Na are twice the moles of Na₂SO₄

0,502 moles of Na₂SO₄ × $\frac{2molesNa}{1moleNa_{2}SO_{4}}$ × 22,99 g/mole = 23,08 g of Na

Molar percent of sodium in original mixture is:

$\frac{23,08g}{26,08g}*100$ = <em>88,50% </em>

I hope it helps

4 0

3 years ago