All categories

3 years ago

Which compound does not form a precipitate when reacted with sodium hydroxide?

Chemistry

1 answer:

Alla [95]3 years ago

5 0

Answer:

NaNO₃

Explanation:

A precipitate is a compound or a salt formed from a precipitation reaction and does not dissolve in water and therefore will exist in solid state.

From the choices given precipitation reaction will occur between;

Fe(NO₃)₃(aq) + 3NaOH(aq) → Fe(OH)₃(s) + 3NaNO₃(aq)
Cu(NO₃)₂(aq) + 2NaOH(aq) → Cu(OH)₂(s) + 2NaNO₃(aq)
FeSO₄(aq) + 2NaOH(aq) → Fe(OH)₂(s) + Na₂SO₄(aq)

Fe(OH)₃, Cu(OH)₂, and Fe(OH)₂ are precipitates.

From the rules of solubility, hydroxides are insoluble except Ca(OH)₂ which is slightly soluble and hydroxides of ammonium and alkali metals.

You might be interested in

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

4 0

3 years ago

In the following reaction, what element is gaining mass?

vagabundo [1.1K]

Mg gained mass because it went from being a single element (on the reactant side) to being a molecule (on the product side).

8 0

3 years ago

The dissociation of a/an ________ releases hydrogen ions and increases the concentration of hydrogen ions in a solution.

Crank

Acid...hope it helps

5 0

3 years ago

Rank these acids according to their expected pKa values.

givi [52]

Answer:

According to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH is the weakest acid, so the highest pKa value.

Explanation:

The pKa values are the negative logarithm of dissociation constant. It represents the relative strengths of the acids. Stronger acids show smaller pKa values and weak acids present larger pKa value. The stronger the acid, the weaker it's the conjugate base. The larger the pKa of the conjugate base, the stronger the acid. The strength of an acid is inversely related to the strength of its conjugate.

Conjugate bases are the substance that has one less proton than the parent acid. The conjugate base of the acid presented in the problem are:

ClCH2COOH -> ClCH2COO- + H+

ClCH2CH2COOH -> ClCH2CH2COO- + H+

CH3CH2COOH -> CH3CH2COO- + H+

Cl2CHCOOH -> Cl2CHCOO - + H+

Cl2CHCOOH. The negative charge presented on its conjugate base is by resonance and inductive effect. This is the strongest acid.

ClCH2COOH. A negative charge is stabilized by resonance and electron-withdrawing but only one atom is present. So this acid is less strong than the first one.

ClCH2CH2COOH. The negative charge is stabilized by resonance and electron-withdrawing atom but the effect is less compared to the two acids showed previously.

CH3CH2COOH. The negative charge is stabilized by resonance and destabilized due to CH3 group. This is the weakest acid among the problem.

Stronger acids have smaller pKa values and weak acids have larger pKa values. Due to the information present in this problem, Cl2CHCOOH is the strongest acid and the lowest pKa. CH3CH2COOH is the weakest acid, so the highest pKa value.

Finally, we can conclude that according to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH is the weakest acid, so the highest pKa value.

3 0

3 years ago

What is the pressure of 5.0 mol nitrogen gas in a 2.0 L container at 286 K

g100num [7]

The "Ideal Gas Law Equations" is
$PV=nRT$
P= Pressure (in Pascals)
V=Volume (in Liters)
n=amount, or number (in moles)
R= 8.3145 $\frac{Pascals*Liters}{Moles*Kelvin}$ or $\frac{PaL}{molK}$
T= Temperature (In Kelvin)

$P= \frac{nRT}{V}$

Plug into the equation and you're good!
$P=\frac{nRT}{V}$
$P =\frac{(5mol)(8.3145)(286K)}{2L}$
$P=5944.8675Pa$

If your teacher cares about sig figs,
2 sig figs (significant figures)
$P=5900Pa$

For other units of pressure,
1 atm = 760 mmHG = 760 Torr = 101326 Pa = 1.01325 bar

5 0

3 years ago