Help me please I need to pass my exam

LiRa [457]

i think helium hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

8 0

3 years ago

The chemist discovers a more efficient catalyst that can produce ethyl butyrate with a 78.0% yield. How many grams would be prod

Vedmedyk [2.9K]

The grams that would be produced from 7.70 g of butanoic acid and excess ethanol is 7.923grams

calculation

Step 1: write the chemical equation for the reaction

CH3CH2CH2COOH + CH3CH2OH → CH3CH2CH2COOCH2CH3 +H2O

step 2: find the moles of butanoic acid

moles= mass/ molar mass

= 7.70 g/ 88 g/mol=0.0875 moles

Step 3: use the mole ratio to determine the moles of ethyl butyrate

moles ratio of CH3CH2CH2COOH :CH3CH2CH2COOCH2CH3 is 1:1 therefore the moles of CH3CH2CH2COOCH2CH3 = 0.0875 x78/100=0.0683moles

step 4: find mass = moles x molar mass

= 0.0683 moles x116 g/mol=7.923grams

5 0

3 years ago

Read 2 more answers

2000 minutes to hours dimensional analysis

elena-14-01-66 [18.8K]

Answer:

33.33 h

Explanation:

You know that

1 h = 60 min

If you divide both sides by 60 min, you get the conversion factor: 1 h/60 min = 1.

If you divide both sides by 1 h, you get the conversion factor: 1 = 60 min/1 h.

Both are conversion factors because they both equal one and multiplying a measurement by one does not change its value.

You choose the conversion factor that gives you the correct dimensions for your answer. It must have the correct dimensions on top (in the numerator),

Thus, to convert 2000 min to hours, you use the conversion factor with “h” on the top.

$\text{Time} = \text{2000 min} \times \dfrac{\text{1 h}}{\text{60 min}} = \textbf{33.33 h}$

5 0

3 years ago

Was the creation of flubber a physical or chemical reaction

azamat

I'm pretty sure it's a chemical reaction.

7 0

4 years ago

. Helium is stored at 293 K and 500 kPa in a 1-cm-thick, 2-m-inner-diamater spherical container made of fused silica. The area w

Ivahew [28]

Answer:

(a) 45.17×10^-14 kg/s

(b) since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.

Explanation:

Helium gas at temperature T=293k

Helium gas at pressure P= 500kPa

The inner diameter of spherical tank is $D_1 = 2m$

The inner radius of spherical tank is : $r_1 = \frac{D_1}{2}$

= $\frac{2}{2}$

=1m

Thickness of the container r = 1cm =0.01m

Outer radius of the spherical tank is ;

t = $r_2 - r_1$

$-r_2 = -t - r-1$

multiplying through with (-) we have ;

$r_2 = t + r_1$

$r_2 = 1 + 0.01m$

$r_2 = 1.01m$

From table of binary diffusion coefficients of solids, the diffusion coefficients of helium in silica is noted as

$D_A_B =4.0 ×10^-14 \frac{m^2}{s}$

From table molar mass and gas constant, the molecular weight of helium is:

M = 4.003kg/kmol

The solubility of helium in fused silica is determined from Table of Solubility of selected gases and soilids.

$S_He$ = 0.00045 kmol/m³. bar

Considering total molar concentration as constant, the molar concentration of helium inside the container is determined as

$C_B_I = S_H_e×P$

= 0.00045kmol/m³. bar × (5)

$C_B_I = 2.25×10^-3 kmol/m³$

From one dimensional mass transfer through spherical layers is expressed as:

$N_di_f_f= 4πr_1 r_2 D_A_B \frac{C_B_I - C_B_2}{r_2 - r_1}$

substituting all the values in the above relation, we have;

$M_di_f_f= 4π(1) (1.01) (4.0×10^-14) \frac{2.25 × 10^-3 -0}{1.01-1}$

$M_di_f_f=11.42×10^-14kmol/s$

(a) The mass flow rate is expressed as

$M_di_f_f = MN_diff$

$M_di_f_f=4.003×11.42×10^-14$

$M_di_f_f=45.71×10^-14kg/s$

(b) The pressure drop in the tank after a week;

For one week the mass flow rate of helium is

$N_di_ff =11.42×10^-14kmol/s$

$N_di_ff= 11.42×10^-14×7×24×3600 kmol/week$

$N_di_f_f=6.9×10^-8kmol/week$

The volume of the spherical tank is $V=\frac{4}{3} πr_1^3$

$V=\frac{4}{3}π×1^3$

V = 4.189m³

The initial mass of helium in the sphere is determined from the ideal gas equation:

PV=NRT

where R is the universal gas constant and its value is R = 8.314 KJ/Kmol.k

N= PV/RT

N= 500 × 4.189/ 8.314 × 293

N= 0.86kmol

The number of moles of helium gas remaining in the tank after one week is:

$N_di_f_f-final = 0.86 - 6.9 × 10^-8 kmol/week$

$N_di_f_f-final ≅ 0.86$

therefore, since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.

3 0

3 years ago