Encuentras un árbol de 29 pies de altura, ¿qué tan alto es el árbol en pulgadas? Hay 12 pulgadas por pie.

A 0.50 molar aqueous solution of hydrochloric acid (HCl) flows into a process unit at 25°C and a rate of 1.25 m3/min. If the spe

Lerok [7]

Answer:

a) 18,23 kgHCl/m³

b) 0,38 kg HCl/s

c) 0,018 kgHCl/kgsolution

d) 8,75x10⁻³ molHCl/molH₂O

e) 0,010 kmolHCl/s

Explanation:

The HCl solution is 0,50 mol/L; it flows in a rate of 1,25 m³/min; Its density is 1,03 kg/L. HCl weights 36,46 g/mol

a) The HCl concentration in kg/m³ is:

$\frac{0,50molHCl}{L}$ × $\frac{36,46gHCl}{mol}$ × $\frac{1kg}{1000g}$ × $\frac{1000L}{1m^3}$ = 18,23 kg HCl/m³

b) As you have a concentration of 18,23 kgHCl/m³

$\frac{18,23 kgHCl}{m^3}$ × $\frac{1,25m^3}{min}$ × $\frac{1min}{60s}$ = 0,38 kg HCl/s

c) The mass fraction in kgHCl/kgsolution is:

$\frac{18,23kgHCl}{m^3}$ × $\frac{1m^3}{1000L}$ × $\frac{1L}{1,03kg}$ = 0,018 kgHCl/kgsolution

d) Supposing L of solution are just of water:

$\frac{0,50 moleHCl}{Lsolution}$ × $\frac{1L}{1,03kg}$ × $\frac{1kg}{1000g}$ × $\frac{18,02gH_{2}O}{1mole}$ = 8,75x10⁻³ molHCl/molH₂O

e) As the flow is 0,38kg HCl/s:

$\frac{0,38 kg}{s}$ × $\frac{1kmole}{36,46kgHCl}$ = 0,010 kmolHCl/s

I hope it helps!

8 0

3 years ago

g Imagine that you found a piece of carved wood that has been embedded in glacial ice. The mass of the wood is 50 grams, and 30%

astra-53 [7]

Answer:

13414 years

Explanation:

From;

0.693/ t1/2 = 2.303/t log No/N

t1/2 = half life of C-14

t = age of the sample

No= initial amount of C-14

N = present amount of C-14

From the question, N = 0.2 No

Hence;

0.693/5700 = 2.303/t log No/0.2No

1.2 * 10^-4 = 2.303/t log 1/0.2

1.2 * 10^-4 = 1.6097/t

t = 1.6097/1.2 * 10^-4

t = 13414 years

4 0

2 years ago

Convert 1.5 mol Na to mass

Elina [12.6K]

Answer:

19.49

Explanation:

1.5 times the mass of Na which is 12.99 is 19.485

4 0

3 years ago

Does substitution reactions cause haloalkanes?

ivanzaharov [21]

Answer:

No, that sort of reaction do not form named compounds.

Explanation:

6 0

3 years ago

Please help me this is my fourth attempt.

Vaselesa [24]

Explanation:

CH4 + 4S ---> CS2 + 2H2S

4) 0.75 mol S × (1 mol CS2/4 mol S) = 0.19 mol CS2

5) 3 mol H2S × (1 mol CH4/2 mol H2S) = 1.5 mol CH4

Fe2O3 + 2Al ---> 2Fe + Al2O3

6) 25 g FeO3 × (1 mol Fe2O3/159.69 g Fe2O3) = 0.16 mol Fe2O3

0.16 mol Fe2O3 × (2 mol Al/1 mol Fe2O3) = 0.32 mol Al

0.32 mol Al × (26.98 g Al/1 mol Al) = 8.6 g Al

7) Given:

45 g Al × (1 mol Al/26.98 g Al) = 1.6 mol Al

85 g Fe2O3 ×(1 mol Fe2O3/159.69 g Fe2O3)

= 0.53 mol Fe2O3

Let's look at how much Fe each reactant will produce:

1.6 mol Al × (2 mol Fe/2 mol Al) = 1.6 mol Fe

0.53 mol Fe2O3 × (2 mol Fe/1 mol Fe2O3) = 1.1 mol Fe

Note that the given amount of Fe2O3 will give us fewer Fe. Therefore, Fe2O3 is the limiting reactant.

8) Al will produce 1.6 mol Fe × (55.845 g Fe/1 mol Fe)

= 89 g Fe

Fe2O3 will produce 1.1 mol Fe × (55.845 Fr/1 mol Fe)

= 61 g Fe

9) Since Fe2O3 is the limiting reactant, the ideal yield of Fe for the reaction is 61 g. If the actual reaction only gave us 25 g Fe. then the percent yield of Fe is

%yield = (25 g Fe/61 g Fe) × 100% = 41%

10) If we only got 25 g Fe, then the amount of Al actually used in the reaction is

25 g Fe × (1 mol Fe/55.845 g Fe) = 0.45 mol Fe

0.45 mol Fe × (2 mol Al/2 mol Fe) = 0.45 mol Al

0.45 mol Al × (26.98 g Al/1 mol Al) = 12 g

Therefore, the leftover amount of Al is

25 g Al - 12 g Al = 13 g Al

8 0

3 years ago