Burning plants contributes to global warming because:

The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maxi

weqwewe [10]

Answer: 72.41% and 26.90% respectively.

Explanation:

At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.

We can calculate recovery as:

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%$

So the answer to the first question is 72.41%.

For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%$

So the answer to the second question is 26.90%.

3 0

3 years ago

in a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH2. In several experiments the rat

Inga [223]

Answer:

Ea= -175.45J

A= 3.5×10^14

k=3.64 ×10^14 s^2.

Explanation:

From

ln k= -(Ea/R) (1/T) + ln A

This is similar to the equation of a straight line:

y= mx + c

Where m= -(Ea/R)

c= ln A

y= ln k

a)

Therefore

21.10 3 104= -(Ea/8.314)

Ea=-( 21.10 3 104×8.314)

Ea= -175.45J

b) ln A= 33.5

A= e^33.5

A= 3.5×10^14

c)

k= Ae^-Ea/RT

k= 3.5×10^14 × e^ -(-175.45/8.314×531)

k = 3.64 ×10^14 s^2.

8 0

3 years ago

48 When an atom of the unstable isotope Na-24 decays, it becomes an atom of Mg-24 because the Na-24 atom spontaneously releases(

Crank

(1) a beta particle is your answer. Na-24 decays through beta decay, turning a neutron into a proton, electron (beta particle), and an neutrino.

4 0

3 years ago

The product of Pb(NO3)2 (aq) + HNO3 (aq)

Scrat [10]

Answer:

Pb(NO3)2 – Lead(II) nitrate sourceaà

Other names: Lead nitrate source Plumbous nitrate source Lead dinitrate source

Appearance: White colourless crystals asource:White or colourless crystals source.

H2O – Water, oxidane source

Other names: Water (H2O) source: Hydrogen hydroxide (HH or HOH) source: Hydrogen oxide source

Appearance: White crystalline solid, almost colorless liquid with a hint of blue, colorless gas source.

8 0

3 years ago

PLEASE HELP!!!!!!!

dsp73

Answer: 1. $2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu$

2. 3 moles of $CuCl_2$ : 2 moles of $Al$

3. 0.33 moles of $CuCl_2$ : 0.92 moles of $Al$

4. $CuCl_2$ is the limiting reagent and $Al$ is the excess reagent.

5. Theoretical yield of $AlCl_3$ is 29.3 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles$

$\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles$

The balanced chemical equation is:

$2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu$

According to stoichiometry :

3 moles of $CuCl_2$ require = 2 moles of $Al$

Thus 0.33 moles of $CuCl_2$ will require= $\frac{2}{3}\times 0.33=0.22moles$ of $Al$

Thus $CuCl_2$ is the limiting reagent as it limits the formation of product and $Al$ is the excess reagent.

As 3 moles of $CuCl_2$ give = 2 moles of $AlCl_3$

Thus 0.33 moles of $CuCl_2$ give = $\frac{2}{3}\times 0.33=0.22moles$ of $AlCl_3$

Theoretical yield of $AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3$

Thus 29.3 g of aluminium chloride is formed.

5 0

2 years ago